Worked-out probability questions answers are given here
step-by-step to get the clear explanation to the student.

1. Out of 300 students in a school, 95 play cricket only,
120 play football only, 80 play volleyball only and 5 play no games. If one
student is chosen at random, find the probability that

(i) he plays volleyball

(ii) he plays either cricket or volleyball

(iii) he plays neither football nor volleyball.

Solution:

Total number of trials = 300 (Since there are 300 students all together).

Number of times a cricket player is chosen = 95 (Since 95 students play cricket).

Number of times a football player is chosen = 120.

Number of times a volleyball player is chosen = 80.

Number of times a student is chosen who plays no games = 5.

(i) Therefore, the probability of getting a player who plays volleyball

= Number of Times a Volleyball Player can be ChosenTotal Number of Trials

Number of Times a Volleyball Player can be ChosenTotal Number of Trials

80300

80300

415

415.(ii) The probability of getting a player who plays either cricket or volleyball

= Number of Times a Cricket or a Volleyball Player can be ChosenTotal Number of Trials

Number of Times a Cricket or a Volleyball Player can be ChosenTotal Number of Trials

95+80300

95+80300

175300

15300

712

712.

(iii) The probability of getting a player who plays neither football nor volleyball = Number of Times a Student can be Chosen who do not Play Football or VolleyballTotal Number of Trials

Number of Times a Student can be Chosen who do not Play Football or VolleyballTotal Number of Trials

30012080300

300–120−80300

100300

100300

13

13.

2. The blood group of 60 students of a class recorded as below.

 Blood Group A B AB O Number of Students 12 20 10 18

A student of the class is selected at
random.

(i) What is the probability that the
selected student has blood group O?

(ii) What is the probability that the
selected student does not have blood group O?

Solution:

(i) Total number of trials = 60. [Since,
there are 60 different students).

Number of students having blood group O =
18.

So, the probability of the student’s blood
group being O

=
Frequency of Favourable
Trials
Total Number of Trials

Frequency of FavourableTrialsTotal Number of Trials

=
Number of Students having Blood
Group O
Total Number of Trials

Number of Students having BloodGroup OTotal Number of Trials

1860

1860

310

310.

(ii) Total number of trials = 60. [Since,
there are 60 different students).

Number of students not having blood group O
= Total Number of Students – Number of Students having Blood Group O = 60 – 18
= 42.

So, the probability of the student’s blood
group being different from O

=
Number of Students not having
Blood Group O
Total Number of Trials

Number of Students not havingBlood Group OTotal Number of Trials

4260

4260

710

710.

Note: Probability of the student’s blood group not being O = 310

310 + 710

710
= 1.

3. In a test of 100 marks, the marks scored
by the students of a class are given below.

A student of the class is selected at random. Find the probability that the student has scored

(i) less than 30

(ii) at least 60

(iii) not less than 80.

Solution:

(i) Total number of trials = 80 [Since there are 80 students all together].

Number of students scoring less than 30 = Cumulative frequency of the interval 20 – 30 = 1 + 3 + 6 = 10.

So, the probability of the student scoring less than 30

= Frequency of Favourable TrialsTotal Number of Trials

Frequency of Favourable TrialsTotal Number of Trials

= Number of Students Scoring Less than 30Total Number of Students

Number of Students Scoring Less than 30Total Number of Students

1080

1080

18

18.

(ii) Total number of trials = 80.

Number of students scoring at least 60 = Number of students scoring more than or equal to 60

= 12 + 5 + 4 + 1 (= Sum of frequencies of the intervals 60 – 70, 70 – 80, 80 – 90, 90 – 100)

= 22.

So, the required probability = Frequency of Favourable TrialsTotal Number of Trials

Frequency of Favourable TrialsTotal Number of Trials

= Number of Students Scoring at least 60Total Number of Students

Number of Students Scoring at least 60Total Number of Students

2280

2280

1140

1140.

(iii) Total number of trials = 80.

Number of students scoring at not less than 80 = Number of students scoring to least 80

= 4 + 1 (= Sum of frequencies of the intervals 80 – 90 and 90 – 100)

= 5.

Therefore, the required probability = Number of Students Scoring not less than 80Total Number of Students

Number of Students Scoring not less than 80Total Number of Students

580

580

116

116.

4. A bag contains 8 red balls and some white balls. If the probability of drawing a white ball is half of the probability of drawing a red ball then find the number of white balls in the bag.

Solution:

Let the number of white balls be n.

The number of red balls = 8.

Therefore, the possible number of outcomes = n + 8.

Now, the probability of drawing a white ball = nn+8

nn+8.

The probability of drawing a red ball = 8n+8

8n+8.

Now, from question,

nn+8

nn+8 = 12

12 ∙ nn+8

nn+8

or, n = 4.

So, number of white balls in the bag is 4.

5. A box contains 90 discs numbered 1 to 90. One disc is drawn at random from the box. What is the probability that is bears

(i) a two-digit number

(ii) a perfect square

(iii) a multiply of 5

(iv) a number divisible by 3 and 5.

Solution:

(i) Total number of possible outcomes = 90 (Since the disks are numbered from 1 to 90).

Number of favourable outcomes of the event E

= Number of two-digit numbers from 1 to 90

= 90 – 9 = 81 (Since all are two-digit numbers except 1 to 9)

Therefore, P(E) = Number of Favourable Outcomes of the Event ETotal Number of Possible Outcomes

Number of Favourable Outcomes of the Event ETotal Number of Possible Outcomes

8190

8190

910

910.

(ii) Total number of possible outcomes = 90.

Number of favourable outcomes of the event F

= Number of perfect squares from 1 to 90

= 9   [Since 1, 4, 9, 16, 25, 36, 49, 64 and 81 are perfect squares).

Therefore, by definition, P(F) = 990

990

= 110

110.

(iii) Total number of possible outcomes = 90.

Number of favourable outcomes of the event G

= Number of multiples of 5 among numbers from 1 to 90

= 18   [Since 5 × 1, 5 × 2, 5 × 3, ….,  5 × 18 are multiple of 5).

Therefore, by definition, P(G) = 1890

1890

= 15

15.

(iv) Total number of possible outcomes = 90.

Number of favourable outcomes of the event H

= Number of numbers divisible by 3 and 5 from 1 to 90

= Number of numbers divisible by 15 from 1 to 90

= 6  [Since 15 × 1, 15 × 2, 15 × 3, ….,  15 × 6 are divisible by 15).

Therefore, by definition, P(H) = 690

690

= 115

115.