## CBSE Class 10 Maths Chapter 15: Probability Important Questions: Free PDF Download

The **probability class 10 important questions** are available for students to help them get a thorough knowledge of the concept. It is advised that students should practice from the given **important questions** to score better marks in the examination. These essential questions from the probability chapter can also be downloaded in PDF version for the convenience of the students. The more they refer to these questions, the better grasp they will have on this topic. All these questions are framed by subject experts following the latest guidelines and patterns of the CBSE Boards. Even, there are solutions provided with the questions so that students can review their answers. Besides, they can also make notes using these questions.

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## Download CBSE Class 10 Maths Important Questions 2023-24 PDF

Also, check CBSE Class 10 Maths Important Questions for other chapters:

CBSE Class 10 Maths Important Questions | ||

Sl.No | Chapter No | Chapter Name |

1 | Chapter 1 | |

2 | Chapter 2 | |

3 | Chapter 3 | |

4 | Chapter 4 | |

5 | Chapter 5 | |

6 | Chapter 6 | |

7 | Chapter 7 | |

8 | Chapter 8 | |

9 | Chapter 9 | |

10 | Chapter 10 | |

11 | Chapter 11 | |

12 | Chapter 12 | |

13 | Chapter 13 | |

14 | Chapter 14 | |

15 | Chapter 15 | Probability |

Important Questions for CBSE Class 10 Maths Chapter 15 - Probability

## Study Important Questions for Class 10 Maths Chapter 15 – Probability

1. An integer is chosen at random from the first two hundreds digit. What is the probability that the integer chosen is divisible by ${6}$ or ${8}$.

Ans: First $200$ integers that are multiples of $6$ are listed as

\[\text{6, 12, 18, }\underline{\text{24}}\text{, 30, 36, 42, }\underline{\text{48}}\text{, 54, 60, 66, }\underline{\text{72}}\text{, 78, 84, 90, }\underline{\text{96}}\text{, 102, 108, 114,}\]\[\underline{\text{120}}\text{, 126, 132, 138, }\underline{\text{144}}\text{, 150, 156, 162, }\underline{\text{168}}\text{, 174, 180, 186, }\underline{\text{192}}\text{, 198}\].

First $200$ integers that are multiples of $8$ are listed as

\[\text{8, 16, }\underline{\text{24}}\text{, 32, 40, }\underline{\text{48}}\text{, 56, 64, }\underline{\text{72}}\text{, 80, 88, }\underline{\text{96}}\text{, 104, 112, }\underline{\text{120}}\text{, 128, 136, }\underline{\text{144}}\text{,}\]\[\text{152, 160, }\underline{\text{168}}\text{, 176, 184, }\underline{\text{192}}\text{, 200}\].

Thus, there are total $50$ numbers which are multiples of $6$ or $8$.

So, the probability that the integer chosen is divisible by $6$ or $8$ is $=\frac{50}{200}=\frac{1}{4}$.

2. A box contains ${12}$ balls out of which ${x}$ are black. if one ball is drawn at random from the box what is the probability that it will be a black ball? If ${6}$ more black balls are out in the box. the probability of drawing a black ball is now double of what it was before. Find ${x}$.

Ans: When some balls are drawn randomly, then it ensures equal likely outcomes.

There are a total $12$ balls.

So, $12$ possible outcomes can occur.

Now, let there are $x$ balls, which are black.

Thus, the probability that a ball drawn is black $\begin{align} & =\frac{\text{Number of favourable outcomes}}{Total\text{ n}umber\text{ of possible outcomes}} \\ & =\frac{x}{12} \\ \end{align}$

Since, $6$ more black balls are out in the box, so there will be a total $x+6$ black balls out of the total balls $12+6=18$.

By the given condition, the probability obtained for drawing black ball in the second case is $=2\times $ the probability obtained for drawing of black ball in the first case.

$\frac{x+6}{18}=2\times \frac{x}{12}$

$\Rightarrow \frac{x+6}{18}=\frac{x}{6}$

$\Rightarrow 6x+36=18x$

$\Rightarrow x=3$

Thus, there are $3$ black balls in the box.

3. A bag contains ${8}$ red balls and ${x}$ blue balls, the odd against drawing a blue ball are \[{2}:{5}\]. What is the value of ${x}$?

Ans: Since, there are $x$ blue balls and $8$ red balls, so the total number of balls $=x+8$.

Therefore, the probability that a blue ball is drawn $=\frac{x}{x+8}$

The probability that a red ball is drawn $=\frac{8}{x+8}$.

So, by the given condition, we have

$\frac{8}{x+8}:\frac{x}{x+8}=2:5$

$\begin{align} & \Rightarrow 2\left( \frac{x}{x+8} \right)=5\left( \frac{8}{x+8} \right) \\ & \Rightarrow 2x=40 \\ & \Rightarrow x=20 \\ \end{align}$

Thus, the value of $x$ is $20$.

4. A card is drawn from a well shuffled deck of cards.

i) What are the odds in favour of getting spade?

Ans. There are $52$ cards in a deck and among these $13$ are spades.

So, the number of cards remaining is $39$.

Therefore, the odds in favour of getting spades is $\frac{13}{52}:\frac{39}{52}=1:3$.

ii) What are the odds against getting a spade?

Ans. The odds against getting a spade are $39$.

iii) What are the odds in favour of getting a face card?

Ans. The odds of obtaining a face card are $12$ to $52$. The odds of not getting a face card are \[40\] to $52$.

The odds in favour of getting a face card $=\frac{12}{52}:\frac{40}{52}=3:10$.

iv) What are the odds in favour of getting a red king

Ans: The odds of obtaining a red king are $2$ to $52$.

The odds of not getting a red king are $50$ to $52$.

Thus, the odds in favour of getting a red king are $=\frac{2}{52}:\frac{50}{52}=1:25$.

5. A die is thrown repeatedly until a six comes up. What is the sample space for this experiment? HINT: \[{A}=\left\{ {6} \right\},\text{ }{B}=\left\{ {1},{2},{3},{4},{5}, \right\}\].

Ans: The sample space for the given experiment is \[\left\{ \text{A, BA, BBA, BBBA, BBBBA} \right\}\].

6. Why is tossing a coin considered to be a fair wav of deciding which team should get the ball at the beginning of a football match?

Ans: Tossing a coin gives equally likely outcomes since they are mutually exclusive events. That’s why tossing a coin is considered to be a fair wav of deciding which team should get the ball.

7. A bag contains ${5}$ red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball. determine the number of blue balls in the bag.

Ans: Suppose that the number of blue balls in the bag is $x$.

Therefore, the number of total balls contained in the bag $=x+5$.

Then, the probability that a blue ball is drawn, $=\frac{x}{x+5}$.

Also, the probability that a red ball is drawn, $=\frac{5}{x+5}$.

Now, by the given condition,

$\begin{align} & \Rightarrow \frac{x}{x+5}=2\cdot \frac{5}{x+5} \\ & \Rightarrow x=10 \\ \end{align}$

Thus, the number of blue balls in the bag is $10$.

8. A box contains \[{12}\] balls out of which \[{x}\] are black. If one ball is drawn at random from the box. What is the probability that it will be a black ball? If ${6}$ more black balls are out in the box the probability of drawing a black ball is now double of what it was before. Find ${x}$?

Ans: The total number of possible outcomes is $12$.

Suppose that, the number of favourable outcomes on the event of drawing black ball is $x$.

Thus, the probability of getting a black ball $=\frac{x}{12}$.

If $6$ more black balls are removed from the box, then the number of possible outcomes \[=\text{12+6}=\text{18}\].

Then, the number of black balls $=x+6$.

Thus, the probability of drawing a black ball is $\frac{x+6}{18}$.

Now, by the given condition,

$\frac{x+6}{18}=2\left( \frac{x}{12} \right)$

Therefore, $x=3$.

9. If \[{65}%\] of the populations have black eyes. \[{25}%\] have brown eyes and the remaining have blue eyes. What is the probability that a person selected at random has

(i) Blue eves

Ans. The number of black eyes $=65$.

The number of Brown eyes $=25$

The number of blue eyes $=10$.

Thus, there are a total of $180$ eyes.

Therefore, the probability of having blue eyes $=\frac{10}{100}=\frac{1}{10}$.

(ii) Brown or black eves

Ans. The probability of having brown or black eyes $=\frac{90}{100}=\frac{9}{10}$

(iii) Blue or black eves

Ans. The probability of having blue or black eyes $=\frac{10}{100}=\frac{1}{10}$.

(iv) neither blue nor brown eves

Ans: The probability of having neither blue nor brown eyes $=\frac{65}{100}=\frac{13}{20}$.

10. Find the probability of having ${53}$ Sundays in

i) a non-leap year

Ans. We know that there are $365$ days in an ordinary year containing \[52\] weeks and $1$ day.

This day may be any one of the $7$ days of the week.

Thus, the probability that this day is Sunday $=\frac{1}{7}$.

Hence, the probability that an ordinary year has $53$ Sunday $=\frac{1}{7}$.

ii) a leap year

Ans: It is known that, there are $366$ days in a leap year containing $52$weeks and $2$ days.

These two days can be \[\left( MT \right),\left( TW \right),\left( W\text{ }Th \right),\left( Th\text{ }F \right),\left( FS \right),\left( SS \right),\left( SM \right)\]. Now, since, $52$ weeks contain $52$ Mondays, so the remaining Monday can be obtained if the two days are \[\left( MT \right)\] or \[\left( SM \right)\], that is, $2$ out of $7$ cases. Thus, the probability of having $53$ Mondays in a leap year $=\frac{2}{7}$.

11. Five cards - the ten, Jack, queen, king and ace, are well shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

Ans. The number of total possible outcomes is $5$.

Since, the number of Queen is $1$, so

the number of favourable outcomes$=1$.

Thus, the probability of getting a Queen card $=\frac{1}{5}$.

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Ans. Since, one card is put aside, so now, the number of total possible outcomes is $5-1=4$.

(a) Since, there is only one ace card, so the number of favourable outcomes is $1$.

Thus, the probability of getting an ace card $=\frac{1}{4}$.

(b) Since, there is no queen card left after the first pick up, so the number of favourable outcomes is $0$.

Thus, the probability of getting a Queen card $=\frac{0}{4}=0$.

12. A number x is chosen at random from the numbers\[{-3, -2, -1, 0},{ 1, 2, 3}\]. What is the probability that \[\left| {x} \right|<{2}\].

Ans: $x$ can have the value among the $7$ given values.

Now, for \[\left| x \right|<2\], $x$ can be $-3,-2,-1,0,1$.

So, the required probability, $P\left( \left| x \right|<2 \right)=\frac{5}{7}$.

13. A number ${x}$ is selected from the numbers \[{1},{2},{3}\] and then a second number ${y}$ is randomly selected from the numbers \[{1},{4},{9}\], What is the probability that the product ${xy}$ of the two numbers will be less than ${9}$?

Ans: Number $x$ can be chosen in three different ways, and number $y$ can be chosen in three different ways as well. So, the two numbers can be chosen in $9$ ways such that,

\[\left( 1,1 \right),\left( l,4 \right),\left( 2,l \right),\left( 2,4 \right),\left( 3,1 \right)\].

Therefore, the number of favourable occurrences is $5$.

Thus, the probability that the product will be less than $9$ is $=\frac{5}{9}$.

14. In the adjoining figure a dart is thrown at the dart board and lands in the interior of the circle. What is the probability that the dart will land in the shaded region?

(Image will be uploaded soon)

Ans: It is given that,

\[\text{AB = CD = 8}\] and \[\text{AD = BC = 6}\].

Then, by the Pythagorean Theorem on the triangle\[ABC\], gives

\[\text{A}{{\text{C}}^{\text{2}}}=\text{A}{{\text{B}}^{\text{2}}}+\text{B}{{\text{C}}^{\text{2}}}\]

\[\Rightarrow \text{A}{{\text{C}}^{\text{2}}}=\text{82}+\text{62}=\text{100}\]

$\Rightarrow AC=10$

$\Rightarrow OA=OC=5$ (since, the point $O$ is the midpoint of \[AC\])

Therefore, the area of the circle is $=\pi {{\left( OA \right)}^{2}}=25\pi $ sq. units.

Also, the area of $ABCD=AB\times BC=8\times 6=48$ sq. units.

Thus, the area of the shaded region $=$ Area of circle $-$ Area of the rectangle \[ABCD\]

That is, Area of shaded region $=\left( 25\pi -48 \right)$ sq. units.

Thus, the probability dart will land in the shaded region $\begin{align} & =\frac{\text{area of shaded region}}{area\text{ of circle}} \\ & =\frac{25\pi -48}{25\pi } \\ \end{align}$

15. In the fig points ${A},{B},{C},$ and ${D}$ are the centres of four circles. each having a radius of ${1}$ unit. If a point is chosen at random from the interior of a square \[{ABCD}\]. What is the probability that the point will be chosen from the shaded region?

Ans: It is given that the radius of the circle is $1$ unit.

Therefore, area of the circle $=$ Area of $4$ sector.

Thus, the side of the square \[ABCD\] is \[2\] units.

Therefore, the area of square \[=\text{2 }\!\!\times\!\!\text{ 2}=\text{4}\] units.

So, the area of the shaded region

$=$ area of square $-\text{ }4\times $ area of the sectors.

$=4-\pi $.

Hence, the required probability $=\frac{4-\pi }{4}$.

16. In the adjoining figure \[{ABCD}\] is a square with sides of length ${6}$ units points ${P}$ & ${Q}$ are the mid points of the sides \[{BC}\] & \[{CD}\]respectively. If a point is selected at random from the interior of the square what is the probability that the point will be chosen from the interior of the triangle \[{APQ}\].

(Image will be uploaded soon)

Ans: The area of the $\Delta PQC$$=\frac{1}{2}\times 3\times 3=\frac{9}{2}=4.5$ sq. units.

The area of the $\Delta ABP$$=\frac{1}{2}\times 6\times 3=9$ sq. units.

The area of the $\Delta ADQ=\frac{1}{2}\times 6\times 3=9$ sq. units.

Therefore, the area of the $\Delta APO$$=$ area of the square $-$(Area of the $\Delta PQC$ $+$ area of the $\Delta ABP$$+$ area of the$\Delta ABP$)

\[=36-\left( 18+4.5 \right)\]

$\begin{align} & =36-22.5 \\ & =13.5 \\ \end{align}$

Hence, the probability that the point will be selected from the interior of $\Delta APQ$$=\frac{13.5}{36}=\frac{135}{360}=\frac{3}{8}$.

17. In a musical chair, the person playing the music has been advised to stop playing the music at any time within ${2}$ minutes after she starts playing. What is the probability that the music will stop within the half minute after starting?

Ans: All the numbers between $0$ and $2$ are conceivable outcomes here. As illustrated in the picture, this is the section of the number line from $0$ to $2$.

Let $A$ be the event in which "the music is turned off during the first half minute." Then there are all points on the number line that are favourable to occurrence \[A\]. i.e.

(Image will be uploaded soon)

The points on the number line from Q to P represent the total number of outcomes. That is, from $0$ to $2$.

The required probability, $P\left( A \right)=\frac{\text{length of OQ}}{length\text{ of OP}}=\frac{\frac{1}{2}}{2}=\frac{1}{4}$.

18. A jar contains \[{54}\] marbles each of which is blue. green or white. The probability of selecting a blue marble at random from the jar is $\frac{{1}}{{3}}$ and the probability of selecting a green marble at random is $\frac{{4}}{{9}}$. How many white marbles does the jar contain?

Ans: Suppose that $b,g,$ and $w$ denotes the number of blue, green and white marbles respectively in the jar.

Therefore, $b+g+w=54$. …… (i)

So, the probability of choosing a blue marble $=\frac{b}{54}$.

Since, the probability of choosing a blue marble $=\frac{1}{3}$, so

$\begin{align} & \frac{1}{3}=\frac{b}{54} \\ & \Rightarrow b=18 \\ \end{align}$

Again, the probability of choosing a green marble $=\frac{4}{9}$.

Therefore,

$\begin{align} & \frac{g}{54}=\frac{4}{9} \\ & \Rightarrow g=24 \\ \end{align}$

Now, putting the obtained values of $b$ and $g$ into the equation (i), yields \[\text{18 + 24 + w}=\text{54}\]

$\Rightarrow w=12$.

Hence, there are a total of $12$ white marbles in the jar.

1 Marks Questions

1. Complete the statements:

(i) Probability of event ${E}$ $+$ Probability of event “not ${E}$” $=$_________.

Ans. Probability of event $E+$Probability of event “not $E$” $=1$.

(ii) The probability of an event that cannot happen is _________. Such an event is called ________.

Ans. The probability of an event that cannot happen is $0$. Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is________. Such an event is called __________.

Ans. The probability of an event that is certain to happen is $1$. Such an event is called a sure or certain event.

(iv) The sum of the probabilities of all the elementary events of an experiment is _________.

Ans. The sum of the probabilities of all the elementary events of an experiment is $1$.

(v) The probability of an event is greater than or equal to ________ and less than or equal to __________.

Ans. The probability of an event is greater than or equal to $0$ and less than or equal to $1$.

2. Which of the following cannot be the probability of an event:

(A) $\frac{{2}}{{3}}$

(B) $-{1}.{5}$

(C) \[15%\]

(D) \[{0}{.7}\]

Ans. By the definition of probability, the maximum and minimum value of probability are $1$ and $0$ respectively.

Therefore, $-1.5$ cannot be the probability of an event.

Thus, option (B) is the correct answer.

3. If \[{P}\left( {E} \right)={0}.{05}\], what is the probability of ‘not\[{E}\]’?

Ans. It is known that, \[\text{P}\left( \text{E} \right)+\text{P}\left( \text{not E} \right)=\text{1}\]

$\begin{align} & \Rightarrow P\left( not\text{ }E \right)=1P\left( E \right) \\ & =10.05 \\ & =0.95 \\ \end{align}$

4. It is given that in a group of ${3}$ students, the probability of ${2}$ students not having the same birthday is \[{0}.{992}\]. What is the probability that the ${2}$ students have the same birthday?

Ans. Suppose that $E$ represents the event that the students have the same birthday.

Then, \[\text{P}\left( \text{E} \right)=\text{0}\text{.992}\].

It is known that, $P\left( E \right)+P\left( \overline{E} \right)=1$

$\begin{align} & \Rightarrow P\left( E \right)=1-P\left( \overline{E} \right) \\ & =1-0.992 \\ & =0.008 \\ \end{align}$

Thus, the probability that the $2$ students have the same birthday is $0.008$.

5. ${12}$ defective pens are accidently mixed with \[{132}\] good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Ans. The number of total favourable outcomes \[=\text{132+12}=\text{144}\].

There are $132$ favourable outcomes.

Thus, the probability of getting a good pen $=\frac{132}{144}=\frac{11}{12}$.

6. Which of the following is polynomial?

(a) ${{{x}}^{{2}}}-{6}\sqrt{{x}}+{2}$

(b) $\sqrt{{x}}+\frac{{1}}{\sqrt{{x}}}$

(c) $\frac{{5}}{{{{x}}^{{2}}}-{3x}+{1}}$

(d) none of these

Ans. (d) none of these.

7. Polynomial ${2}{{{x}}^{{4}}}{+3}{{{x}}^{{3}}}{-5}{{{x}}^{{2}}}-{5}{{{x}}^{{2}}}{+9x+1}$ is a

(a) linear polynomial

(b) quadratic polynomial

(c) cubic polynomial

(d) bi-quadratic polynomial

Ans. Since, the degree of the polynomial $\text{2}{{\text{x}}^{\text{4}}}\text{+3}{{\text{x}}^{\text{3}}}\text{-5}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{2}}}\text{+9x+1}$ is $4$, so it is a bi-quadratic polynomial.

Thus, option (d) is the correct answer.

8. If ${\alpha }$ and ${\beta }$ are zeros of ${{{x}}^{{2}}}+{5x}+{8}$, then the value of $\left( {\alpha +\beta } \right)$ is

(a) \[{5}\]

(b) \[-{5}\]

(c) \[{8}\]

(d) \[-{8}\]

Ans. Since, $\alpha $ and $\beta $ are the zeroes of ${{x}^{2}}+5x+8$, so the value of $\alpha +\beta =-5$.

Thus, option (b) is the correct answer.

9. The sum and product of the zeros of a quadratic polynomial are ${2}$ and \[-{15}\] respectively. The quadratic polynomial is

(a) ${{{x}}^{{2}}}{-2x+15}$

(b) ${{{x}}^{{2}}}{-2x}-{15}$

(c) ${{{x}}^{{2}}}+{2x}-{15}$

(d) ${{{x}}^{{2}}}+{2x+15}$

Ans. The polynomial with the sum of zeroes $2$ and product of zeroes $-15$ is given by ${{x}^{2}}-\left( 2 \right)x+\left( -15 \right)$, that is, ${{x}^{2}}-2x-15$.

So, option (b) is the correct answer.

10. Cards each marked with one of the numbers \[{4},{5},{6},\ldots ,{20}\] are placed in a box and mixed thoroughly. One card is drawn at random from the box, what is the probability of getting an even prime number?

(a) ${0}$

(b) \[{1}\]

(c) \[{2}\]

(d) \[{3}\]

Ans. The number of possible outcomes is $17$.

Now, the only number which is even and prime is $2$, but it does not belong to the list $4,5,6,...,20$.

So, the probability of getting an even prime $=\frac{0}{17}=0$.

Thus, option (a) is the correct answer.

11. A bag contains \[{5}\] red and \[{4}\] black balls. A ball is drawn at random from the bag. What is the probability of getting a black ball?

(a) $\frac{{1}}{{3}}$

(b) $\frac{{2}}{{9}}$

(c) $\frac{{4}}{{9}}$

(d) None of these

Ans. There are a total of $5+4=9$ bags and from them, $4$ are black.

Therefore, the probability of getting a black ball $=\frac{4}{9}$.

Thus, option (c) is the correct answer.

12. A dice is thrown once, what is the probability of getting a prime number?

(a) ${1}$

(b) $\frac{{1}}{{2}}$

(c) $\frac{{3}}{{2}}$

(d) ${0}$

Ans. The number of possible outcomes while throwing one dice is $6$.

Now, the list of $3$ prime numbers is $2,3,5$.

Therefore, the probability of getting a prime number is $=\frac{3}{6}=\frac{1}{2}$.

Thus, option (b) is the correct answer.

13. What is the probability that a number selected from the numbers ${1},{2},{3},\ldots ,{15}$ is a multiple of ${4}$?

(a) $\frac{{1}}{{5}}$

(b) $\frac{{1}}{{2}}$

(c) $\frac{{2}}{{3}}$

(d) ${1}$

Ans. The number of total possible outcomes is $15$.

The list of $3$ numbers from $1$ to $15$,that are multiple of $4$ is given by $4,8,12$.

Therefore, the probability of selecting a number that is multiple of $4$ $=\frac{3}{15}=\frac{1}{5}$.

Thus, option (a) is the correct answer.

14. Cards marked with the numbers ${2}$ to \[{51}\] are placed in a box and mixed thoroughly. One card is drawn from this box, find the probability that the number on the card is an even number.

(a) $\frac{{1}}{{2}}$

(b) ${1}$

(c) $\frac{{3}}{{2}}$

(d) None of these

Ans. From $2$ to $51$, there are total $50$ numbers and so the total number of outcomes is $50$.

The number of events from $2$ to $51$ is $\frac{50}{2}=25$.

Thus, the probability of getting a card with an even number $=\frac{25}{50}=\frac{1}{2}$.

Hence, option (a) is the correct answer.

15. The king, queen and jack of clubs are removed from a deck of \[{52}\] playing cards and then well shuffled. One card is selected from the remaining cards, find the probability of getting a king.

(a) $\frac{{3}}{{49}}$

(b) ${1}$

(c) $\frac{{7}}{{17}}$

(d) none of these

Ans. Since, from the deck of $52$ cards, one king, queen, and jack are removed, so the total cards remaining is $52-3=49$.

In $49$ cards, there are a total of $3$ king cards.

Thus, the probability of getting a king $=\frac{3}{49}$.

Hence, option (a) is the correct answer.

16. What is the probability of getting a number less than ${7}$ in a single throw of a die?

(a) $\frac{{1}}{{2}}$

(b) ${0}$

(c) ${1}$

(d) none of these

Ans. The number of total possible outcomes is $6$.

All the numbers less than $7$ are $1,2,3,4,5,6$.

Thus, the probability of getting a number less than $7$ in a single throw is $=\frac{6}{6}=1$.

Hence, option (c) is the correct answer.

17. One card is drawn from a well shuffled deck of ${52}$ cards. Find the probability of drawing ‘${10}$’ of a black suit.

(a) $\frac{{1}}{{26}}$

(b) ${1}$

(c) $\frac{{1}}{{2}}$

(d) ${0}$

Ans. The number of possible outcomes is $52$.

The number of cards with $10$ of black suit is $2$.

Thus, the probability of drawing ‘$10$’ of a black suit $=\frac{2}{52}=\frac{1}{26}$

Hence, option (a) is the correct answer.

18. Cards each marked with one of the numbers \[{4,5,6,\ldots ,20}\] are placed in a box and mixed thoroughly. One card is drawn at random from the box, what is the probability of getting an even prime number?

(a) \[{0}\]

(b) \[{1}\]

(c) \[{2}\]

(d) \[{3}\]

Ans. The number of possible outcomes is $17$.

Now, the only number which is even and prime is $2$, but it does not belong to the list $4,5,6,...,20$.

So, the probability of getting an even prime $=\frac{0}{17}=0$.

Thus, option (a) is the correct answer.

19. A bag contains ${5}$ red and ${4}$ black balls. A ball is drawn at random from the bag. What is the probability of getting a black ball?

(a) $\frac{{1}}{{3}}$

(b) $\frac{{2}}{{9}}$

(c) $\frac{{4}}{{9}}$

(d) None of these

Ans. There are a total of $5+4=9$ bags and from them, $4$ are black.

Therefore, the probability of getting a black ball $=\frac{4}{9}$.

Thus, option (c) is the correct answer.

20. A dice is thrown once, what is the probability of getting a prime number?

(a) ${1}$

(b) $\frac{{1}}{{2}}$

(c) $\frac{{3}}{{2}}$

(d) ${0}$

Ans. The number of possible outcomes while throwing one dice is $6$.

Now, the list of $3$ prime numbers is $2,3,5$.

Therefore, the probability of getting a prime number is $=\frac{3}{6}=\frac{1}{2}$.

Thus, option (b) is the correct answer.

21. What is the probability that a number selected from the numbers \[{1,2,3,\ldots ,15}\] is a multiple of ${4}$?

(a) $\frac{{1}}{{5}}$

(b) $\frac{{1}}{{2}}$

(c) $\frac{{2}}{{3}}$

(d) ${1}$

Ans. The number of possible outcomes is $15$.

Now, the list of the $3$ numbers between $1$ and $15$, which are multiple of $4$ are $4,8,12$.

Therefore, the required probability $=\frac{3}{15}=\frac{1}{5}$.

Thus, option (a) is the correct answer.

22. If \[{E}\] be any event, then value of \[{P}\left( {E} \right)\] lie in between

(a) ${0}<{P}\left( {E} \right)<{1}$

(b) ${0}\le {P}\left( {E} \right)<{1}$

(c) ${0}\le {P}\left( {E} \right)\le {1}$

(d) ${0}\le {P}\left( {E} \right)\le {2}$

Ans. By the definition of probability, the maximum and minimum value of probability are $1$ and $0$ respectively.

So, $0\le P\left( E \right)\le 1$.

Thus, option (c) is the correct answer.

23. Maximum and minimum value of probability is

(a) $\left( {1},{1} \right)$

(b) \[\left( {1},{0} \right)\]

(c) \[\left( {0},{1} \right)\]

(d) none of these

Ans. By the definition of probability, the maximum and minimum value of probability are $1$ and $0$ respectively.

Thus, the option (b) is the correct answer.

24. An unbiased die is thrown. What is the probability of getting an even number or a multiple of \[{3}\]?

(a) $\frac{{2}}{{3}}$

(b) $\frac{{3}}{{2}}$

(c) ${1}$

(d) none of these

Ans. The number of possible outcomes while throwing an unbiased die is $6$ and the number of events is $3$ (which are $2,4,6$).

Also, the numbers, which are multiple of $3$ are $3,6$.

Thus, the $4$ numbers which are even or multiple of $3$ are $2,3,4,6$.

Hence, the probability of getting an even number or a multiple of $3$ is $=\frac{4}{6}=\frac{2}{3}$.

So, option (a) is the correct answer.

25. Let \[{E}\] be any event, then the value of \[{P}\left( {E} \right)+{P}\left( {not}\text{ }{E} \right)\] equals to

(a) ${1}$

(b) ${0}$

(c) \[{3}\]

(d) $\frac{{1}}{{2}}$

Ans. By the definition of probability,

if $E$ is an event, then the value of \[P\left( E \right)+P\left( \overline{E} \right)=1\], that is,

$P\left( E \right)+P\left( \text{not E} \right)=1$

So, option (a) is the correct answer.

26. Degree of polynomial ${{{y}}^{{3}}}{-2}{{{y}}^{{2}}}{-}\sqrt{{3y}}{+}\frac{{1}}{{2}}$ is

(a) $\frac{{1}}{{2}}$

(b) ${2}$

(c) ${3}$

(d) $\frac{{3}}{{2}}$

Ans. It is known that the degree of a polynomial is the highest power of the variable contained in that polynomial.

Therefore, the degree of the polynomial ${{y}^{3}}-2{{y}^{2}}-\sqrt{3y}+\frac{1}{2}$ is $3$.

Thus, option (c) is the correct answer.

27. Zeros of \[{P}\left( {x} \right)={2}{{{x}}^{{2}}}+{9x}{35}\] are

(a) ${7}$ and $\frac{{5}}{{2}}$

(b) \[-{7}\] and $\frac{{5}}{{2}}$

(c) \[{7}\] and ${5}$

(d) ${7}$ and \[{2}\]

Ans. The zeros of the polynomial $P\left( x \right)=2{{x}^{2}}+9x-35$ are $-7$ and $\frac{5}{2}$, because $P\left( -7 \right)=2{{\left( -7 \right)}^{2}}+9\left( -7 \right)-35=0$ and

$P\left( \frac{5}{2} \right)=2{{\left( \frac{5}{2} \right)}^{2}}+9\left( \frac{5}{2} \right)-35=0$.

Thus, option (b) is the correct answer.

28. The quadratic polynomial whose zeros are ${3}$ and \[-{5}\] is

(a) \[{{{x}}^{{2}}}+{2x}{15}\]

(b) \[{{{x}}^{{2}}}+{3x}{8}\]

(c) \[{{{x}}^{{2}}}{ 5x15}\]

(d) None of these

Ans. The polynomial with the zeroes $3$ and $-5$ is given by

${{x}^{2}}-\left( 3-5 \right)x+3\left( -5 \right)$

$={{x}^{2}}+2x-15$.

So, option (a) is the correct answer.

29. If ${\alpha }$ and ${\beta }$ are the zeros of the quadratic polynomial\[{P}\left( {x} \right)={{{x}}^{{2}}}{px+q}\], then the value of ${{{\alpha }}^{{2}}}{+}{{{\beta }}^{{2}}}$ is equal to

(a) \[{{{p}}^{{2}}}{2q}\]

(b) $\frac{{p}}{{q}}$

(c) \[{{{q}}^{{2}}}{2p}\]

(d) none of these

Ans. Since, $\alpha $ and $\beta $ are the roots of the polynomial $P\left( x \right)={{x}^{2}}-px+q$, so $\alpha +\beta =p$ …… (i)

and $\alpha \beta =q$.

Therefore,

$\begin{align} & {{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \\ & ={{p}^{2}}-2q \\ \end{align}$

So, option (a) is the correct answer.

2 Marks Questions

1. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Ans. “A driver tries to start a car” in the experiment. We cannot presume that each event is equally likely to occur since the car starts or does not start. As a result, there are no equally likely outcomes in the experiment.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Ans. “A player tries to shoot a basketball,” says the experiment. We cannot assume that each result is equally likely to occur, whether she shoots or misses the shot. As a result, no equally likely outcomes are possible in the experiment.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Ans. During the test, “A true-false question is asked in a trial. The response is correct or incorrect.” We know with certainty that the outcome will be one of two likely outcomes: right or wrong. We can reasonably expect each event, correct or wrong, to occur in the same way.

As a result, the likelihood of doing it right or wrong are equal.

(iv) A baby is born. It is a boy or a girl.

Ans. “A baby is born, it is a boy or a girl,” says the experiment. We know with certainty that the outcome will be one of two likely outcomes: a boy or a girl. We have reason to believe that each event, boy or girl, is equally likely to occur. As a result, both boy and female outcomes are equally likely.

2. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Ans. Because we know that a coin toss can only land in one of two ways – head up or tail up – the tossing of a coin is regarded as a fair manner of selecting which team should have the ball at the start of a football game. It is reasonable to infer that either event, whether head or tail, has the same probability of occurring as the other, i.e., the outcomes head and tail are equally likely to occur. So, the outcome of a coin toss is absolutely unexpected.

3. A bag contains lemon flavoured candles only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:

(i) an orange flavoured candy?

Ans. Consider the occurrence with the experiment of removing an orange-flavoured candy from a bag of lemon-flavored candies.

Note that, there are not any outcomes that represent an orange flavoured candy.

Hence, the event is impossible and so its probability is $0$.

(ii) a lemon flavoured candy?

Ans. Consider the event of taking a lemon flavoured candy from a bag that contains only lemon flavoured candies.

This event represents a certain event. So, its probability is $1$.

4. A bag contains ${3}$ red balls and \[{5}\] black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

(i) red?

Ans. The number of total balls in the bag is $3+5=8$. Therefore, there are a total of $8$ possible outcomes.

Since the bag contains 3 red balls, the number of favourable outcomes is $=3$.

Thus, the probability of getting a red ball $=\frac{3}{8}$.

(ii) not red?

Ans. There are $5$ balls, which are not red.

So, the number of favourable outcomes is $5$.

Thus, the probability of obtaining a ball that is not red $=\frac{5}{8}$.

5. A box contains ${5}$ red marbles, \[{8}\] white marbles and \[{4}\] green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be:

(i) red?

Ans. The number of total marbles in the box \[=\text{5+8+4}=\text{17}\].

So, the number of possible outcomes is $17$.

The number of red marbles in the box is $5$.

Therefore, the number of favourable outcomes is $5$.

Hence, the probability of getting a red marble $=\frac{5}{17}$.

(ii) white?

Ans. The number of white marbles in the box is $8$.

So, the number of favourable outcomes $=8$.

Thus, the probability of getting a white marble $=\frac{8}{17}$.

(iii) not green?

Ans. The number of marbles which are not green is \[\text{5+8}=\text{13}\].

So, the number of favourable outcomes is $13$.

Thus, the probability of not obtaining a green marble $=\frac{13}{17}$.

6. A piggy bank contains hundred ${50}$ p coins, fifty Re. ${1}$ coins, twenty Rs. \[{2}\] coins and ten Rs. \[{5}\] coins. If it is equally likely that of the coins will fall out when the bank is turned upside down, what is the probability that the coin:

(i) will be a \[{50}\] p coin?

Ans. The number of total coins in a piggy bank \[=\text{100+50+20+10}=\text{180}\].

That is, the number of total possible outcomes is $180$.

Now, since, in the piggy bank, the number of \[50\] coins are $100$, so there are $100$ favourable outcomes.

Thus, the probability of falling out of a \[50\] p coin $=\frac{100}{180}=\frac{5}{9}$.

(ii) will not be a Rs. ${5}$ coin?

Ans. Except the Rs. $5$ coin, there are \[\text{100+50+20}=\text{170}\] coins.

So, the number of favourable outcomes is $170$.

Thus, the probability of falling out of a coin other than Rs. $5$ coin $=\frac{170}{180}=\frac{17}{18}$.

7. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing ${5}$ male fishes and ${8}$ female fishes. What is the probability that the fish taken out is a male fish?

Ans. The total number of fish (male and female) in the tank \[=\text{5 + 8}=\text{13}\].

So, the total number of possible events is $13$.

Now, since, the number of male fishes in the tank is $5$, so the number of favourable outcomes is $5$.

Thus, the probability of taking out a male fish $=\frac{5}{13}$.

8. Five cards – then ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

Ans. The number of total possible outcomes is $5$.

Since, the number of Queen is $1$, so

the number of favourable outcomes$=1$.

Thus, the probability of getting a Queen card $=\frac{1}{5}$.

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Ans. Since, one card is put aside, so now, the number of total possible outcomes is $5-1=4$.

(a) Since, there is only one ace card, so the number of favourable outcomes is $1$.

Thus, the probability of getting an ace card $=\frac{1}{4}$.

(b) Since, there is no queen card left after the first pick up, so the number of favourable outcomes is $0$.

Thus, the probability of getting a Queen card $=\frac{0}{4}=0$.

9. (i) A lot of ${20}$ bulbs contain ${4}$ defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

Ans. The number of total possible outcomes is $20$.

The number of defective bulbs is $4$.

Therefore, the favourable outcomes are $4$.

Thus, the probability of getting a defective bulb $=\frac{4}{20}=\frac{1}{5}$.

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Ans. Since, one defective bulb is replaced, so the number of favourable outcomes \[=\text{20 -- 1}=\text{19}\].

Now, there are total $19-4=15$ non-defective bulbs.

That is, the number of favourable outcomes is $15$.

Thus, the probability of getting a non-defective bulb is $=\frac{15}{19}$.

10. A box contains \[{90}\] discs which are numbered from ${1}$ to \[{90}\]. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

Ans. Since, there are a total of $90$discs, so the number of possible outcomes is $90$. Now, the number of two-digit numbers from $1$ to $90$ is \[\text{90 -- 9 }=\text{81}\].

So, the number of favourable outcomes is $81$.

Thus, the probability of getting a disc with a two-digit number $=\frac{81}{90}=\frac{9}{10}$.

(ii) a perfect square numbers.

Ans. From 1 to 90, the perfect squares are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

Favourable outcomes = 9

Hence P (getting a perfect square) =

(iii) a number divisible by ${5}$.

Ans. The numbers divisible by 5 from 1 to 90 are 18.

Favourable outcomes = 18

Hence P (getting a number divisible by 5) =

11. A child has a die whose six faces show the letters as given below: \[{A},{B},{C},{D},{E},{A}\].

The die is thrown once. What is the probability of getting:

(i) ${A}$?

Ans. The number of total possible outcomes $=6$.

Since, there are $2$ $A's$, so the number of possible outcomes $=2$.

Thus, the probability of getting a letter $A$ $=\frac{2}{6}=\frac{1}{3}$.

(ii) ${D}$?

Ans. Since, there is only one $D$, so the number of favourable outcomes is $1$.

Therefore, the probability of getting a letter $D$$=\frac{1}{6}$

12. Suppose you drop a die at random on the rectangular region shown in the figure given on the next page. What is the probability that it will land inside the circle with diameter ${1}$ m?

(Image will be uploaded soon)

Ans. The area of rectangle (in the given figure) $=3\times 2=6\text{ }{{\text{m}}^{2}}$

Also, the area of the circle inside the rectangle,

$\begin{align} & =\pi {{r}^{2}} \\ & =\pi \left( {{\frac{1}{2}}^{2}} \right) \\ & =\frac{\pi }{4}\text{ }{{\text{m}}^{2}} \\ \end{align}$

Thus, the probability that the die will land inside the circle $=\frac{\frac{\pi }{4}}{6}=\frac{\pi }{24}$.

13. A lot consists of \[{144}\] ball pens of which \[{20}\] are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:

(i) she will buy it

Ans. The total number of ball pens is $144$.

There are $20$ defective pens.

So, the number of good pens \[=\text{144 --20}=\text{124}\].

That is, there are $124$ favourable outcomes.

Hence, the probability that she will buy $=\frac{124}{144}=\frac{31}{36}$.

(ii) she will not buy it?

Ans. There are $20$ favourable outcomes.

Thus, the probability that she will not buy $=\frac{20}{144}=\frac{5}{36}$.

14. A bag contains ${5}$ red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Ans. Suppose that the number of blue balls in the bag is $x$.

Therefore, the number of total balls contained in the bag $=x+5$.

Then, the probability that a blue ball is drawn, ${{P}_{1}}=\frac{x}{x+5}$.

Also, the probability that a red ball is drawn, ${{P}_{2}}=\frac{5}{x+5}$.

Now, by the given condition,

$\begin{align} & {{P}_{1}}=2{{P}_{2}} \\ & \Rightarrow \frac{x}{x+5}=2\cdot \frac{5}{x+5} \\ & \Rightarrow \frac{x}{{x+5}}=2\cdot \frac{5}{{x+5}} \\ & \Rightarrow x=10 \\ \end{align}$

Thus, the number of blue balls in the bag is $10$.

15. A box contains ${12}$ balls out of which ${x}$ are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?

If ${6}$ more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find ${x}$.

Ans. It is given that the box contains $12$ balls.

So, there are a total of $12$ possible outcomes.

Now, let there are $x$ favourable outcomes while drawing a black ball.

Then, the probability of getting a black ball, ${{P}_{1}}=\frac{x}{12}$.

Now, if \[6\] more balls are put in the box, then the number of possible outcomes becomes \[=\text{12 + 6}=\text{18}\].

Then, there are $x+6$ favourable outcomes.

Therefore, now, the probability of getting a black ball, ${{P}_{2}}=\frac{x+6}{18}$.

So, the given condition,

$\begin{align} & {{P}_{2}}=2{{P}_{1}} \\ & \Rightarrow \frac{x+6}{18}=2\cdot \frac{x}{12} \\ & \Rightarrow \frac{x+6}{18}=\frac{x}{6} \\ & \Rightarrow 6x+36=18x \\ & \Rightarrow 12x=36 \\ & \Rightarrow x=3 \\ \end{align}$

Thus, the value of $x$ is $3$.

16. A jar contains ${24}$ marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is $\frac{{2}}{{3}}$. Find the number if blue marbles in the jar.

Ans. Since, the jar contains $24$ marbles, so the number of favourable outcomes is $=24$.

Suppose that there are $x$ green marbles.

Therefore, the number of favourable outcomes is $x$.

So, the probability of getting a green marble is $\frac{x}{24}$.

It is given that the probability that the marble drawn is green $=\frac{2}{3}$.

That is, $\frac{x}{24}=\frac{2}{3}$

$\Rightarrow x=16$

So, the number of green marbles is $16$.

Hence, the number of blue marbles \[=\text{24 -- 16}=\text{8}\].

17. Why is tossing a coin considered is the way of deciding which team should get the ball at the beginning of a football match?

Ans. When a coin is tossed, the probability of getting head $=\frac{1}{2}$ and

probability of getting a tail $=\frac{1}{2}$.

So, both probabilities are the same.

That’s why, tossing a coin is considered to be a fair way of deciding which team should get the ball.

18. An unbiased die is thrown, what is the probability of getting an even prime number?

Ans. When an unbiased die is thrown, the list of $6$ outcomes is given by \[\text{1,2,3,4,5,6}\].

Also, the number of favourable outcomes is $1$ (Since, $2$ is the only number which is even and prime.)

Thus, the probability of getting an even prime number is $\frac{1}{6}$.

19. Two unbiased coins are tossed simultaneously, find the probability of getting two heads.

Ans. The list of $4$ outcomes is given by \[\text{HH, HT, TH, TT}\].

The only favourable outcome is \[HH\] [Two heads]

Thus, the probability of getting two heads while tossing two unbiased coins simultaneously $=\frac{1}{4}$.

20. One card is drawn from a well shuffled deck of ${52}$ cards. Find the probability of getting a jack of hearts.

Ans. The number of total outcomes is $52$.

The only favourable case is $1$. (since, there exists only one jack of hearts)

Hence, the probability of getting a jack of hearts $=\frac{1}{52}$.

21. A game consists of tossing a one-rupee coin ${3}$ times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three fails and loses otherwise. Calculate the probability that Hanif will lose the game.

Ans. It is given that a coin is tossed thrice.

So, the list of possible outcomes is given by \[\left\{ \text{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} \right\}\].

Therefore, the number of possible outcomes is $8$.

The outcomes with \[3\] heads and \[3\] tails are \[\text{HHH,TTT}\].

So, the number of favourable outcomes is $2$.

Thus, the probability that Hanif will win the game $\frac{2}{8}=\frac{1}{4}$

The probability that Hanif will lose the game $=1-\frac{1}{4}=\frac{3}{4}$.

22. Gopy buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing \[~{5}\] male fish and \[{8}\] female fish. What is the probability that the fish taken out is a male fish?

Ans. The number of total fishes \[=\text{5+8}=\text{13}\].

There are $5$ male fishes.

Thus, the probability that the fish taken out is a male $=\frac{5}{13}$.

23. A lot consists of \[{144}\] ball pens of which \[{20}\] are defective and the others are food. Arti will buy a pen if it is good but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) she will buy it

Ans. The total number of ball pens is $144$.

There are $20$ defective pens.

So, the number of good pens \[=\text{144 --20}=\text{124}\].

The probability that she will buy $=\frac{124}{144}=\frac{31}{36}$.

(ii) she will not buy it?

Ans. The probability that she will not buy $=\frac{20}{144}=\frac{5}{36}$.

24. Harpreet tosses two different coins simultaneously (say one is of Rs ${1}$ and other is Rs ${2}$), what is the probability that she gets “at least one head”?

Ans. The list of $4$ possible outcomes is given by \[\left\{ \text{HH,TT,TH,HT} \right\}\].

The list of $3$ outcomes when at least one head is obtained is given by $\left\{ \text{TH,HT,HH} \right\}$.

Thus, the probability of getting at least one head $=\frac{3}{4}$.

25. Why is tossing a coin considered is the way of deciding which team should get the ball at the beginning of a football match?

Ans. When a coin is tossed, the probability of getting head $=\frac{1}{2}$ and

probability of getting a tail $=\frac{1}{2}$.

That is, both the probabilities are the same.

That’s why, tossing a coin is considered to be a fair way of deciding which team should get the ball.

26. Two unbiased coins are tossed simultaneously, find the probability of getting two heads.

Ans. The list of $4$ outcomes while tossing two unbiased coins is given by \[\text{HH, HT, TH, TT}\].

Therefore, the only favourable outcome is $HH$.

Hence, the probability of getting two heads while tossing two unbiased coins $=\frac{1}{4}$.

27. One card is drawn from a well shuffled deck of ${52}$ cards. Find the probability of getting a jack of hearts.

Ans. The number of total outcomes is $52$.

The number of favourable cases is $1$. (since, in a deck of playing cards, there is only one jack of hearts)

Hence, the probability of getting a jack of hearts is $=\frac{1}{52}$.

28. If two dice are thrown once, find the probability of getting ${9}$.

Ans. The number of total possible outcomes while throwing two dice is $6\times 6=36$.

The possible outcomes for getting $9$ are \[\left( \text{3+6} \right)\text{,}\left( \text{4+5} \right)\text{,}\left( \text{5+4} \right)\text{,}\left( \text{6+3} \right)\].

That is, the favourable outcomes are $4$.

Hence, the probability of getting $9$ while throwing two dice is $=\frac{4}{36}=\frac{1}{9}$.

29. A card is drawn from a well shuffled deck of playing cards. Find the probability of getting a face card.

Ans. It is known that the deck of playing cards contains $52$ cards. So, the number of total possible outcomes is $52$.

The number of favourable outcomes is \[=\text{4+4+4}=\text{12}\], where $4$ cards are jack, $4$ cards are queen and $4$ cards are king.

Thus, the probability of getting a face card while drawing a random card from the well shuffled deck is $=\frac{12}{52}=\frac{3}{13}$.

30. What is the probability of having ${53}$ Mondays in a leap year?

Ans. It is known that, there is $366$ days in a leap year containing $52$weeks and $2$ days.

These two days can be \[\left( MT \right),\left( TW \right),\left( W\text{ }Th \right),\left( Th\text{ }F \right),\left( FS \right),\left( SS \right),\left( SM \right)\]. Now, since, $52$ weeks contain $52$ Mondays, so the remaining Monday can be obtained if the two days are \[\left( MT \right)\] or \[\left( SM \right)\], that is, $2$ out of $7$ cases. Thus, the probability of having $53$ Mondays in a leap year $=\frac{2}{7}$.

31. Cards bearing numbers ${3}$ to \[{20}\] are placed in a bag and mixed thoroughly. A card is taken out from the bag at random, what is the probability that the number on the card taken out is an even number?

Ans. The number of total outcomes is \[=203=17\].

The list of $9$ even numbers marked on the cards is given by\[\text{4,6,8,10,12,14,16,18,20}\].

So, there are a total of $9$ favourable outcomes.

Thus, the probability of getting a card marked with an even number is $=\frac{9}{17}$.

3 Marks Questions

1. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers \[{1,2,3,4,5,6,7,8}\] (see figure) and these are equally likely outcomes. What is the probability that it will point at:

(Image will be uploaded soon)

(i) ${8}$?

Ans. An arrow can point to any of the $8$ numbers in eight different ways.

Therefore, there are $8$ possible outcomes.

So, the number $8$ can appear only one way.

Thus, the probability that the arrow points at $8$ is $=\frac{1}{8}$.

(ii) an odd number?

Ans. In the circle, there are only $4$ odd numbers, viz., $1,3,5,7$.

So, there are $4$ favourable outcomes.

Thus, the probability that the arrow points at an odd number is $=\frac{4}{8}=\frac{1}{2}$.

(iii) a number greater than ${2}$?

Ans. In the circle, the numbers greater than $2$ are $3,4,5,6,7,8$.

So, there are a total of $6$ favourable outcomes.

Thus, the probability that the arrow points at a number greater than $2$ is $=\frac{6}{8}=\frac{3}{4}$.

(iv) a number less than ${9}$?

Ans. In the circle, the numbers that are less than $9$ are $1,2,3,4,5,6,7,8$.

So, there are $8$ possible outcomes.

Thus, the probability that the arrow points at a number less than $9$, is $=\frac{8}{8}=1$.

2. A dice is thrown once. Find the probability of getting:

(i) a prime number.

Ans. The total number of favourable outcomes from a dice roll is $6$.

The prime numbers that can appear in the dice are \[\text{2, 3 and 5}\].

Thus, the number of possible outcomes is $3$.

So, the probability of getting a prime number is $=\frac{3}{6}=\frac{1}{2}$

(ii) a number lying between ${2}$ and ${6}$.

Ans. On a dice, the numbers which lie between $2$ and $6$ are \[\text{3, 4, 5}\].

So, there are a total of $3$ favourable outcomes.

Thus, the probability of getting a number lying between $2$ and $6$ is $=\frac{3}{6}=\frac{1}{2}$.

(iii) an odd number.

Ans. The odd numbers on a dice are \[\text{1, 3 and 5}\].

So, there are a total of $3$ favourable outcomes.

Thus, the probability of obtaining an odd number is $=\frac{3}{6}=\frac{1}{2}$.

3. A game consists of tossing a one-rupee coin ${3}$ times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e., three heads or three tails and loses otherwise. Calculate the probability that Hanif will lose the game.

Ans. When a coin is tossed three times, then the outcomes are: \[\text{HHH, HHT, HTH, THH, TTH, HTT, THT, TTT}\].

So, there are a total of $8$ possible outcomes.

The outcomes when he loses the game are: $HHT,\text{ HTH, THH, TTH, HTT, THT}$.

Therefore, there are a total of $6$ favourable outcomes.

Thus, the probability of losing the game is $=\frac{6}{8}=\frac{3}{4}$.

4. A die is numbered in such a way that its faces show the numbers \[{1,2,2,3,3,6}\]. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

+ | 1 | 2 | 2 | 3 | 3 | 6 |

1 | 2 | 3 | 3 | 4 | 4 | 7 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

2 | 5 | |||||

3 | ||||||

3 | 5 | 9 | ||||

6 | 7 | 8 | 8 | 9 | 9 | 12 |

What is the probability that the total score is:

(i) even

Ans. The following describes the complete table:

+ | 1 | 2 | 2 | 3 | 3 | 6 |

1 | 2 | 3 | 3 | 4 | 4 | 7 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

6 | 7 | 8 | 8 | 9 | 9 | 12 |

There are total $6$ rows and $6$ columns in the table, so $6\times 6=36$ possible outcomes can be obtained.

Now, the number of favourable outcomes of obtaining a total score that is even is $18$.

Thus, the probability of obtaining a total score even is $=\frac{18}{36}=\frac{1}{2}$.

(ii) ${6}$

Ans. The favourable outcomes of obtaining a total score $6$ are $4$.

Thus, the probability of obtaining a total score of $6$ is $=\frac{4}{36}=\frac{1}{9}$.

(iii) at least ${6}$?

Ans. The favourable outcomes of obtaining a total score at least $6$ are \[15\]. Thus, the probability of obtaining the total score at least $6$ is $=\frac{15}{36}=\frac{5}{12}$.

5. \[{18}\] cards numbered \[{1,2,3,\ldots ,18}\] are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probabilities that the card bears

(i) an even number.

Ans. From $1$ to $18$, there are $18$ cards.

So, the number of possible outcomes is $18$.

Now, the list of the $9$ even numbers marked on the cards is given by \[\text{2,4,6,8,10,12,14,16,18}\].

Thus, the probability of drawing a card marked with an even number $=\frac{9}{18}=\frac{1}{2}$.

(ii) a number divisible by ${2}$ or ${3}$.

Ans. The list of $12$ numbers that are divisible by $2$ or $3$ are \[\text{2,3,4,6,8,9,10,12,14,15,16,18}\].

So, the number of favourable outcomes is $12$.

Thus, the probability of drawing a card marked with a number divisible by $2$ or $3$ is $=\frac{12}{18}=\frac{2}{3}$.

6. A bag contains \[{5}\] red balls, \[{4}\] green balls and \[{7}\] white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is

(a) white

Ans. The bag contains a total of $5+4+7=16$ balls.

So, there are a total of $16$ possible outcomes.

Now, the number of white balls in the bag is $7$, that is, the number of favourable outcomes is $7$.

Thus, the probability of drawing a white ball is $=\frac{7}{16}$.

(b) neither red nor white

Ans. The number of balls that are neither red nor white $=$ the number of green balls $=4$.

So, the number of possible outcomes is $4$.

Hence, the probability of drawing a ball that is neither red nor white $=\frac{4}{16}=\frac{1}{4}$.

7. A box contains \[{20}\] balls bearing numbers \[{1,2,3,4,\ldots ,20}\]. A ball is drawn at random from the box, what is the probability that the number on the ball is

(i) an odd number

Ans. From $1$ to $20$, there are a total $20$ balls, that is, the number of total outcomes is $20$.

Now, the list of $10$ odd numbers from $1$ to $20$ is given by \[\text{1,3,5,7,9,11,13,15,17,19}\].

Thus, the probability of drawing a ball bearing an odd number is $=\frac{10}{20}=\frac{1}{2}$.

(ii) divisible by \[{2}\] or \[{3}\]

Ans. The list of $10$ numbers that are divisible by $2$ is given by

\[\text{2,4,}\underline{\text{6}}\text{,8,10,}\underline{\text{12}}\text{,14,16,}\underline{\text{18}}\text{,20}\].

The list of $6$ numbers that are divisible by $3$ is given by

\[\text{3,}\underline{\text{6}}\text{,9,}\underline{\text{12}}\text{,15,}\underline{\text{18}}\].

Now, the $3$ numbers that are divisible by both $2$ and $3$ are \[\text{6,12,18}\].

So, the numbers divisible by $2$ or $3$ are $=\text{10+6 --3}=\text{13}$.

Thus, there are a total of $13$ possible favourable outcomes.

Hence, the probability of drawing a ball bearing a number that is divisible by $2$ or $3$ is $=\frac{13}{20}$.

(iii) prime number

Ans. The list of $8$ prime numbers between $1$ and $20$ is given by \[\text{2,3,5,7,11,13,17,19}\].

So, the number of favourable outcomes is $8$.

Thus, the probability of drawing a ball bearing a prime number is $=\frac{8}{20}=\frac{2}{5}$.

8. A bag contains ${5}$ red and some blue balls,

(i) if probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag.

Ans. Suppose that there are $x$ blue balls in the bag.

So, there are a total of $x+5$ balls in the bag.

Therefore, the probability of drawing a blue ball $=\frac{x}{x+5}$ and

the probability of drawing a red ball $=\frac{5}{x+5}$.

Now, by the given condition,

$\frac{x}{{x+5}}=2\cdot \frac{5}{{5+x}}$ $\Rightarrow x=10$

Thus, there are a total of $10$ blue balls in the bag.

(ii) if probability of drawing a blue ball from the bag is four times that of a red ball, find the number of blue balls in the bag.

Ans. According to the given condition,

$\frac{x}{5+x}=4\cdot \frac{5}{5+x}$

$\Rightarrow \frac{x}{{5+x}}=4\cdot \frac{5}{{5+x}}$

$\Rightarrow x=20$

Thus, there are a total of $20$ blue balls in the bag.

9. A box contains ${3}$ blue marbles, ${2}$ white marbles. If a marble is taken out at random from the box, what is the probability that it will be a white one? Blue one? Red one?

Ans. The number of total possible outcomes is $3=2+4=9$.

There are a total of $2$ white marbles.

So, the probability of taking a white marble $=\frac{2}{9}$.

There are a total of $3$ blue marbles.

Therefore, the probability of taking a blue marble $=\frac{3}{9}=\frac{1}{3}$.

Also, there are a total of $4$ red marbles.

Thus, the probability of getting a red marble $=\frac{4}{9}$.

10. The integers from 1 to 30 inclusive are written on cards (one number on one card). These card one put in a box and well mixed. Joseph picked up one card. What is the probability that his card has

(i) number ${7}$

Ans. From $1$ to $30$, there are a total of $30$ integers.

So, there are a total of $30$ possible outcomes.

Thus, the probability of getting the number $7$ is $=\frac{1}{30}$

(ii) an even number

Ans. The list of $15$ even numbers from $1$ to $30$ is given by \[\text{2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}\].

So, there are a total of $15$ favourable outcomes.

Hence, the probability of picking up an even number $=\frac{15}{30}=\frac{1}{2}$.

(iii) a prime number

Ans. The list of $10$ prime numbers from $1$ to $30$ is given by \[\text{2,3,5,7,11,13,17,19,23,29}\].

So, there are a total of $10$ favourable outcomes.

Hence, the probability of picking up a prime number $=\frac{10}{30}=\frac{1}{3}$.

11. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavoured candy?

Ans. It is given that the bag has only lemon flavoured candies.

So, the probability of getting an orange flavoured candy $=\frac{0}{1}=0$.

(ii) a lemon flavoured candy?

Ans. It is provided that the bag contains only the lemon flavoured candies.

So, the probability of getting an orange flavoured candy $=\frac{1}{1}=1$.

12. A bag contains ${6}$ red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red, find the number of blue balls in the bag.

Ans. Let there be $x$ blue balls in the bag.

So, the bag contains a total of $x+6$ balls.

Therefore, the probability of drawing a blue ball $=\frac{x}{x+6}$ and

the probability of drawing a red ball $=\frac{6}{x+6}$.

Now, by the given condition,

$\frac{x}{{x+6}}=2\cdot \frac{6}{{6+x}}$

$\Rightarrow x=12$

Thus, there are a total of $12$ blue balls in the bag.

13. A bag contains \[{5}\] red, \[{4}\] black and \[{3}\] green balls. A ball is taken out of the bag at random, find the probability that the selected ball is

(i) of red colour.

Ans. The bag contains total \[5+4+3=12\] balls, that is, there are a total of $12$ outcomes.

Also, the number of red balls contained in the bag is $5$.

Hence, the probability of selecting a red ball is $=\frac{5}{12}$.

(ii) not of green colour.

Ans. The number of balls except the green ones \[=\text{12--3}=\text{9}\].

So, the number of favourable outcomes is $9$.

Thus, the probability of selecting a ball that is not green is $=\frac{9}{12}=\frac{3}{4}$.

14. From a well shuffled pack of \[{52}\] cards, black aces and black queens are removed. From the remaining cards a card is drawn at random, find the probability of drawing a king or a queen.

Ans. A pack of cards contains $52$ cards. From them, there are only $2$ black ace cards and $2$ black queen cards.

Therefore, the number of cards left $=52-2-2=48$.

So, there are $48$ equally likely cases available.

Also, in $48$ cards, there are a total \[4+2=6\] queens and kings left. Thus, the number of favourable cases is $6$.

Hence, the probability of drawing a king or a queen is $=\frac{6}{48}=\frac{1}{6}$.

15. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Ans. When a driver tries to start a car, the likelihood of the car starting or not starting are not equal.

(ii) A player attempts to shoot a basketball, she/he shoots or misses the shot.

Ans. When a player tries to shoot a basketball, the likelihood of hitting the target or missing the shot are not equal.

(iii) A baby is born. It is a boy or a girl.

Ans. A baby is born; whether it is a boy or a girl, is an equally likely event.

16. Find the probability that a number selected at random from the numbers \[{1,2,3,\ldots ,35}\] is a

(i) prime number

Ans. The list of $11$ prime numbers between $1$ and $35$ is given by\[\text{2,3,5,7,11,13,17,19,23,29,31}\].

Between $1$ and $35$, there are total $35$ numbers, so the number of total outcomes is $35$.

Thus, the probability that the chosen number is a prime $=\frac{11}{35}$.

(ii) multiple of ${7}$

Ans. The list of $5$ numbers between $1$ and $35$ that are multiple of $7$ is given by \[\text{7,14,21,28,35}\].

So, the number of favourable outcomes is $5$.

Thus, the probability of choosing a number that is multiple of $7$ $=\frac{5}{35}=\frac{1}{7}$

(iii) multiple of ${3}$ or ${5}$.

Ans. The list of $11$ numbers between $1$ and $35$ which are multiple of $3$ is given by \[3,6,9,12,\underline{15},18,21,24,27,\underline{30},33\].

Also, the list of $7$ numbers between $1$ and $35$ which are multiple of $5$ is given by \[5,10,\underline{15},20,25,\underline{30},35\].

Thus, the numbers that are multiple of both $3$ and $5$ are \[15,30\].

So, the numbers which are multiple of $3$ or $5$ is given by

$\begin{align} & =11+7-2 \\ & =16 \\ \end{align}$

Thus, the probability of getting a number that is multiple of $3$ or $5$ is

$=\frac{16}{35}$.

4 Marks Questions

1. One card is drawn from a well-shuffled deck of ${52}$ cards. Find the probability of getting:

(i) a king of red colour.

Ans. Total number of outcomes \[=52\].

There are two suits of red cards, that is, diamond and heart.

Each of the suits has one king.

So, there is only one favourable outcome.

Thus, the probability of getting a king of red colour $=\frac{2}{52}=\frac{1}{26}$.

(ii) a face card

Ans. The number of face cards in a pack is \[12\].

So, the number of favourable outcomes is $12$.

Thus, the probability of getting a face card is $=\frac{12}{52}=\frac{3}{13}$.

(iii) a red face card

Ans. It is known that there are $2$ suits of red cards, that is, diamond and heart and each of the suits contain $3$ face cards.

Therefore, there are a total of $6$ favourable outcomes.

Thus, the probability of getting a red face card $=\frac{6}{52}=\frac{3}{26}$.

(iv) the jack of hearts

Ans. Recall that, in a deck of $52$ cards, there is only one jack of heart.

So, the number of favourable outcomes is $1$.

Thus, the probability of getting the jack of hearts $=\frac{1}{52}$.

(v) a spade

Ans. It is known that there are \[13\] cards of spades.

So, the number of favourable outcomes is $13$.

Thus, the probability of getting a spade is $=\frac{13}{52}=\frac{1}{4}$.

(vi) the queen of diamonds.

Ans. Note that, there is only one queen of diamonds.

So, the number of favourable outcomes is $1$.

Thus, the probability of getting the queen of diamonds $=\frac{1}{52}$.

2. A die is thrown twice. What is the probability that:

(i) ${5}$ will not come up either time?

Ans. The following are the outcomes of the experiment where a dice is thrown twice:

\[\left( \text{1, 1} \right)\text{ }\left( \text{1, 2} \right)\text{ }\left( \text{1, 3} \right)\text{ }\left( \text{1, 4} \right)\text{ }\left( \text{1, 5} \right)\text{ }\left( \text{1, 6} \right)\]

\[\left( \text{2, 1} \right)\text{ }\left( \text{2, 2} \right)\text{ }\left( \text{2, 3} \right)\text{ }\left( \text{2, 4} \right)\text{ }\left( \text{2, 5} \right)\text{ }\left( \text{2, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{3, 1} \right)\text{ }\left( \text{3, 2} \right)\text{ }\left( \text{3, 3} \right)\text{ }\left( \text{3, 4} \right)\text{ }\left( \text{3, 5} \right)\text{ }\left( \text{3, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{4, 1} \right)\text{ }\left( \text{4, 2} \right)\text{ }\left( \text{4, 3} \right)\text{ }\left( \text{4, 4} \right)\text{ }\left( \text{4, 5} \right)\text{ }\left( \text{4, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{5, 1} \right)\text{ }\left( \text{5, 2} \right)\text{ }\left( \text{5, 3} \right)\text{ }\left( \text{5, 4} \right)\text{ }\left( \text{5, 5} \right)\text{ }\left( \text{5, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{6, 1} \right)\text{ }\left( \text{6, 2} \right)\text{ }\left( \text{6, 3} \right)\text{ }\left( \text{6, 4} \right)\text{ }\left( \text{6, 5} \right)\text{ }\left( \text{6, }\!\!~\!\!\text{ 6} \right)\]

So, there are a total $36$ outcomes.

Now let $A$ denotes the event that first throw shows $5$ and

let $B$ denote the event that the second throw shows $5$.

Thus, there are a total $6$ favourable outcomes in each of the cases.

Therefore, $P\left( A \right)=\frac{6}{36}$ and $P\left( B \right)=\frac{6}{36}$

$\Rightarrow P\left( \overline{A} \right)=1-\frac{6}{36}=\frac{30}{36}=\frac{5}{6}$ and $P\left( \overline{B} \right)=\frac{5}{6}$.

Hence, the probability that $5$ will not come in either time

$\begin{align} & =\frac{5}{6}\times \frac{5}{6} \\ & =\frac{25}{36} \\ \end{align}$

(ii) \[{5}\] will come up at least once?

Ans. Suppose that, \[S\] is the sample space related to the random experiment of throwing a die two times. So, \[n\left( S \right)=36\].

Therefore, \[A\cap B=\]first and second throw shoe $5$, that is, obtaining $5$ in each throw.

Then, \[\text{A=}\left\{ \left( \text{5,1} \right),\left( \text{5,2} \right),\left( \text{5,3} \right),\left( \text{5,4} \right),\left( \text{5,5} \right),\left( \text{5,6} \right) \right\}\]

and \[\text{B=}\left\{ \left( \text{1,5} \right),\left( \text{2,5} \right),\left( \text{3,5} \right),\left( \text{4,5} \right),\left( \text{5,5} \right),\left( \text{6,5} \right) \right\}\].

\[P\left( A \right)=\frac{6}{36}\],\[P\left( B \right)=\frac{6}{36}\] and \[P\left( A~\cap B \right)=\frac{1}{36}\].

Thus, the probability that $5$ will come at least once in throwing a dice twice, $P\left( A\cap B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cup B \right)$

$\begin{align} & =\frac{6}{36}+\frac{6}{36}-\frac{1}{36} \\ & =\frac{11}{36} \\ \end{align}$

3. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on the following days?

(i) the same day?

Ans. The list of favourable outcomes related with two consumers visiting a specific shop in the same week (Tuesday to Saturday) is:

\[\left( \text{T,T} \right)\text{ }\left( \text{T,W} \right)\text{ }\left( \text{T,TH} \right)\text{ }\left( \text{T,F} \right)\text{ }\left( \text{T,S} \right)\]

\[\left( \text{W,T} \right)\text{ }\left( \text{W,W} \right)\text{ }\left( \text{W,TH} \right)\text{ }\left( \text{W,F} \right)\text{ }\left( \text{W,S} \right)\]

\[\left( \text{TH, T} \right)\text{ }\left( \text{TH, W} \right)\text{ }\left( \text{TH, TH} \right)\text{ }\left( \text{TH, F} \right)\text{ }\left( \text{TH, S} \right)\]

\[\left( \text{F, T} \right)\text{ }\left( \text{F, W} \right)\text{ }\left( \text{F, TH} \right)\text{ }\left( \text{F, F} \right)\text{ }\left( \text{F, S} \right)\]

\[\left( \text{S,T} \right)\text{ }\left( \text{S, W} \right)\text{ }\left( \text{S, TH} \right)\text{ }\left( \text{S, F} \right)\text{ }\left( \text{S S} \right)\]

Therefore, there are a total of $25$ favourable outcomes.

Now, the outcomes of visiting on the same day can be listed as

\[\left( \text{T,T} \right)\text{,}\left( \text{W,W} \right)\text{,}\left( \text{TH,TH} \right)\text{,}\left( \text{F,F} \right)\] and \[\left( S,S \right)\].

Thus, the probability that both Shyam and Ekta will visit the shop on the same day is $=\frac{5}{25}=\frac{1}{5}$.

(ii) consecutive days?

Ans. The list of favourable outcomes of visiting the shop on consecutive days are \[\left( \text{T, W} \right)\text{,}\left( \text{W, T} \right)\text{,}\left( \text{W, TH} \right)\text{,}\left( \text{TH, W} \right)\text{,}\left( \text{TH, F} \right)\text{,}\left( \text{F, TH} \right)\text{,}\left( \text{S, F} \right)\] and \[\left( F,\text{ }S \right)\].

So, there are a total of $8$ favourable outcomes.

Thus, the probability that both Shyam and Ekta will visit the shop on consecutive days $=\frac{8}{25}$.

(iii) different days?

Ans. There are total \[255=20\] favourable outcomes of visiting on different days.

So, there are a total of $20$ favourable outcomes.

Thus, probability that both Shyam and Ekta will visit the shop on different days $=\frac{20}{25}=\frac{4}{5}$.

4. A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is

(i) a card of spades of an ace

Ans. The number of cards in a deck \[=52\].

There are a total of $13$ spades and $4$ aces in a deck of playing cards.

There is one card which is common [i.e., ace of spade]

So, the number of favourable outcomes \[=13+41=16\].

Thus, the probability of getting a card of spades of an ace $=\frac{16}{52}=\frac{4}{13}$.

(ii) a red king

Ans. The number of red kings \[=2\].

Thus, the probability of getting a red king is $=\frac{2}{52}=\frac{1}{26}$.

(iii) neither a king nor a queen.

Ans. The total number of kings and queens \[=4+4=8\].

The number of cards that are neither king nor a queen \[=528=44\].

Thus, the probability of getting neither a king nor a queen is $=\frac{44}{52}=\frac{11}{13}$.

(iv) either a king or a queen

Ans. The total number of kings and queens \[=4+4=8\].

Thus, the probability of getting either a king or a queen is $=\frac{8}{52}=\frac{2}{13}$.

(v) a face card.

Ans. The number of face cards \[=4+4+4=12\] (Jack, queen and king are the face cards)

Therefore, the probability of getting a face card is $=\frac{12}{52}=\frac{3}{13}$.

(vi) cards which is neither king nor a red card.

Ans. The number of cards that are neither red card nor king \[=\text{ }52\left( 26+42 \right)\]

\[\begin{align} & =5228 \\ & =24 \\ \end{align}\]

Thus, the probability of getting a card that is neither a king nor a red card $\begin{align} & =\frac{24}{52} \\ & =\frac{6}{13} \\ \end{align}$

5. Cards marked with numbers \[{1},{2},{3},\ldots ,{25}\] are placed in a box and mixed thoroughly and one card is drawn at random from the box, what is the probability that the number on the card is

(i) a prime number?

Ans. Total number of outcomes that are possible $=25$

There are $9$ favourable cases, which can be listed as \[2,3,5,7,11,13,17,19,23\].

Thus, the probability of getting a prime number $=\frac{9}{25}$.

(ii) a multiple of \[{3}\] or \[{5}\]?

Ans. The numbers that are multiple of $3$ or $5$ can be listed as\[3,5,6,9,10,12,15,18,20,21,24,25\].

So, there are a total of $12$ favourable cases.

Therefore, the probability of getting a multiple of $3$ and $5$ is $=\frac{12}{25}$.

(iii) an odd number?

Ans. The odd numbers can be listed as \[1,3,5,7,9,11,13,15,17,19,21,23,25\].

Therefore, there are a total $13$ favourable cases.

Thus, the probability of getting an odd number $=\frac{13}{25}$.

(iv) neither divisible by \[{5}\] nor by \[{10}\]?

Ans. The numbers that are neither divisible by $5$ nor by $10$ can be listed as \[1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24\].

So, there are a total of $20$ favourable cases.

Thus, the probability of getting a number that is neither divisible by $5$ nor by $10$ is $=\frac{20}{25}=\frac{4}{5}$.

(v) perfect square?

Ans. The perfect square numbers can be listed as \[1,4,9,16,25\].

Therefore, there are a total of $5$ favourable cases.

Thus, the probability of getting a perfect square number $=\frac{5}{25}=\frac{1}{5}$.

(vi) a two-digit number?

Ans. All two-digit numbers can be listed as \[10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25\].

So, there are a total $16$ favourable cases.

Thus, the probability of getting a two-digit number $=\frac{16}{25}$.

6. From a pack of ${52}$ playing cards, jacks, queens, kings and aces of red colour are removed. From the remaining a card is drawn at random. Find the probability that the card drawn is

(i) a black queen

Ans. Total number of cards $=52$.

Number of cards that are removed $=2+2+2+2=8$ ($2$ jack, $2$ queen, $2$ king and $2$ aces of red colour)

The number of cards remains $=52-8=44$

Therefore, the total number of outcomes $=44$

The number of favourable outcomes $=2$ (Since, the number of black queens is $2$)

Thus, the probability of getting a black queen $=\frac{2}{44}=\frac{1}{22}$

(ii) a red card

Ans. The number of favourable outcomes $=$ the number of remaining red cards$=26-8=18$.

Thus, the probability of getting a red card is $=\frac{18}{44}=\frac{9}{22}$.

(iii) a black jack

Ans. The number of favourable outcomes $=$ The number of black jacks$=2$

Therefore, the probability of getting a black jack $=\frac{2}{44}=\frac{1}{22}$.

(iv) a picture cards

Ans. The number of picture cards remains $=2+2+2=6$ [King, queen, jack are picture cards]

Therefore, the probability of getting a picture card is $=\frac{6}{44}=\frac{3}{22}$.

(v) a honourable card

Ans. The number of Honourable cards remains $=2+2+2+2=8$ (ace, jack, queen and king are honourable cards)

Thus, the probability of getting a honourable card $=\frac{8}{44}=\frac{2}{11}$.

## CBSE Class 10 Maths Probability Important Questions

### Explain the Concept of Probability Mentioned in Important Questions of Probability Class 10

Probability is a branch of mathematics that deals with numerical explanations, determining the outcome of whether an event is true or not. This possibility explains the concept of probability. The numerical value of probability is expressed from zero to one. Probability has been introduced in mathematics to explain the likelihood of an event that is about to take place. This is also helpful in the probability distribution, where one can know the outcome of a random event. The probability of all events in a given sample space adds up to 1. Class 10 probability questions are crucial for students who are appearing for their Class 10th boards.

### The Probability Formula is Depicted as Follows

Probability of an event about to occur P(E) = Number of favourable outcomes/Total number of outcomes

### Class 10 Maths Chapter 15 Important Questions Based on Exercise?

Q. Which of the given experiments have equally likely outcomes? Explain.

(i) A driver tries to start a car. The car starts or may not start.

(ii) A player strives to shoot a football. He misses or shoots the football.

(iii) A trial is made to get a solution: a true-false question. The solution may be right or wrong.

(iv) A child is born. Whether it is a girl or a boy.

Solution:

(i) This statement doesn’t have a likely outcome, as the car may begin to operate or it may not.

(ii) This statement also doesn’t have a likely outcome as the player may shoot or miss the ball.

(iii) This statement has a likely outcome, as the solution has to be either right or wrong.

(iv) This statement has a likely outcome because the newborn child has to be a girl or a boy.

Q. Pull a random card from a pack of cards. What is the probability that the card pulled has a feminine face?

Solution:

A standard deck of cards has 52 cards.

Total number of outcomes = 52

Number of favorable events = 4 x 1 = 4 (considering Queen only)

Therefore Probability, P = Number of Favourable Outcomes/Total Number of Outcomes = 4/52= 1/13.

Q. If P(E) = 0.25, what is the probability of ‘not E’?

Solution:

We already know that

P(E) + P(not E) = 1

It is provided that, P(E) = 0.25

So, P(not E) = 1 – P(E)

P(not E) = 1 – 0.25

Hence, P(not E) = 0.75

Q. If 10 defective balls are accidentally mixed with 144 good ones. It is not possible to just look at a ball and tell whether or not it is defective. One ball is drawn out at random from this set of balls. Determine the probability that the ball pulled out is a good one.

Solution:

Numbers of balls = Numbers of defective balls + Numbers of good balls

∴ Total number of pens = 144 + 10 = 154 pens

P(E) = (Number of favourable results) / (Total number of results)

P(picking a good ball ) = 144/154 = 72/77 = 0.935

Q. One card is taken out from a well-shuffled deck of 52 cards. State the probability that the card will

(i) be a king,

(ii) not be a king.

Solution:

Well-shuffling of cards ensures fairly possible outcomes.

(i) Card drawn is a king

There are a total of four kings in a deck of cards.

Let A be the event ‘the card is a king’.

The number of outcomes favourable to A = n(A) = 4

The number of possible results = Total number of cards n(S) = 52

Hence, P(A) = n(A)/n(S) = 4/52 = 1/13

(ii) Card drawn is not a king

Let B be the event ‘card drawn is not a king’.

The number of outcomes favourable to the event B = n(B) = 52 – 4 = 48

Hence, P(B) = n(B)/n(S) = 48/52 = 12/13

### Probability Important Questions Class 10- State the Different Types of Probability?

Here is the answer to probability class 10 most important questions. There are mainly three types of probability.

Theoretical Probability

Experimental Probability

Axiomatic Probability

### Theoretical Probability

It depends on the possible chances of something that is about to occur. The theoretical probability concept originates from the reasoning behind probability. For example, if a dice is rolled, the theoretical probability of getting a sixer will be ⅙.

### Experimental Probability

It depends on the basis of observation of an experiment. The experimental probability is generally calculated by the total number of possible outcomes by the total number of trials taken. For example, if a coin is tossed 10 times and out of that head comes 4 times, then the experimental probability of a head is 4/10 or ⅖.

### Axiomatic Probability

A set of rules or axioms are established, which are applied to all types in an axiomatic probability. These axioms are set by Kolmogorov and are popularly called the Kolmogorov’s three axioms. In Kolmogorov's axiomatic approach to probability, the chances of non-occurrence and occurrence of an event can be quantified.

### Explain Events and Outcomes in Probability Important Questions Class 10?

An outcome is a result obtained through a random experiment. For example, when we toss a coin, the tail is the outcome. An event refers to the set of outcomes. For example, when we roll a dice, the probability of getting a result less than four is an event. This is the answer to important questions of probability class 10.

## What are Various Types of Events Based on Probability Class 10 Important Questions?

### Here Are the Following Types of Events in Probability

### Elementary Events

An event that has only one outcome of an event is referred to as an elementary event. For example: take a coin and toss it in the air for ‘n’ number of times. After the trial of this experiment, it will possibly have two outcomes- Heads and Tails. So, for any individual toss in a coin, the outcome has to be between the head and tail.

In elementary events, the sum of probabilities of all events in an experiment is one. For example- the tossing of a coin experiment P(Heads) + P (Tails)

= (1/2)+ (1/2) =1

### Impossible Events

The event that has no chance of happening or occurring is termed as an impossible event, is called as an impossible event, i.e P(E)=0. For example, the probability of getting an eight on a dice is zero. This is because the number 8 can never appear on a dice.

### Sure Event

The event where there is a 100% chance of occurrence or happening is termed as a sure event. The probability of occurrence in a sure event is one. For example, the probability of throwing a number less than 7 in dice is 1. So, P(E) = P(Obtaining a number less than 7) = 6/6= 1.

### Complementary Event

In complementary events, there are only two outcomes of an event and these are the only two possible outcomes. A simple example is that of a coin where there are only two possible outcomes, the heads, and tails. P(E) + P (È) = 1, where E and È are two complementary events. The event È represents the complement of the event E.

### What is Probability Density Function?

The probability density function is an** important question of probability class 10**. It is represented as a density of a continuous random variable that comes between a certain range of values. It explains the normal distribution and how the mean distribution occurs. The standard normal distribution helps create the statistics and database, which are applicable in science for representing the real-valued variable, whose distribution is unknown.

### How Class 10 Probability Questions are Crucial for Preparation?

By referring to **CBSE class 10 maths probability important questions**, students can stay ahead of the competition.

They can practice important questions and recognize their mistakes.

Students can make notes and highlight the vital information based on these questions.

By solving the important questions, the students can revise the probability topic very well.

These crucial questions are also valuable to help solve their homework problems.

The questions will allow students to understand various sums given in this probability chapter.

Overall, these important questions serve as a resourceful study guide for the students.

## Important Related Links for CBSE Class 10 Maths

CBSE Class 10 Maths Study Materials |

## Conclusion

Vedantu's Probability Class 10 Notes for CBSE Maths Chapter 15 offer a comprehensive and well-structured resource for students seeking to master this fundamental concept. Through their free PDF download, Vedantu provides a valuable aid to learners, simplifying complex probability principles with clarity and precision. These notes enable students to understand probability's theoretical foundations and practical applications, enhancing their problem-solving skills. The user-friendly format and engaging examples ensure an enjoyable learning experience. Vedantu's commitment to providing high-quality educational materials empowers students to excel in their studies, instilling confidence and proficiency in probability, a skill that extends beyond mathematics into various real-world scenarios. By making these notes freely accessible, Vedantu exemplifies its dedication to fostering academic excellence for all.

## FAQs on Important Questions for CBSE Class 10 Maths Chapter 15 - Probability

1. What is the difference between theoretical probability and experimental probability?

Theoretical probability is the probability of an event occurring based on the assumptions that the events are equally likely and that the experiment is repeated an infinite number of times. Experimental probability is the probability of an event occurring based on the results of a finite number of trials.

2. What are the addition and multiplication theorems of probability?

The addition theorem of probability states that the probability of two events occurring is equal to the sum of the probabilities of each event occurring, if the events are mutually exclusive. Mutually exclusive events are events that cannot occur at the same time. The multiplication theorem of probability states that the probability of two events occurring is equal to the product of the probabilities of each event occurring, if the events are not mutually exclusive. Not mutually exclusive events are events that can occur at the same time.

3. What are some applications of probability in real life?

Probability is used in a variety of fields, such as gambling, insurance, weather forecasting, marketing, and medicine.

Gambling: Probability is used to calculate the odds of winning a particular bet. For example, in a game of roulette, the probability of winning a bet on red is 1/2.

Insurance: Probability is used to calculate the risk of a particular event occurring and to set premiums accordingly. For example, an insurance company might charge a higher premium for a house in a flood-prone area than for a house in a low-risk area.

Weather forecasting: Probability is used to predict the likelihood of different weather conditions occurring. For example, a weather forecaster might say that there is a 60% chance of rain tomorrow.

Marketing: Probability is used to target advertising campaigns to specific groups of people. For example, an advertiser might target their ads to people who are likely to be interested in their product based on their age, gender, location, and other factors.

Medicine: Probability is used to diagnose diseases and to assess the risks and benefits of different treatments. For example, a doctor might use a probability model to determine the likelihood that a patient has a particular disease based on their symptoms.

4. What are some tips for solving probability problems?

Here are some tips for solving probability problems:

Read the problem carefully and identify the relevant information.

List the possible outcomes of the experiment.

Identify the favorable outcomes.

Calculate the probability of the event by dividing the number of favorable outcomes by the total number of possible outcomes.

Check your answer to make sure it makes sense.

5. What is the meaning of "sample space"?

The sample space is the set of all possible outcomes of an experiment. For example, if you roll a die, the sample space is the set of numbers 1, 2, 3, 4, 5, and 6.

6. What is an elementary event?

An elementary event is a single outcome of an experiment. For example, if you roll a die, one elementary event is the outcome of getting a 1.

7. What is a favorable outcome?

A favorable outcome is an outcome of an experiment that satisfies a particular condition. For example, if you roll a die and you want to get a number greater than 3, the favorable outcomes are 4, 5, and 6.

8. What is the probability of an event?

The probability of an event is the likelihood of that event occurring. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For example, the probability of getting a head when a coin is tossed is 1/2.

11. A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number which is a perfect square.

Total number of cards = 50 – 6 + 1 = 45

Perfect square numbers are 9, 16, 25, 36, 49, i.e.,

5 numbers

∴ P(a prefect square) = $\dfrac{5}{45}$ = $\dfrac{1}{9}$

12. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:

1. A black king.

2. A card of red colour.

3. A card of black colour.

Removed red colour cards = 3×2 = 6

Remaining cards = 52 – 6 = 46

1. Number of black kings = 2

P(a black king) = $\dfrac{2}{46}$=$\dfrac{1}{23}$

2. Number of red colour cards = 26

Remaining red colour cards = 26 – 6 = 20

P(a card of red colour) =$\dfrac{20}{46}$=$\dfrac{10}{23}$

3. Number of black cards = 26

P(a black colour card) = $\dfrac{26}{46}$=$\dfrac{13}{23}$

13. Three distinct coins are tossed together. Find the probability of getting:

I. at least 2 heads

II. at most 2 heads.

When three coins are tossed following is the sample space:

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

I. P(at least 2 heads) = $\dfrac{4}{8}$ =$\dfrac{1}{2}$

II. P(at most 2 heads) = $\dfrac{7}{8}$

14. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5.

Total number of tickets = 40

Multiples of 5 are 5,10,15,20,25,30,35,40

Number of favourable tickets = 8

$\therefore$ P(a number which is a multiple of 5) =$\dfrac{8}{40}$=$\dfrac{1}{5}$

15. How students can score good marks in Mathematics class 10.

Students can score good marks in mathematics class 10 by revising each chapter 2-3 times. Also, they need to solve sample papers to improve their time management and accuracy. Students can solve sample papers on Vedantu site.