Unit Test (Solutions): Probability | Mathematics (Maths) Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: The probability, which is a non-leap year selected at random, will contain 53 Sundays is (1 mark)

(a) 1/7
(b) 2/7
(c) 3/7
(d) 5/7

Ans: (a)

A non-leap year has 365 days and thus 52 weeks and 1 day. This 1 day may be Sunday, Monday, Tuesday, Wednesday or Thursday or Friday, or Saturday.
Hence, out of 7 possibilities, 1 favorable event is the event that the one day is Sunday.
The required probability = 1/7
So, the correct answer is an option (a).

Q2: A card has been selected from a deck of 52 cards. The probability for the being the red face card is (1 mark)

(a) 3/26
(b) 3/13
(c) 2/13
(d) 1/2

Ans: (a)

In a deck for the 52 cards, there are 12 face cards that are 6 red and 6 black cards. Hence, probability for getting a red face card = 6/52 = 3/26.

Q3:  When the probability for the event is p, the probability of the complementary event will be (1 mark)

(a) p – 1
(b) p
(c) 1 – p
(d) 1-1/p

Ans: (c)

As the probability for the event + probability for the complimentary event = 1 
It can be written as 
Probability for the complimentary event = (1 – Probability for the an event) = 1 – p

Q4:  Which among the following cannot be the probability of an event? (1 mark)

(a) 1/3
(b) 0.1
(c) 3%
(d) 17/16

Ans: (d)

The probability for the event always lies between 0 and 1. Probability for the event cannot be more than 1 or negative as (17/16) > 1

Q5:  If the probability of winning a game is 0.07, what is the probability of losing it? (1 mark)

Ans: Given that the probability of winning a game = 0.07
We know that the events of winning a game and losing the game are complementary events. 
Thus, P(winning a game) + P(losing the game) = 1
So, P(losing the game) = 1 – 0.07 = 0.93

Q6: The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap? (2 marks)

Ans: Given,
Total number of apples in the heap = n(S) = 900
Let E be the event of selecting a rotten apple from the heap.
Number of outcomes favourable to E = n(E)
P(E) = n(E)/n(S)
0.18 = n(E)/900
⇒ n(E) = 900 × 0.18
⇒ n(E) = 162
Therefore, the number of rotten apples in the heap = 162

Q7: A die is thrown once. What is the probability of getting a number less than 3? (2 marks)

Ans: Given that a die is thrown once.
Total number of outcomes = n(S) = 6
i.e. S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number less than 3.
n(E) = Number of outcomes favourable to the event E = 2
Since E = {1, 2}
Hence, the required probability = P(E) = n(E)/n(S)
= 2/6 = 1/3

Q8: 12 defective pens are accidentally 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 is taken out is a good one. (2 marks)

Ans: Numbers of pens = Numbers of defective pens + Numbers of good pens
∴ Total number of pens = 132 + 12 = 144 pens
P(E) = (Number of favourable outcomes) / (Total number of outcomes)
P(picking a good pen) = 132/144 = 11/12 = 0.916

Q9:  An integer is chosen between 0 and 100. What is the probability that it is (3 marks)

(i) divisible by 7?
(ii) not divisible by 7?

Ans: Number of integers between 0 and 100 = n(S) = 99

(i) Let E be the event ‘integer divisible by 7’
Favorable outcomes to the event E = 7, 14, 21.…., 98
Number of favorable outcomes = n(E) = 14
Probability = P(E) = n(E)/n(S) = 14/99

(ii) Let F be the event ‘integer not divisible by 7’
Number of favorable outcomes to the event F = 99 – Number of integers divisible by 7
= 99-14 = 85
Hence, the required probability = P(F) = n(F)/n(S) = 85/99

Q10: Cards marked with the number 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that the number on the card is: (3 marks)

(i) An even number
(ii) A number less than 14
(iii) A number is perfect square

Ans: Total even no from 2 to 101=50

(i) Probability of drawing an even no= total even no/ total no
= 50/100
= 1/2=0.5

(ii) no less than 14 = 13
Probability = 13/100
= 0.13

(iii) total perfect square=9
Probability = 9/100
= 0.09

Q11: The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 1/5. The probability of selecting a black marble at random from the same jar is 1/4. If the jar contains 11 green marbles, find the total number of marbles in the jar. (3 marks)

Ans: Given that,
P(selecting a blue marble) = 1/5
P(selecting a black marble) = 1/4
We know that the sum of all probabilities of events associated with a random experiment is equal to 1.
So, P(selecting a blue marble) + P(selecting a black marble) + P(selecting a green marble) = 1
(1/5) + (1/4) + P(selecting a green marble) = 1
P(selecting a green marble) = 1 – (1/4) – (1/5)
= (20 – 5 – 4)/20
= 11/20
P(selecting a green marble) = Number of green marbles/Total number of marbles
11/20 = 11/Total number of marbles {since the number of green marbles in the jar = 11}
Therefore, the total number of marbles = 20

Q12: Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown, and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately. (5 marks)

Ans: Number of total outcome = n(S) = 36
(i) Let E₁ be the event ‘getting sum 2’
Favorable outcomes for the event E₁ = {(1,1),(1,1)}
n(E₁) = 2
P(E₁) = n(E₁)/n(S) = 2/36 = 1/18

(ii) Let E₂ be the event ‘getting sum 3’
Favorable outcomes for the event E₂ = {(1,2),(1,2),(2,1),(2,1)}
n(E₂) = 4
P(E₂) = n(E₂)/n(S) = 4/36 = 1/9

(iii) Let E₃ be the event ‘getting sum 4’
Favorable outcomes for the event E₃ = {(2,2)(2,2),(3,1),(3,1),(1,3),(1,3)}
n(E₃) = 6
P(E₃) = n(E₃)/n(S) = 6/36 = 1/6

(iv) Let E₄ be the event ‘getting sum 5’
Favorable outcomes for the event E₄ = {(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}
n(E₄) = 6
P(E₄) = n(E₄)/n(S) = 6/36 = 1/6

(v) Let E₅ be the event ‘getting sum 6’
Favorable outcomes for the event E5 = {(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)}
n(E₅) = 6
P(E₅) = n(E₅)/n(S) = 6/36 = 1/6

(vi) Let E₆ be the event ‘getting sum 7’
Favorable outcomes for the event E6 = {(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}
n(E₆) = 6
P(E₆) = n(E₆)/n(S) = 6/36 = 1/6

(vii) Let E₇ be the event ‘getting sum 8’
Favorable outcomes for the event E₇ = {(5,3),(5,3),(6,2),(6,2)}
n(E₇) = 4
P(E₇) = n(E₇)/n(S) = 4/36 = 1/9

(viii) Let E₈ be the event ‘getting sum 9’
Favorable outcomes for the event E8 = {(6,3),(6,3)}
n(E₈) = 2
P(E₈) = n(E₈)/n(S) = 2/36 = 1/18

Q13: A bag consists of 10 red, 5 blue, and 7 green balls. A ball is drawn randomly. Find out the probability that this ball is a (5 marks)

(i) red ball
(ii) green ball
(iii) not a blue ball

Ans: No. having a red ball = 10
No. having a blue ball = 5
No. having green balls = 7
When the ball is drawn out of 22 balls (5 blue + 7 green + 10 red), the total number having outcomes is  n(S) = 22.

(i) Let E₁ = Event for getting a red ball
n(E₁) = 10
So, Required probability=n(E₁)/n(S)=10/22=5/11

(ii) Let E₂ = Event for getting a green ball n(E₂) = 7
So, Required probability=n(E₂)/n(S)=7/22

(iii) Let E₃ = event for getting a red ball or a green ball..
n(E₃) = (10 + 7) = 17
Required probability=n(E3)/n(S)=17/22
Therefore, the required probability of not  getting a blue ball is 17/22