Important Questions for CBSE Class 9 Maths Chapter 15 - Probability

1 Mark Questions

1. Out of \[35\] students Participating in a debate \[10\] are girls. The Probability that winner is a boy is:

${\text{(a) }}1$

${\text{(b) }}\dfrac{2}{7}$

${\text{(c) }}\dfrac{3}{7}$

${\text{(d) }}\dfrac{5}{7}$

Ans: \[\left( {\text{d}} \right)\,\,\dfrac{5}{7}\]

2. There are \[5\] balls, each of the colours white, blue, green, red and yellow in a bag. If \[1\] balls is drawn from the bag, then the Probability that the ball drawn is red is

${\text{(a) }}\dfrac{4}{5}$

${\text{(b) }}\dfrac{1}{4}$

${\text{(c) }}\dfrac{1}{5}$

${\text{(d) }}\dfrac{1}{{20}}$

Ans: ${\text{(c)}}\,\,\dfrac{{\text{1}}}{{\text{5}}}$

3. If ${\text{P(e)}}\,\,{\text{ = }}\,\,{\text{0}}{\text{.25}}$ what is the value of ${\text{P}}$ (not${\text{E}}$)

${\text{(a) }}0.5$

${\text{(b) }}1$

${\text{(c) }}0$

${\text{(d) }}0.75$

Ans: \[\left( {\text{d}} \right){\text{ 0}}{\text{.75}}\]

4. Sum of the probabilities of all events of a trial is

\[\left( {\text{a}} \right)\]less than \[1\]

\[\left( {\text{b}} \right)\]greater than \[1\]

\[\left( {\text{c}} \right)\]lies between \[0\] and \[1\]

\[\left( {\text{d}} \right){\text{ 1}}\]

Ans: \[\left( {\text{d}} \right){\text{ 1}}\]

5. A four digit number is to be formed by using the digits \[2,4,7,8.\] The probability that the number will start with \[7\] is

${\text{(a) }}\dfrac{3}{4}$

${\text{(b) }}\dfrac{1}{4}$

${\text{(c) }}\dfrac{1}{3}$

${\text{(d) }}\dfrac{1}{7}$

Ans: $\left( {\text{b}} \right)\,\,\dfrac{1}{4}$

6. The probability of an event of a trial :

\[\left( {\text{a}} \right)\]is \[1\]

\[\left( {\text{b}} \right)\]lies between \[0\] and \[1\] (both inclusive)

\[\left( {\text{c}} \right)\]is \[0\]

\[\left( {\text{d}} \right)\]is greater than \[1\]

Ans: \[\left( {\text{b}} \right)\]lies between \[0\] and \[1\] (both inclusive)

7. A die is thrown once, the probability of getting a prime number on the die is :

${\text{(a) }}\dfrac{1}{6}$

${\text{(b) }}\dfrac{1}{3}{\text{ }}$

${\text{(c) }}\dfrac{1}{2}$

${\text{(d) }}\dfrac{2}{3}$

Ans: ${\text{(a) }}\dfrac{1}{6}$

8. If two coins are tossed, then the probability of getting no tail is :

${\text{(a) }} - \dfrac{1}{4}$

${\text{(b) }}\dfrac{1}{4}$

${\text{(c) }}\dfrac{1}{5}$

${\text{(d) }}\dfrac{3}{4}$

Ans: ${\text{(b) }}\dfrac{1}{4}$

9. If a dice is thrown once what is the probability of getting an even prime number.

${\text{(a) }}\dfrac{1}{6}$

${\text{(b) }}\dfrac{1}{2}$

${\text{(c) }}\dfrac{2}{3}$

${\text{(d) }}1$

Ans: ${\text{(a) }}\dfrac{1}{6}$

10. A card is drawn from a pack of \[52\] cards what is the probability of getting an non ace card.

\[{\text{(a) }}\dfrac{1}{{13}}\]

\[{\text{(b) }}\dfrac{{12}}{{13}}\]

\[{\text{(c) }}\dfrac{1}{4}\]

${\text{(d)}}$none of these

Ans:  $\dfrac{{12}}{{13}}$

11. The minimum value of probability is

${\text{(a) }}1$

\[\left( {\text{b}} \right)\;\,\,\dfrac{1}{2}\]

\[\left( {\text{c}} \right){\text{ 0}}\]

\[\left( {\text{d}} \right)\]none of these

Ans: \[\left( {\text{c}} \right){\text{ 0}}\]

12. Performing an experiment once is called

\[\left( {\text{a}} \right)\]Trial 

\[\left( {\text{b}} \right)\] Event 

\[\left( {\text{c}} \right)\]Probability 

\[\left( {\text{d}} \right)\]None of these

Ans: \[\left( {\text{a}} \right)\]Trial 

13. What is the probability of a number greater than \[6\] for a single throw of a die?

\[{\text{(a) }}0\]

\[{\text{(b) }}1\]

${\text{(c) }}\dfrac{1}{2}$

${\text{(d)}}$none of these

Ans: \[{\text{(a) }}0\]

14. If ${\text{P(E)}}\,{\text{ = }}\,\dfrac{{\text{3}}}{{\text{4}}}$ what is value of ${\text{P}}$ (not${\text{E}}$).

${\text{(a) }}\dfrac{3}{4}$

${\text{(b) }}\dfrac{1}{4}$

${\text{(c) }}1$

${\text{(d)}}$none of these

Ans: ${\text{(b) }}\dfrac{1}{4}$

15. A card is drawn from a pack of \[52\] playing cards. What is the probability of getting a king of black colour

${\text{(a) }}\dfrac{1}{{26}}$

${\text{(b) }}\dfrac{4}{{52}}$

${\text{(c) }}\dfrac{1}{4}$

${\text{(d)}}$none of these

Ans: ${\text{(a) }}\dfrac{1}{{26}}$

16. A coin is tossed \[2\] times what is the probability of getting at most \[2\] heads.

${\text{(a) }}\dfrac{3}{4}$

${\text{(b) }}\dfrac{1}{2}$

${\text{(c) }}\dfrac{1}{4}$

${\text{(d)}}$none of these

Ans: ${\text{(d)}}$none of these

2 Marks Questions

1. A teacher analyses the performance of two sections of students in a mathematics 

test of \[100\] marks given in the following table:

(i) Find the probability that a student obtained less than \[20\% \] in the mathematics test.

Ans: Number of students obtaining marks less than \[20\] out of\[100\]

That is, $20\%  = 7$

Total students in the class $ = 90$

Thus, ${\text{P(}}$ A student obtained less than $20\% ) = \dfrac{7}{{90}}$

(ii) Find the probability that a student obtained \[60\] or above.

Ans: Number of students obtaining \[60\] or above is \[23\]

Total students in the class $ = 90$

Thus, ${\text{P}}$ (A student obtained marks \[60\] or above $) = \dfrac{{23}}{{90}}$

2. To know the opinion of the students about the subject statistics, a survey of \[200\] students was conducted. The data is recorded in the following table:

Find the probability that a student chosen at random: 

(i) likes statistics 

Ans: Number of students who like statistics $ = 135$

Total students $ = 200$

Thus, ${\text{P}}({\text{ a student likes statistics }}) = \dfrac{{135}}{{200}} = \dfrac{{27}}{{40}}$

(ii) dislikes it.

Ans: Number of students who like statistics $ = 65$

Total students $ = 200$

Thus, ${\text{P}}({\text{ a student dislikes statistics }}) = \dfrac{{65}}{{200}} = \dfrac{{13}}{{40}}$

3. A die is thrown once. Find the probability of getting

(i) a prime number

Ans: When a die is thrown, the outcomes are \[1,2,3,4,5,6\]

Prime numbers are\[ = 2,3,5\]

Frequency of happening prime number is $3$

The probability of getting prime number $ = \dfrac{3}{6} = \dfrac{1}{2}$

(ii) a number less than $5$

Ans: When a die is thrown, the outcomes are \[1,2,3,4,5,6\]

Numbers less than $5$ are \[1,2,3,4\]

Frequency of happening of a no. less than $5$ is $4$

Probability of getting a number less than $5 = \dfrac{4}{6} = \dfrac{2}{3}$

4. Activity: Note the frequency of two wheelers, three wheelers and four wheelers 

going past during a time interval, in front of your school gate. Find the probability that 

any one vehicle out of the total vehicles you have observed is a two wheeler. 

Ans: Let you noted the frequency of types of wheelers after school time (that is

\[{\text{3 pm to 3}}{\text{.30 pm}}\]) for half an hour. Let the following table shows the frequency of 

wheelers.

Total number of vehicles \[ = 125 + 45 + 30 = 200\]

Number of Two wheelers\[ = 125\]

The probability that any one vehicle out of the total vehicles is a two wheeler  \[ = \dfrac{{125}}{{200}} = \,\dfrac{5}{8}.\]

5. Activity: Ask all the students in your classroom to write a \[3\]-digit number. Choose any student from the room at random. What is the probability that the number written by him is divisible by \[3\]if the sum of its digits is divisible by \[3\].

Ans: Let the number of students in your class is \[24\].

Let \[3\]-digit number written by each of them is as follows:

$837,172,643,371,124,512,432,948,311,252,999,557,784,928,867,798,665,245,107,463,267,523,944,314$Numbers divisible by \[3\] are $ = 837,432,948,252,999,867,798$ and \[267\]

Number of \[3\]-digit numbers divisible by $3 = 8$

${\text{P}}(3 - $digit numbers divisible by \[3\]$) = \dfrac{8}{{24}} = \dfrac{1}{3}$

6. Eleven bags of wheat flour, each marked 5 kg, actually contained the following 

weights of four (in kg):\[4.97,{\text{ }}5.05,{\text{ }}5.08,{\text{ }}5.03,{\text{ }}5.00,{\text{ }}5.06,{\text{ }}5.08,{\text{ }}4.98,{\text{ }}5.04,{\text{ }}5.07,{\text{ }}5.00\]

Find the probability that any of these bags chosen at random contains more than \[5\] kg of flour.

Ans: Number of bags containing more than  of wheat flour \[ = 7\]

Total number of wheat flour bags $ = 11$

${\text{P}}$(a bag containing more than $5\;{\text{kg}}$ of wheat flour) $ = \dfrac{7}{{11}}$

7. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for \[30\] days is as follows: 

0.03 0.08 0.08 0.09 0.04 0.17 

0.16 0.05 0.02 0.06 0.18 0.20 

0.11 0.08 0.12 0.13 0.22 0.07 

0.08 0.01 0.10 0.06 0.09 0.18 

0.11 0.07 0.05 0.07 0.01 0.04

Using this table, find the probability of the concentration of sulphur dioxide in the interval $0.12 - 0.16$ on any of these days.

Ans: From the above frequency distribution table we observe that 

Number of days during which the concentration of sulphur dioxide lies in interval 

$0.12 - 0.16 = 2$

Total number of days during which concentration of sulphur dioxide recorded \[ = 30\]

${\text{P}}$(day when concentration of sulphur dioxide (in ppm) lies in $0.12 - 0.16) = \dfrac{2}{{30}} = \dfrac{1}{{15}}$

8. The blood group of \[30\] students of class VIII are recorded as follows:

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, 

A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

Represent this data in the form of a frequency distribution table. 

Use this table to determine the probability that a student of this class selected at random has blood group \[{\text{AB}}.\]

Ans: From the above frequency distribution table we observe that

Number of students having blood group ${\text{AB}} = 3$

Total number of students whose blood group were recorded \[ = 30\]

Therefore, \[{\text{P}}\]( a student having blood group ${\text{AB}}) = \dfrac{3}{{30}} = \dfrac{1}{{10}}$

9. A die is thrown \[1000\] times with the frequencies for the outcomes \[1,2,3,4,5\] and \[6\] as 

given in the following table:

Find the probability of getting each outcome.

Ans: Total frequency\[ = 1000\]

Number of outcome getting the number $1 = 179$

Probability of outcome $1$ is,

${\text{P(1) = }}\dfrac{{{\text{179}}}}{{{\text{1000}}}}{\text{ = 0}}{\text{.179}}$

Number of outcome getting the number $2 = 150$

Probability of outcome $2$is,

${\text{P(2) = }}\dfrac{{{\text{150}}}}{{{\text{1000}}}}{\text{ = 0}}{\text{.15}}$

Number of outcome getting the number $3 = 157$

Probability of outcome $3$is,

${\text{P(3) = }}\dfrac{{{\text{157}}}}{{{\text{1000}}}}{\text{ = 0}}{\text{.157}}.$

Number of outcome getting the number $4 = 149$

Probability of outcome $4$is,

${\text{P(4) = }}\dfrac{{{\text{149}}}}{{{\text{1000}}}}{\text{ = 0}}{\text{.149}}$

Number of outcome getting the number $5 = 175$

Probability of outcome $5$is,

${\text{P(5) = }}\dfrac{{{\text{175}}}}{{{\text{1000}}}}{\text{ = 0}}{\text{.175}}$

Number of outcome getting the number $6 = 190$

Probability of outcome $6$

${\text{P(6) = }}\dfrac{{{\text{190}}}}{{{\text{1000}}}}{\text{ = 0}}{\text{.19}}$

10. Two coins are tossed \[729\] times and the out comes are:

No tail:\[189\], One tail:\[297\], Two tails: \[243\]

Find the Probability of the occurrence of each of these events.

Ans: Number of total trials $ = 729$

\[{{\text{E}}_1}{\text{,}}\,\,{{\text{E}}_2}\]and \[{{\text{E}}_3}\] are events getting no tail, one tail and two tails, then

${\text{P}}\left( {{{\text{E}}_{\text{1}}}} \right){\text{ = }}\dfrac{{{\text{189}}}}{{{\text{729}}}}{\text{ = }}\dfrac{{\text{7}}}{{{\text{27}}}}$

${\text{P}}\left( {{{\text{E}}_{\text{2}}}} \right){\text{ = }}\dfrac{{{\text{297}}}}{{{\text{729}}}}{\text{ = }}\dfrac{{{\text{11}}}}{{{\text{27}}}}$

${\text{P}}\left( {{{\text{E}}_{\text{3}}}} \right){\text{ = }}\dfrac{{{\text{243}}}}{{{\text{729}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}$

11. A bag contains \[15\] cards bearing numbers $1,2,3,4, \ldots  \ldots  \ldots ..14,15$. A card is drawn from the bag. Find the Probability that it bears:

(i) a Prime number 

Ans: Total number of cards $ = 15$

Number of total trials $ = 15$

Among $1,2,3,4, \ldots  \ldots .,14,15$, prime number are \[2,3,5,7,11,13\]

Number of favorable outcomes $ = 6$

Thus, the probability that it bears a Prime number is,

${\text{P}}$(Prime number) $ = \dfrac{6}{{15}} = \dfrac{2}{5}$

(ii) A number divisible by \[2\]

Ans: Total number of cards $ = 15$

Number of total trials $ = 15$

Among $1,2,3,4, \ldots  \ldots  \ldots  \ldots .,14,15$,

Numbers divisible by \[2\] are \[2,4,6,8,10,12,14\]

Now, Number of outcomes $ = 7$

Thus, the probability that it bears a number divisible by \[2\]is,

${\text{P}}\,{\text{(}}$number divisible by \[2\]$) = \dfrac{7}{{15}}$

12. A coin is tossed \[400\] times and outcomes are

Tail:\[230\] Head:\[170\]. Find the Probability to get

(i) a head

Ans: Total outcomes $ = 400$

Head $ = 170$

The probability to get a head is,

${\text{P(H) = }}\dfrac{{{\text{170}}}}{{{\text{400}}}}{\text{ = }}\dfrac{{{\text{17}}}}{{{\text{40}}}}$

(ii) a tail

Ans: Total outcomes $ = 400$

Tail \[ = 230\]

The probability to get a tail is,

${\text{P(T) = }}\dfrac{{{\text{230}}}}{{{\text{400}}}}{\text{ = }}\dfrac{{23}}{{{\text{40}}}}$

13. A survey of \[200\] students was conducted to check the opinion of students about the topic geometry. It was found that \[175\] students do not like geometry. Find the probability of the students who like geometry.

Ans: Total number of Students $ = 200$

The number of students do not like geometry $ = 175$

The number of students who like geometry $ = 200 - 175 = 25$

Now, the probability of the students who like geometry is

${\text{P}}$(Number of students who like geometry) $ = \dfrac{{25}}{{200}} = \dfrac{5}{{40}} = \dfrac{1}{8}$

14. Three coins are tossed simultaneously \[200\] times with the following frequencies of different outcomes.

Compute the probability of $2$ heads coming up.

Ans: Total number of tosses $ = 200$

Number of outcomes of $2$ heads $ = 72$

Thus, the probability of $2$ heads coming up is

${\text{P}}(2{\text{ heads }}) = \dfrac{{72}}{{200}} = \dfrac{9}{{25}}$

15. The heights of 70 students are given in the following table.

Find the probability that a student has height.

(i) $169\;{\text{cm}}$ 

Ans: The total number of students $ = 70$

Then, Number of total trial $ = 70$

The number of students has height $169\;{\text{cm}} = 0$

${\text{P(}}$ a student has a height $169\;{\text{cm}}) = \dfrac{0}{{70}} = 0$

(ii) Less than $150\;{\text{cm}}$

Ans: The total number of students $ = 70$

Number of students has height less than $150\;{\text{cm}} = 0$

${\text{P(}}$a student has a height less than $150\;{\text{cm}}) = \dfrac{0}{{70}} = 0$

16. A bag contains \[20\] cards numbered from $1$ to \[20\] one card is drawn from the bag. Find the probability that it bears a prime number.

Ans: The total number of cards $ = 20$

Numbers of marks on the cards are $1,2,3,4,5,6, \ldots  \ldots  \ldots  \ldots  \ldots ,20$

Prime numbers are $\{ 2,3,5,7,11,13,17,19\} $

Total prime numbers $ = 8$

Therefore, the probability that it bears a prime number is

${\text{P}}$(a prime number) $ = \dfrac{8}{{20}} = \dfrac{2}{5}$

17. Three coins are tossed simultaneously \[200\] times with the following frequencies of different outcomes

Compute the probability of $2$ heads coming up.

Ans: The total number of tosses $ = 200$

Number of outcomes of $2$ heads $ = 72$

Thus, the probability of $2$ heads coming up is

${\text{P}}(2{\text{ heads }}) = \dfrac{{72}}{{200}} = \dfrac{9}{{25}}$.

18. A die is thrown once. Find the probability of getting 

(i) an odd number 

Ans: If a die is thrown once, total possible outcomes are ${\text{S}}\,{\text{ = }}\,{\text{\{ 1,2,3,4,5,6\} }}$

Here the odd numbers are, ${\text{E}}\,{\text{ = }}\,{\text{\{ 1,3,5\} }}$

Total odd numbers, ${\text{(E)}}\,{\text{ = }}\,{\text{3}}$

Total numbers, ${\text{n(S)}}\,\,{\text{ = }}\,\,{\text{6}}$

The probability of getting an odd number is,

${\text{P(E)}}\,{\text{ = }}\dfrac{{{\text{n(E)}}}}{{{\text{n(S)}}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{6}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}$

(ii) a number greater than one

Ans: If a die is thrown once, total possible outcomes are ${\text{S}}\,{\text{ = }}\,{\text{\{ 1,2,3,4,5,6\} }}$,

Here the numbers greater than one are, ${\text{E}}\,{\text{ = }}\,{\text{\{ 2,3,4,5,6\} }}$

Total numbers greater than one are, ${\text{(E)}}\,{\text{ = }}\,5$

Total numbers, ${\text{n(S)}}\,\,{\text{ = }}\,\,{\text{6}}$

The probability of getting a number greater than one is,

${\text{P(E)}}\,{\text{ = }}\dfrac{{{\text{n(E)}}}}{{{\text{n(S)}}}}{\text{ = }}\dfrac{5}{{\text{6}}}$

19. Two coins are tossed \[340\] times and the outcomes are

Two tail $ = 115$ 

One tail $ = 100$ 

No tail $ = 125$

Find the probability of occurrence of 

(i) one tail 

Ans: Total possible outcomes $ = 340$

Occurrence of one tail = 100

Thus,the probability of occurrence of one tail is,

${\text{P(E)}}\,{\text{ = }}\dfrac{{{\text{100}}}}{{{\text{340}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{17}}}}$

(ii) three tail 

Ans: Total possible outcomes $ = 340$

Occurrence of three tails$ = 0$

Thus, the probability of occurrence of three tail is,

${\text{P(E)}}\,{\text{ = }}\dfrac{{\text{0}}}{{{\text{340}}}}{\text{ = }}\,{\text{0}}$.

20. To know the option of the students about the subject mathematics a survey of \[200\] students was conducted. The obtained data is given below.

Find the probability that a student chosen at random

(i) like mathematics

Ans: Number of students who like mathematics $ = 135$

Total students $ = 200$

Thus, ${\text{P}}($a student like mathematics) $ = \dfrac{{135}}{{200}} = \dfrac{{27}}{{40}}$

(ii) dislikes it.

Ans: Number of students who like mathematics $ = 65$

Total students $ = 200$

Thus, ${\text{P}}($ a student dislike mathematics) $ = \dfrac{{65}}{{200}} = \dfrac{{13}}{{40}}$

21. Out of \[17\] boys and \[13\] girls of a class, \[1\] student is to be selected. Find the probability of selecting a girl

Ans: Total number of students $ = 17 + 13 = 30$

Number of girls $ = 13$

Thus, the probability of selecting a girl is

${\text{P}}$(a student selecting a girl) $ = \dfrac{{13}}{{30}}$.

22. A bag contains \[5\] white, \[4\]red and \[3\] black balls. A ball is drawn from the bag, find the probability that it is not black

Ans: Number of white balls $ = 5$

Number of red balls $ = 4$

Number of black balls \[ = 3\]

Total number of balls $ = 5 + 4 + 3 = 12$

${\text{P}}($Black balls$) = \dfrac{3}{{12}} = \dfrac{1}{4}$

Therefore, the probability that the ball not black is

${\text{P}}$(the ball is not black) $ = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.

23. A card is drawn from a \[52\] pack of cards. Find the probability that it is a queen

Ans: Total number of cards $ = 52$

Number of queens $ = 4$

Thus, the probability that it is a queen is

\[{\text{P}}\](getting a queen) $ = \dfrac{4}{{52}} = \dfrac{1}{{13}}$.

24. There are \[500\] tickets of a lottery out of which \[10\] are prize winning tickets. A person buys one ticket. Find the probability that he gets a prize winning ticket.

Ans: Total number of lottery tickets $ = 500$

Number of prize winning tickets $ = 10$

Thus, the probability that he gets a prize winning ticket is

${\text{P}}$(Prize winning ticket) $ = \dfrac{{10}}{{500}} = \dfrac{1}{{50}}$.

25. The central Board of secondary education has a waiting list of examinations of \[150\] Persons. Out of these, \[60\]are women and \[90\] are men. One examiner is selected to replace an examiner who has not reported at the centre . Find the probability that the examiner selected is a :

(i) woman 

Ans: Number of trials \[ = 150\]

Number of women $ = 60$

${\text{P(}}$The examiner selected is a woman $) = \dfrac{{60}}{{150}} = \dfrac{2}{5}$

(ii) man

Ans: Number of men $ = 90$

Number of men $ = 60$

${\text{P(}}$The examiner selected is a man $) = \dfrac{{90}}{{150}} = \dfrac{3}{5}$

3 Marks Questions

1. \[1500\] families with \[2\] children were selected randomly and the following data were recorded:

Compute the probability of a family, chosen at random, having: 

(i) \[2\] girls 

Ans: Total number of families \[ = 1500\]

Number of families having \[2\] girls $ = 475$

Thus, the probability of a family, chosen at random having \[2\] girls is

${\text{P(}}$Family having \[2\] girls$) = \dfrac{{475}}{{1500}} = \dfrac{{19}}{{60}}$

(ii) $1$ girl

Ans: Total number of families \[ = 1500\]

Number of families having $1$ girl $ = 814$

Thus, the probability of a family, chosen at random having \[1\] girl is

${\text{P(}}$Family having $1$ girl$) = \dfrac{{814}}{{1500}} = \dfrac{{407}}{{750}}$

(iii) No girl

Ans: Total number of families \[ = 1500\]

Number of families having no girl $ = 211$

Thus, the probability of a family, chosen at random having no girl is

${\text{P(}}$Family having no girl\[)\] $ = \dfrac{{211}}{{1500}}$

Also check whether the sum of these probabilities is \[1\].

Ans: Sum of all probabilities $ = \dfrac{{19}}{{60}} + \dfrac{{407}}{{750}} + \dfrac{{211}}{{1500}}$

$ = \dfrac{{475 + 814 + 211}}{{1500}}$

$ = \dfrac{{1500}}{{1500}}$

$ = 1$

Therefore, the sum of probabilities is \[1\].

2. An organization selected \[2400\] families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is:

(i) earning Rs. $10000 - 13000$ per month and owning exactly \[2\] vehicles.

Ans: ${\text{P}}$ (earning Rs. $10000 - 13000$ per month and owning exactly \[2\] vehicles$) = \dfrac{{29}}{{2400}}$

(ii) earning Rs. \[16000\]or more per month and owning exactly \[1\] vehicle.

Ans: ${\text{P}}$ (earning Rs. \[16000\] or more per month and owning exactly \[1\] vehicles) $ = \dfrac{{579}}{{2400}}$

(iii) earning less than Rs. \[7000\] per month and does not own any vehicle.

Ans: ${\text{P}}$ (earning Rs. \[7000\] per month and does not own any vehicles) $ = \dfrac{{10}}{{2400}} = \dfrac{1}{{240}}$

(iv) earning Rs. $13000 - 16000$ per month and owning more than \[2\] vehicles.

Ans: \[{\text{P}}\] (earning Rs. $13000 - 16000$ per month and owning more than \[2\] vehicles)$ = \dfrac{{25}}{{2400}} = \dfrac{1}{{96}}$

(v) not more than \[1\] vehicle.

Ans: ${\text{P}}$ (owning not more than \[1\] vehicle) =$\dfrac{{2062}}{{2400}} = \dfrac{{1031}}{{1200}}$

3. The marks obtained by \[30\] students is given in the following table:

(i) \[60\] marks

Ans: Total number of students\[ = 30\]

Number of students securing \[60\] marks\[ = 4\]

${\text{P}}$(Students securing \[60\] marks $) = \dfrac{4}{{30}} = \dfrac{2}{{15}}$

(ii) \[75\] marks

Ans: Total number of students\[ = 30\]

Number of students securing \[75\] marks $ = 2$

${\text{P}}($Students securing \[75\] marks$) = \dfrac{2}{{30}} = \dfrac{1}{{15}}$

(iii) Less than \[60\] marks

Ans: Total number of students\[ = 30\]

Number of students securing less than \[60\] marks $ = 5 + 7 = 12$

${\text{P}}$(Students securing less than \[60\] marks$) = \dfrac{{12}}{{30}} = \dfrac{2}{5}$

4. A tyre manufacturing company kept a record of the distance covered shows the

 results of \[1000\] tyres

If you buy a tyre of this company, What is the Probability that

(i) it will need to be replaced before it has covered $4000\;{\text{km}}$

Ans: Number of tyres which covered distance less than $4000\;{\text{km}} = 20$

Total number of tyres $ = 1000$

Required probability ${\text{P(E)}}\,{\text{ = }}\dfrac{{{\text{20}}}}{{{\text{1000}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{50}}}}$

(ii) it will last more than $9000\;{\text{km}}$

Ans: Number of tyres needed to replaced more then $9000\;{\text{km}} = 325 + 445 = 770$

Total number of tyres $ = 1000$

Required Probability ${\text{P(E)}}\, = \dfrac{{770}}{{1000}} = \dfrac{{77}}{{100}} = 0.77$

(iii) it will need to be replaced after it has covered somewhere between $4000\;{\text{km}}$ and 

$14000\;{\text{km}}$

Ans: Number of tyres needed to be replaced between $4000\;{\text{km}}$, to $14,000\;{\text{km}}$.

Required probability ${\text{P(E)}} = \dfrac{{535}}{{1000}} = 0.535$

5. The ages of \[30\] workers in a factory are as follows

Age (in years) $21 - 23,\,\,23 - 25,\,\,25 - 27,\,\,27 - 29,\,\,29 - 31,\,\,31 - 33,\,\,33 - 35$ workers \[3,\,4,\,5,\,6,\,5,\,4,3\]

Find the probability that the age of a works lies in the interval

(i) $27 - 29$

Ans: The number of workers lies in the interval $27 - 29$ $ = 6$

Total number of workers $ = 30$

Required probability $ = \dfrac{6}{{30}} = \dfrac{1}{5}$

(ii) $29 - 35$

Ans: Number of workers having age between $29 - 35 = 5 + 4 + 3 = 12$

Total number of workers $ = 30$

Required Probability $ = \dfrac{{12}}{{30}} = \dfrac{2}{5}$

(iii) $21 - 27$

Ans: Number of workers having age between $21 - 27 = 3 + 4 + 5 = 12$

Total number of workers $ = 30$

Required Probability $ = \dfrac{{12}}{{30}} = \dfrac{2}{5}$

6. A die is thrown \[450\] times and outcomes are noted in the frequency distribution 

table given below.

Find the probability of the occurrence of the event.

(i) $4$ 

Ans: Total number of times die thrown $ = 450$

Number of times the number 4 comes up $ = 70$

${\text{P(}}$getting number $4$$) = \dfrac{{70}}{{450}} = \dfrac{7}{{45}}$

(ii) a number $ < 3$ 

Ans: Total number of times die thrown $ = 450$

Number of times the die turns up $1$ or $2 = 90 + 60 = 150$

${\text{P}}(\,$getting $1$ or $2$$) = \dfrac{{150}}{{450}} = \dfrac{1}{3}$

(iii) $7$

Ans: Total number of times die thrown $ = 450$

Number of times the die turn up $7 = 0$

${\text{P(}}$getting ${\text{7}}\,{\text{)}}\,{\text{ = }}\dfrac{{\text{0}}}{{{\text{450}}}}{\text{ = }}\,{\text{0}}$

7. From a well- shuffled pack of \[52\] cards, a card is drawn at random, find the

probability that it is :

(i) A spade 

Ans: Total number of cards $ = 52$

Number of spade $ = 13$

${\text{P}}({\text{ Spade }}) = \dfrac{{13}}{{52}} = \dfrac{1}{4}$

(ii) Black 

Ans: Total number of cards $ = 52$

Number of black cards\[ = 26\]

${\text{P(}}\,{\text{black}}\,{\text{cards}}\,{\text{)}} = \dfrac{{26}}{{52}} = \dfrac{1}{2}$

(iii) Ace of diamond

Ans: Total number of cards $ = 52$

Number of ace of diamond\[ = 1\]

${\text{P(}}$ace of diamond$) = \dfrac{1}{{52}}$

8. Two coins are tossed \[250\]times and the outcomes are:

(i) No head $ = 70$ 

(ii) One head $ = 85$ 

(iii) Two heads $ = 95$

Find the probability of the occurrence of each of these events.

Ans: Total number of times coin tossed $ = 250$

(i) Number of times no head comes up $ = 70$

${\text{P}}\,({\text{No Head}}) = \dfrac{{70}}{{250}} = \dfrac{7}{{25}}$

(ii) Number of times one head comes up\[ = 85\]

${\text{P}}\,({\text{One head) }} = \dfrac{{85}}{{250}} = \dfrac{{17}}{{50}}$

(iii) Number of times two head comes up\[ = 95\]

${\text{P (Two head) }} = \dfrac{{95}}{{250}} = \dfrac{{19}}{{50}}$

9. Out of \[100\] balls in a bag \[25\] are green, \[30\]are yellow and \[45\] are white. Find the Probability that a ball drawn from the bag is 

(i) green 

Ans: Total number of balls are \[100\]

Number of green balls $ = 25$

${\text{P(G)}}\,{\text{ = }}\dfrac{{{\text{25}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}$

(ii) yellow 

Ans: Total number of balls are \[100\]

Number of yellow balls\[ = 30\]

${\text{P(Y)}}\,{\text{ = }}\dfrac{{{\text{30}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{\text{3}}}{{{\text{10}}}}$

(iii) white

Ans: Total number of balls are \[100\]

Number of white balls\[ = 45\]

${\text{P(W)}}\,{\text{ = }}\dfrac{{{\text{45}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{\text{9}}}{{{\text{20}}}}$

10. Fifty seeds were selected at random from each of \[5\] bags of seeds and were kept under standardized conditions favorable to germination. After \[20\] days, the number of seeds which had germinated in each collection were counted and recorded as follows.

Bag: \[1,2,3,4,5\]

No. of seeds germinated \[40,48,42,39,41\]

What is the probability of germination of

(i) More than \[40\] seeds in a bag 

Ans: Number of bags in which more than \[40\] seeds germinated out of \[50\] seeds is \[3\]

Thus, Required probability ${\text{P(E)}}\,{\text{ = }}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = }}\,{\text{0}}{\text{.6}}$

(ii) \[49\]seeds in a bag 

Ans: Number of bags in which \[49\] seeds germinated $ = 0$

Thus, Required probability ${\text{P(E)}}\,{\text{ = }}\dfrac{{\text{0}}}{{\text{5}}}{\text{ = }}\,{\text{0}}$

(iii) More than \[35\] seeds in a bag

Ans: Number of bags in which more than \[35\] seeds germinated $ = 5$

Thus, Required probability ${\text{P(E)}}\,{\text{ = }}\dfrac{{\text{5}}}{{\text{5}}}{\text{ = 1}}$

11. It is known that a box of \[550\] bulbs contain \[22\] defective bulbs. One bulb is taken out at random from the box. Find the probability of getting

(i) Defective bulbs 

Ans: Total number of bulbs $ = 550$

Number of defective bulbs \[ = 22\]

${\text{P}}$(Defective bulbs) $ = \dfrac{{22}}{{550}} = 0.04$

(ii) Good bulbs

Ans: Total number of bulbs $ = 550$

Number of defective bulbs \[ = 22\]

Number of good bulbs $ = 550 - 22 = 528$

${\text{P}}$(Good bulbs) $ = \dfrac{{528}}{{550}} = 0.96$

12. Frequency distribution of marks obtained by \[70\] Students is given below:

Marks obtained $0 - 10,10 - 20,\,20 - 40,40 - 45,45 - 60,60 - 70,70 - 80$

No. of students \[4,8,20,10,12,6,10\]

Find the probability that the marks obtained by a student lies in the internal

(i) $0 - 40$ 

Ans: Total number of students $ = 70$

Number of students getting marks $0 - 40$$ = 4 + 8 + 20 = 32$

${\text{P}}$(students getting marks $0 - 40) = \dfrac{{32}}{{70}} = 0.457$

(ii) $0 - 80$ 

Ans: Total number of students $ = 70$

Number of students getting marks $0 - 80$$ = 4 + 8 + 20 + 10 + 12 + 6 + 10 = 70$

${\text{P}}$(students getting marks $0 - 80) = \dfrac{{70}}{{70}} = 1$

(iii) $80 - 90$

Ans: Total number of students $ = 70$

Number of students getting marks $80 - 90 = 0$

${\text{P}}$(students getting marks $80 - 90) = \dfrac{0}{{70}} = 0$

13. A box contains \[150\] balls of red, blue and white colours out of these \[50\] balls are red, \[40\]balls are blue and \[60\] balls are white. One ball is drawn from the bag. Find the probability that the ball drawn is

(i) Red 

Ans: Total number of balls $ = 150$

Number of red balls $ = 40$

${\text{P}}($red ball$) = \dfrac{{50}}{{150}} = \dfrac{1}{3}$

(ii) blue

Ans: Total number of balls $ = 150$

Number of blue balls $ = 40$

${\text{P}}$(blue ball) $ = \dfrac{{40}}{{150}} = \dfrac{4}{{15}}$

(iii) white

Ans: Total number of balls $ = 150$

Number of white balls $ = 60$

${\text{P}}$(white ball) $ = \dfrac{{60}}{{150}} = \dfrac{2}{5}$

14. A die is thrown \[500\] times. The frequency of the outcomes of the event \[1,2,3,4,5\] and \[6\] are recorded in the following frequency distribution table:

Find the probability of the occurrence of an 

(i) even number 

Ans: Total number of outcomes $ = 500$

Frequency of dice getting even number $75 + 90 + 70 +  = 235$

The probability of the occurrence of an even number is,

${\text{P}}$(even number) $ = \dfrac{{235}}{{500}} = \dfrac{{47}}{{100}}$

(ii) odd number.

Ans: Total number of outcomes $ = 500$

The probability of the occurrence of an odd number is,

${\text{P}}$(odd number) $ = 1 - \dfrac{{47}}{{100}} = \dfrac{{53}}{{100}}$

15. The distance (in km) of \[40\] engineers from their residence to their place of work were found as follows:

5 3 10 20 25 11 13 7 12 31 

19 10 12 17 18 11 32 17 16 2 

7 9 7 8 3 5 12 15 18 3 

12 14 2 9 6 15 15 7 6 12

What is the empirical probability than an engineer lives:

(i) less than 7 km from her place of work?

Ans: Total number of engineers \[ = 40\]

Number of engineers living less than $7\;{\text{km}}$ from her place of work $ = 9$

${\text{P}}$ (Engineer living less than $7\;{\text{km}}$ from her place of work) $ = \dfrac{9}{{40}}$

(ii) more than or equal to $7\;{\text{km}}$ from her place of work

Ans: Total number of engineers \[ = {\text{ }}40\]

Number of engineers living more than or equal to $7\;{\text{km}}$ from her place of work \[ = 31\]

Thus, ${\text{P}}$( Engineer living more than or equal to $7\;{\text{km}}$ from her place of work) $ = \dfrac{{31}}{{40}}$

(iii) within $\dfrac{1}{2}{\text{km}}$ from her place of work?

Ans: Total number of engineers \[ = 40\]

Number of engineers living within km from her place of work \[ = 0\]

Thus, ${\text{P}}$(Engineer living within $\dfrac{1}{2}\;{\text{km}}$ from her place of work $) = \dfrac{0}{{40}} = 0$

4 Marks Questions

1. The weekly pocket expenses of students are given below:

Find the probability that the weekly pocket expenses of a student are

(a) (i) Rs \[59\]

Ans: Number of students $ = 39$

Number of trials $ = 39$

Number of students with weekly pocket expenses of Rs $59 = 10$

${\text{P}}$(the weekly pocket expenses of a student are ${\text{Rs}}\,59) = \dfrac{{10}}{{39}}$

(ii) more than Rs \[59\]

Ans: Number of students $ = 39$

Number of trials $ = 39$

Number of students with weekly pocket expenses of more than Rs $59 = 6 + 1 = 7$

${\text{P}}$(the weekly pocket expenses of a student are more than Rs \[59\]$) = \dfrac{7}{{39}}$

(iii) less than Rs \[59\]

Ans: Number of students $ = 39$

Number of trials $ = 39$

Number of students with weekly pocket expenses of less than Rs $59 = 7 + 4 + 3 + 8 = 22$

${\text{P}}$(the weekly pocket expenses of a student are less than Rs \[59\]$) = \dfrac{{22}}{{39}}$

(b) Find the sum of probabilities computed in (i), (ii), and (iii)

Ans: Sum of probabilities in (i),(ii), and (iii)

$ = \dfrac{{10}}{{39}} + \dfrac{7}{{39}} + \dfrac{{22}}{{39}} = \dfrac{{39}}{{39}} = 1$

2. Cards marked \[{\text{2 to 101}}\] are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that number on the card is 

(i) an even number 

Ans: Total number of cards $ = 100$

Even numbers are $ = 50$

${\text{P}}$(even number) $ = \dfrac{{50}}{{100}} = \dfrac{1}{2}$

(ii) a number less than \[14\]

Ans: Total number of cards $ = 100$

Numbers less than \[14\]\[ = (2,3,4,5,6,7,8,9,10,11,12,13) = 12\]

\[{\text{P}}({\text{ number less than }}14) = \dfrac{{12}}{{100}} = 0.12\]

(iii) a number which is a perfect square

Ans: Total number of cards $ = 100$

Number which is a Perfect square$ = \{ 1,4,9,16,25,36,49,64,81,100\}  = 10$

${\text{P}}($number which is a perfect square $) = \dfrac{{10}}{{100}} = \dfrac{1}{{10}}$

(iv) a prime number less than \[20\]

Ans: Total number of cards $ = 100$

Prime numbers less than \[20\]$ = \{ 2,3,5,7,11,13,17,19\} $$ = 8$

${\text{P}}($ Prime number less than \[20\]$) = \dfrac{8}{{100}} = \dfrac{2}{{25}}$

(v) an odd number

Ans: ${\text{P}}$ (odd number $) = 1 - {\text{P}}$ (an even number)

\[ = 1 - \dfrac{1}{2} = \dfrac{1}{2}\].

Download Important Questions Class 9 Maths Chapter 15

Introduction to Probability

Probability is the measure of how likely an event is to occur or how likely it is that a proposition is true. Any events can’t be predicted with certainty but can be expressed as to how likely it can occur using the idea of Probability.

Probability of an event ranges between 0 and 1, where 0 Probability means the event to be an impossible one and Probability of 1 indicates a certain event.

Formula for Probability

The formula of Probability is defined as the possibility of an event to occur is equal to the ratio of the number of favourable outcomes and the total number of outcomes.

Probability of event  P(E) = Number of favourable outcomes/Total Number of outcomes

Probability of an Event

Assume E be an event, it can occur in r ways out of a sum of n probable or possible equally likely ways. The Probability of happening of the event or its success can be expressed as;

P(E) = r/n

The probability that the event will not occur or known as the failure event is expressed as:

P(E’) = (n-r)/n = 1-(r/n)

E’ represents that the event will not occur.

Therefore, now we can say;

P(E) + P(E’) = 1

This means that the sum of all the probabilities in any random test or experiment is equal to 1.

Equally Likely Events

When the events have the same Probability to occur, then they are called equally likely events. The sample space is equally likely if all of them have the same probability of occurring. For example, if you throw a die, then the probability of getting 1 is 1/6. Similarly, the probability of getting all the numbers on die, one at a time is 1/6. The following are some examples of equally likely events when throwing a die:

• Getting numbers 3 and 5 on throwing a die

• Getting an even number and an odd number on  throwing a die

• Getting numbers 1, 2 or 3 on rolling a die

All of these are equally likely events because the probabilities of each event are equal.

Impossible and Sure Events

If the probability of occurrence of an event is 0, then such event is known as an impossible event. If the probability of occurrence of an event is 1, it is known as a sure event. The empty set ϕ is an example impossible event and the sample space S is a sure event.

Complementary Events

The possibility that there will be only two outcomes of an event which states that an event will occur or not. Daily life examples like a person will come or not come to your house, getting a job or not getting a job, etc are examples of complementary events. Basically, the complement of an event occurring in the exact opposite that the Probability of it is not occurring. Some more examples are:

• It will rain or not rain today

• The student will pass the exam or not pass the exam.

• You win the lottery or you don’t.

Independent Events

If the probability of occurrence of an event A is not affected by the occurrence of another event B, then the events A and B are said to be independent events.

Note: If A and B are two events associated with the same random experiment, then A and B are known as independent events if P(A ∩ B) = P(B) . P(A)

Mutually Exclusive Events

If the occurrence of one event excludes the occurrence of another event, such events are known as mutually exclusive events. In other words, two events don’t have any common point. For example, if S = {1, 2, 3, 4, 5, 6} and E1, E2 are two events such that E1 consists of numbers less than 3 and E2 consists of numbers greater than 4.

So, occurrence of first event  E1 = {1,2} and occurrence of second event E2 = {5,6} .

Therefore, E1 and E2  are mutually exclusive.

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