NCERT Solutions for Class 10 Maths Chapter 14 Probability

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 15
Chapter Name :Probability

Exercise 15.1

Q1 Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = __________
(ii) The probability of an event that cannot happen is ______ Such an event is called____
(iii) The probability of an event that is certain to happen is _____ Such an event is called ______
(iv) The sum of the probabilities of all the elementary events of an experiment is ______
(v) ) The probability of an event is greater than or equal to______and less than or equal to ___

Answer.
(i) 1
(ii) 0, impossible event
(iii) 1,sure event or certain event
(iv) 1
(v) 0,1

Page : 308 , Block Name : Exercise 15.1

Q2 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.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl

Answer.
(i) It is not an equally likely event, as it depends on various factors such as whether the car will
start or not. And factors for both the conditions are not the same.
(ii) It is not an equally likely event, as it depends on the player's ability and there is no
information given about that.
(iii) It is an equally likely event.
(iv) It is an equally likely event.

Page : 308 , Block Name : Exercise 15.1

Q3 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?

Answer. When we toss a coin, the possible outcomes are only two, head Or tail, which are
equally likely outcomes. Therefore, the result of an individual toss is completely
unpredictable.

Page : 308 , Block Name : Exercise 15.1

Q4 Which of the following cannot be the probability of an event?
(A) ⅔
(B) -1.5
(C) 15 %
(D) 0.7

Answer. Probability of an event (E) is always greater than or equal to 0. Also, it is always less
than or equal to one. This implies that the probability of an event cannot be negative or
greater than 1. Therefore, out of these alternatives, -1.5 cannot be a probability of an event.
Hence, (B)

Page : 308 , Block Name : Exercise 15.1

Q5 If P(E) = 0.05, what is the probability of ‘not E’?

Answer. We Know that,
¯
¯¯¯
P(E) = 1 − P(E)
¯
¯¯¯
P(E) = 1 − 0.05

= 0.95

Therefore, the probability of ‘not E’ is 0.95

Page : 308 , Block Name : Exercise 15.1

Q6 A bag contains lemon avoured candies only. Malini takes out one candy without looking
into the bag. What is the probability that she takes out
(i) an orange avoured candy?
(ii) a lemon avoured candy?

Answer.
(i) The bag contains lemon avoured candies only, It does not contain any orange avoured
candies. This implies that every time, she will take out only lemon avoured candies.
Therefore, event that Malini will take out an orange avoured candy is an impossible event.
Hence, P (an orange avoured candy) = 0
(ii) As the bag has lemon avoured candies, Malini will take out only lemon avoured candies.
Therefore, event that Malini will take out a lemon avoured candy is a sure event.

P (a lernon avoured candy) =1

Page : 308 , Block Name : Exercise 15.1

Q7 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?

Answer. Probability that two students are not having same birthday P (E ) = 0.992
¯
¯¯¯

Probability that two students are having same birthday P (E) = 1 − P (E )
¯
¯¯¯

= 1 − 0.992

= 0.008

Page : 308 , Block Name : Exercise 15.1

Q8 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 ? (ii) not red?

Answer.
(i) Total number of balls in the bag= 8
Probability of getting a red ball = Number of favourable outcomes

Number of total possible outcomes

=⅜

(ii) Probability of not getting red ball = 1 - Probability of getting a red ball
=1− 3

8

=
5

8

Page : 308 , Block Name : Exercise 15.1

Q9 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 ?
(ii) white ? (iii) not green?

Answer. Total number of marbles =5+8+4 =17

(i) Number of red marble= Number of favourable outcomes

Number of total possible outcomes

= 5/17

(ii) Number of white marbles=8
Number of white marble=
Number of favourable outcomes

Number of total possible outcomes

=8/17

(iii) Number of green marble = Number of favourable outcomes

Number of total possible outcomes

=4/17

Probability of not getting a green marble = 1 − 4

17
=
13

17

Page : 308 , Block Name : Exercise 15.1

Q10 A piggy bank contains hundred 50p coins, fty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5
coins. If it is equally likely that one 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 ? (ii) will not be a ₹ 5 coin?

Answer. Total number of coins in a piggy bank =100+50+20+10 =180

(i) Probability of getting a 50p coin = Number of favourable outcomes

Number of total possible outcomes

100 5
= =
180 9

(ii) Number of Rs 5 Coins =10
Probability of getting a 5 Rs coin = Number of favourable outcomes

Number of total possible outcomes

10 1
= =
180 18

Probability of not getting a 5Rs coin = 1 − 1

18
17
=
18

Page : 309 , Block Name : Exercise 15.1

Q11 Gopi buys a sh from a shop for his aquarium. The shopkeeper takes out one sh at
random from a tank containing 5 male sh and 8 female sh (see Fig.) . What is the probability
that the sh taken out is a male sh?

Answer. Total number of shes in a tank
= Number of male shes + number of female shes
= 5+8=1
Probability of getting a male sh = Number of favourable outcomes

Number of total possible outcomes

5
=
13

Page : 309 , Block Name : Exercise 15.1

Q12 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 Fig. ), and these are equally likely outcomes. What is the
probability that it will point at

(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Answer. Total number of possible outcomes =8
(i) Probability of getting 8 = Number of favourable outcomes

Number of total possible outcomes
=
1

8

(ii) Total number of odd numbers on spinner= 4
Probability of getting an odd number = Number of favourable outcomes

Number of total possible outcomes

4 1
= =
8 2

(iii) The numbers greater than 2 are 3,4,5,6,7 and 8
Therefore , total numbers greater than 2 =6
Probability of getting a number greater than 2
Number of favourable outcomes 6 3
= = =
Number of total possible outcomes 8 4

(iv) The numbers less than 9 are 1,2,3,4,6,7 and 8
Therefore total numbers less than 9 =8
Probability of getting a number less than 9= 8/8=1

Page : 309 , Block Name : Exercise 15.1

Q13 A die is thrown once. Find the probability of getting (i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number

Answer. The possible outcomes when a dice is thrown ={1, 2, 3, 4, 5, 6}
Number Of possible outcomes of a dice = 6
(i) Prime numbers on a dice are 2, 3, and 5.
Total prime numbers on a dice= 3
Probability of getting a prime number = = 3

6
1

2

(ii) Numbers lying between 2 and 6 =3,4,5
Total numbers lying between 2 and 6=3
Probability of getting a number lying between 2 and 6 = 3

6
=
1

2

(iii) Odd numbers on a dice= 1,3 and 5
Total odd numbers on a dice =3
Probability of getting an odd number = 3

6
=
1

2

Page : 309 , Block Name : Exercise 15.1

Q14 One card is drawn from a well-shuf ed deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds

Answer. (i) Total number of cards in a well-shuf ed deck =52
Total number of kings of red colour =2
P (getting a king of red colour ) = Number of favourable outcomes

Number of total possible outcomes

2 1
= =
52 26

(ii) Total number of face cards =12
P(getting a face cards) = Number of favourable outcomes.

Number of total possible outcomes

12 3
= =
52 13

(iii) Total number of red face cards =6
P(getting a red face card) = Number of favourable outcomes

Number of total possible outcomes

6 3
= =
52 26

(iv) Total number of Jack of hearts =1
p(getting a jack of hearts) =
Number of favourable outcomes

Number of total possible outcomes

=1/52
(v) Total number of spade cards =13
P (getting a spade card) =
Number of favourable outcomes

Number of total possible outcomes

=13/52
=1/4
(vi) Total numbers of queen of diamonds =1
P(getting a queen of diamond) = Number of favourable outcomes

Number of total possible outcomes

=1/52

Page : 309 , Block Name : Exercise 15.1

Q15 Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuf ed with their
face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(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?

Answer. Total numbers of cards =5

Total numbers of queens= 1
P(getting a queen) = Number of favourable outcomes

Number of total possible outcomes

1
=
5

(ii) When the queen is drawn and put aside, the total number of remaining cards will be 4
(a) Total number of aces=1
P(getting an ace) =¼

(b) As queen is already drawn, therefore, the number of queens will be 0.
P(getting a queen) =0/4=0

Page : 309 , Block Name : Exercise 15.1

Q16 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 taken out is a good one

Answer. Total number of pens =12+132=144
Total number of good pens =132
P(getting a good pen) = Number of favourable outcomes

Number of total possible outcomes

132 11
= =
144 12

Page : 309 , Block Name : Exercise 15.1

Q17 (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?
(ii) Suppose the bulb drawn in
(iii) 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 ?

Answer. Total numbers of bulbs=20
Total number of defective bulbs=4
P(getting a defective bulb) = Number of favourable outcomes.

Number of total possible outcomes

4 1
= =
20 5

(ii) Remaining total number of bulbs=19
Remaining total number of non-defective bulbs=16-1=15
P(getting a not defective bulb) =15/19

Page : 309 , Block Name : Exercise 15.1

Q18 A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random
from the box, nd the probability that it bears (i) a two-digit number (ii) a perfect square
number (iii) a number divisible by 5.

Answer. Total number of discs=90
(i) Total number of two digit numbers between 1 and 90 =81
P(getting a two digit number) =
81 9
=
90 10

(ii) Perfect squares between 1 and 90 are 1,4,9,16,25,36,49,64, and 81.
Therefore, total number of perfect squares between 1 and 90 is 9
P(getting a perfect square) =
9 1
=
90 10

(iii) Numbers that are between 1 and 90 and divisible by 5 are
5,10,15,20,25,30,35,40,45,50,55,60,65,70,80,85, and 90.Therefore a total numbers divisible by
5=
18 1
=
90 5

Page : 309 , Block Name : Exercise 15.1

Q19 A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?

Answer. Total number of possible outcomes on the dice=6
(i) Total number of faces having A on it =2
P(getting A) = =
2 1

6 3

(ii) Total number of faces having D on it = 1
P(getting D) =1/6

Page : 310 , Block Name : Exercise 15.1

Q20 Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is
the probability that it will land inside the circle with diameter 1m?

Answer. Area of rectangle = l × b = 3 × 2 = 6m 2

2

Area of circle (of diameter 1 m) = πr 2
= π(
1

2
) = −
1

4
m
2

π

P (die will land inside the circle) = 4

6
=
π

24

Page : 310 , Block Name : Exercise 15.1

Q21 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 ? (ii) She will not buy

it ?

Answer. Total number of pens 144
Total number of defective pens = 20
Total number of good pens 144 − 20 = 124

(i) Probability of getting a good pen = =
124 31
=
144 36

P (Nuri buys a pen) = 31

36

(ii) P (Nuri will not buy a pen) = 1 −
31 5
=
36 36

Page : 310 , Block Name : Exercise 15.1

Q22 Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your
answer.

Answer. (i) It can be observed that,
To get the sum as 2, possible outcomes = (1, 1)
To get the sum as 3, possible outcomes = (2, 1) and (1, 2)
To get the sum as 4, possible outcomes= (3, 1), (1, 3), (2, 2)
To get the sum as 5, possible outcomes= (4, 1), (1, 4), (2, 3), (3, 2)
To get the sum as 6, possible outcomes= (5, 1), (1, 5), (2, 4), (4, 2)(3, 3)
To get the sum as 7, possible outcomes= (6, 1), (1, 6), (2, 5), (5, 2)(3, 4), (4, 3)
To get the sum as 8, possible outcomes= (6, 2), (2, 6), (3, 5), (5, 3)(4, 4)
To get the sum as 9, possible outcomes= (3, 6), (6, 3), (4, 5), (5, 4)
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
To get the sum as 11, possible outcomes = (5, 6), (6, 5)
To get the sum as 12, possible outcomes = (6, 6)

(ii) Probability of each of these sums will not be 1/11 as these sums are not equally likely

Page : 310 , Block Name : Exercise 15.1

Q23 A game consists of tossing a one rupee coin 3 times and noting its outcome each time.
Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses
otherwise. Calculate the probability that Hanif will lose the game

Answer. The possible outcomes are {HHH, TTT, HHT , HTH , THH, TTH ,THT, HTT}
Number of total possible outcomes=8
Numbers of favourable outcomes= 2 { i.e TTT and HHH}
P ( Hanif will win the game) = =
2 1

8 4

P(Hanif will lose the game ) = 1 − 1

4
=
3

4

Q24 A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5
will come up at least once? [Hint : Throwing a die twice and throwing two dice simultaneously
are treated as the same experiment]

Answer. Total number of outcomes = 6 x 6 =>36
(i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5,3) ,
(5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)

Hence, total number of favourable cases 11
P (5 will come up either time) = 11

36

P (5 will not come up either time) =
11

36

(ii)Total number of cases, when 5 can come at least once =11
P (5 will come at least once) =
11

36

Page : 310 , Block Name : Exercise 15.1

Q25 Which of the following arguments are correct and which are not correct? Give reasons for
your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two
tails or one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number.
Therefore, the probability of getting an odd number is 1/2 .

Answer.
(i) Incorrect
When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). It can be
observed that there can be one Of each in two possible ways — (H, T), (T,H)
Therefore, the probability of getting two heads is 4 , the probability of getting two tails is 1/4 ,
and the
probability of getting one of each is 1/2
It can be observed that for each outcome, the probability is not 1/3

(ii) Correct
When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these, 1, 3, 5 are
odd and 2, 4, 6 are even numbers.
Therefore the probability of getting an odd number is ½

Page : 311 , Block Name : Exercise 15.1

Exercise 15.2

Q1 Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to
Saturday). Each is equally likely to visit the shop on any day as on another day. What is the
probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?

Answer. There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the
shop in 5 ways.
Therefore, total number Of outcomes = 5 x 5 = 25

(i) They can reach on the same day in 5 ways.
i.e., (t, t), (w, w), (th, th), (f, f), (s, s)
P (both will reach on same day) =
5 1
=
25 5

(ii) They can reach on consecutive days n these 8 ways - (t, w), (w, th), (th, f), (f, s), (w, t), (th,
w), (f, th), (s, f)
Therefore, P (both will reach on consecutive days) =
8

25

(iii) P (both will reach on same day) = (From (i)]
1

5

P (both will reach on different days) = 1 − 1

5
=
4

5

Page : 311 , Block Name : Exercise 15.2

Q2 A die is numbered in such a way that ts faces show the number 1, 2, 2, 3, 3, 6. It is thrown
two times and the total score in two throws is noted. Complete the following table which gives
a few values of the total score on the two throws:

What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6?

Answer.

Total number of possible outcomes when two dice are thrown == 6 × 6 = 36

(i) Total times when the sum is even 18
P (getting an even number) = =
18

36
1

2

(ii) Total times when the sum is 6 = 4
P (getting sum as 6) = 4
=
36
1

9

(iii) Total times when the sum is at least 6 (i.e., greater than 5) =15
P (getting sum at least 6) = 15

36
=
5

12

Page : 311 , Block Name : Exercise 15.2

Q3 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

Answer. Let the number of blue balls be x
Number of red balls 5
Total number of balls = x + 5
P (getting a red ball) = 5

x+5

P (getting a blue ball) =
x

x+5

Given that,
5 x
2( ) =
x+5 x+5

2
10(x + 5) = x + 5x
2
x − 5x − 50 = 0
2
x − 10x + 5x − 50 = 0

x(x − 10) + 5(x − 10) = 0

Either x − 10 = 0 or x + 5 = 0
x = 10 or x = −5

However, the number of balls cannot be negative.
Hence, number of blue balls =10

Page : 311 , Block Name : Exercise 15.2

Q4 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

Answer. Total number of balls =12
Total number Of black balls = x
P (getting a black ball) = 12
x

If 6 more black balls are put in the box, then
Total number of balls 12 + 6 18
Total number of black balls = x + 6
P (getting a black ball now) =
x+6

18

According to the condition given in the question,
x x+6
2( ) =
12 18

3x = x + 6

2x = 6

x = 3

Page : 311 , Block Name : Exercise 15.2

Q5 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 2 /3 ⋅ Find the number of blue balls in
the jar.

Answer. Total number of marbles = 24
Let the total number of green marbles be x.
Then, total number of blue marbles =24-x
P (getting a given marble) = x

24

According to the condition given in the question,
x 2
=
24 3

x = 16

Therefore, total number of green marbles in the jar = 16
Hence, total number of blue marbles = 24 − x = 24 − 16 = 8

Page : 312 , Block Name : Exercise 15.2