CBSE Class 10 Maths Important Questions - Chapter 14 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

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

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

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

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 $\begin{array}{rl}& =\frac{\text{Number of favourable outcomes}}{Total\text{ n}umber\text{ of possible outcomes}}\\ & =\frac{x}{12}\end{array}$

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

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

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

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: $$A=\left\{6\right\},B=\{1,2,3,4,5,\}$$.

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,

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

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 $$\left(MT\right),\left(TW\right),\left(WTh\right),\left(ThF\right),\left(FS\right),\left(SS\right),\left(SM\right)$$. 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, $P(\left|x\right|<2)=\frac{5}{7}$.

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,

$$(1,1),(l,4),(2,l),(2,4),(3,1)$$.

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?

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 $=(25\pi -48)$ sq. units.

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

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 $-4\times$ 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 $$APQ$$.

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

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

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

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

$$=36-(18+4.5)$$

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

Hence, the probability that the point will be selected from the interior of $\mathrm{\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.

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

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

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

Therefore, 

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

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

$\begin{array}{rl}& \Rightarrow P\left(notE\right)=1P\left(E\right)\\ & =10.05\\ & =0.95\end{array}$

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,  $P\left(E\right)+P\left(\bar{E}\right)=1$

$\begin{array}{rl}& \Rightarrow P\left(E\right)=1-P\left(\bar{E}\right)\\ & =1-0.992\\ & =0.008\end{array}$

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 $2x^4+3x^3-5x^2-5x^2+9x+1$ 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 $x^2-\left(2\right)x+(-15)$, that is, $x^2-2x-15$.

So, option (b) is the correct answer.

10. Cards each marked with one of the numbers $$4,5,6,\dots ,20$$ 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 $1,2,3,\dots ,15$ 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 $$4,5,6,\dots ,20$$ 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 $$1,2,3,\dots ,15$$ 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) $(1,1)$

(b) $$(1,0)$$

(c) $$(0,1)$$

(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 $$P\left(E\right)+P\left(notE\right)$$ 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 $$P\left(E\right)+P\left(\bar{E}\right)=1$$, that is,

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

So, option (a) is the correct answer.

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

Thus, option (c) is the correct answer.

27. Zeros of $$P\left(x\right)=2x^2+9x35$$ 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 $P\left(x\right)=2x^2+9x-35$ are $-7$ and $\frac{5}{2}$, because $P(-7)=2{(-7)}^2+9(-7)-35=0$ 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) $$x^2+2x15$$

(b) $$x^2+3x8$$

(c) $$x^{2}5x15$$

(d) None of these

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

$x^2-(3-5)x+3(-5)$

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

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,

 $\begin{array}{rl}& {\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta \\ & =p^2-2q\end{array}$

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 $2$ $A^{\prime }s$, 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?

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

Also, the area of the circle inside the rectangle,

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

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,

$\begin{array}{rl}& P_1=2P_2\\ & \Rightarrow \frac{x}{x+5}=2\cdot \frac{5}{x+5}\\ & \Rightarrow \frac{x}{x+5}=2\cdot \frac{5}{x+5}\\ & \Rightarrow x=10\end{array}$

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,

$\begin{array}{rl}& P_2=2P_1\\ & \Rightarrow \frac{x+6}{18}=2\cdot \frac{x}{12}\\ & \Rightarrow \frac{x+6}{18}=\frac{x}{6}\\ & \Rightarrow 6x+36=18x\\ & \Rightarrow 12x=36\\ & \Rightarrow x=3\end{array}$

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 $$\left(MT\right),\left(TW\right),\left(WTh\right),\left(ThF\right),\left(FS\right),\left(SS\right),\left(SM\right)$$. 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:

(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.