Class-10-Probability-Ex 15.1

NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1

NCERT Solutions for Maths Chapter 15, Exercise 15.1 involve complete answers for each question in the exercise 15.1. The solutions provide students a strategic methods to prepare for their exam. Class 10 Maths Chapter 15 Probability Exercise 15.1 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems.NCERT Solutions for Chapter 15 Probability Exercise 15.1 prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way.

Question 1. 
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  _______..

Solution:

(i) 1 

(ii) 0, impossible event 

(iii) 1, sure or certain event

(iv) 1 

(v) 0, 1

Question 2. 
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.

Solution;

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 players ability and there is no information given about it.

The experiments (iii) and (iv) have equally likely outcomes.

Question 3.
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?

Sol-

When we toss a coin, the outcomes head and tail are equally likely. So, the result of an individual coin toss is completely unpredictable.

Question 4. 
Which of the following cannot be the probability of an event?
(A) ²/₃ (B) –1.5 (C) 15% (D) 0.7

Probability of an event is always greater than or equal to 0.Also it is always less than or equal to 1.This imply that the probability of an event cannot be negative or greater than 1.

Therefore out of these alternative -1.5 cannot be a probability of an event.

Hence answer is option (B) -1.5.

Question 5. 
If P(E) = 0.05, what is the probability of ‘not E’?

Solution:

We know that,

P(not E)=1-P(E)

P(not E)=1-0.05

P(not E)=0.95

Therefore, the probability of ‘not E’ is 0.95.

Question 6:
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i)An orange flavoured candy?
(ii)Alemon flavoured candy?

Solution:

(i)The bag contains lemon flavoured candies only. It does not contain any orange flavoured candies. Thus, P (an orange flavoured candy) = 0

(ii)P (a lemon flavoured candy) = 1

Question 7:
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?

Solution:

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

Probability that two students are having same birthday P (E) = 1 − P(not E)

= 1 − 0.992

= 0.008

Question 8:
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?

Solution:

(i) Total number of balls in the bag = 8

Probability of getting a red ball = ^{No. of favourable outcomes}/_{No. of total possible outcomes}

Probability of getting a red bal  = ³/₈

(ii)Probability of not getting a red bal  =1-Probability of getting a red bal

                                                                =1- ³/₈ 

                                                                = ⁵/₈

Question 9:
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?

Solution:

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

(i)Number of red marbles = 5

Probability of getting a red marble = ⁵/₁₇

(ii)Number of white marbles = 8

Probability of getting a white marble = ⁸/₁₇

iii)Number of green marbles = 4

Probability of getting a green marble = ⁴/₁₇

Probability of not getting a green marble  = 1 - ⁴/₁₇

                                                                    = ¹³/₁₇

Question 10:
A piggy bank contains hundred 50 p coins, fifty ₹ 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 an ₹ 5 coin?

Solution:

(i) Total number of coins in a piggy bank = 100 + 50 + 20 + 10= 180

Number of 50 p coins = 100

Probability of getting a 50 p coin = ¹⁰⁰/₁₈₀

(ii) Number of ₹ 5 coins = 10

Probability of getting a Rs. 5 p coin = ¹⁰/₁₈₀

                                                            = ¹/₁₈

Probability of not getting an ₹ 5 coin = 1-¹⁷/₁₈

                                 = ¹⁷/₁₈

Question 11.
Gopi 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?

Solution:

Total number of fishes in a tank= Number of male fishes + Number of female fishes

= 5 + 8 = 13

Probability of getting a male fish = = ⁵/₁₃

Question 12:
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 the given figure), and these are equally likely outcomes. What is the probability that it will point at 8?
(i) An odd number?
(ii) A number greater than 2?
(iii) A number less than 9?

Solution:

Total number of possible outcomes = 8

Probability of getting = ¹/₈

(i) Total number of odd numbers on spinner = 4

Probability of getting an odd number = ⁴/₈ = ¹/₂

(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 = ⁶/₈ = ³/₄

(iii)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

Question 13:
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.

Solution:
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

Question 14:
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) A king of red color
(ii) A face card
(iii) A red face card
(iv) The jack of hearts
(v) A spade
(vi) The queen of diamonds

Solution:

Total number of cards in a well-shuffled deck = 52

(i) Total number of kings of red colour = 2

P (getting a king of red color) = ²/₅₂ =¹/₂₆

(ii) Total number of face cards = 12

P (getting a face card) = ¹²/₅₂ = ⁴/₁₃

(iii) Total number of red face cards = 6

P (getting a red face card) = ⁶/₅₂ = ³/₂₆

(iv) Total number of Jack of hearts = 1

P (getting a Jack of hearts) = ¹/₅₂

(v) Total number of spade cards = 13

P (getting a spade card) =¹³/₅₂ = ¹/₄

(vi) Total number of queen of diamonds = 1

P (getting a queen of diamond)=¹/₅₂

Question 15:
Five cards−−the 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?
(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?

Solution:

(i) Total number of cards = 5

Total number of queens = 1

P (getting a queen)=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)=¹/₄

(b) As queen is already drawn, therefore, the number of queens will be 0.

P (getting a queen)=⁰/₄= 0

Question 16:
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.

Solution:

Total number of pens = 12 + 132 = 144

Total number of good pens = 132

P (getting a good pen) = ¹³²/₁₄₄

=¹¹/₁₂

Question 17:
(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 
(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?

Solution:

(i) Total number of bulbs = 20

Total number of defective bulbs = 4

P (getting a defective bulb) = ⁴/₂₀ = ⁴/₅

(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

Question 18:
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
(ii) A perfect square number
(iii) A number divisible by 5.

Solution:

(i) Total number of discs = 90

Total number of two-digit numbers between 1 and 90 = 81

P (getting a two-digit number) = ⁸¹/₉₀ = ⁹/₁₀

(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)=⁹/₉₀=¹/₁₀

(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, 75, 80, 85, and 90. 

Therefore, total numbers divisible by 5 = 18

Probability of getting a number divisible by 5 = ¹⁸/₉₀

=¹/₅

Question 19: 
A child has a die whose six faces shows the letters as given below: 

A	B	C	D	E	A

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

Solution: 

Total number of possible outcomes on the dice = 6 

(i) Total number of faces having A on it = 2 

P (getting A) = ²/₆ = ¹/₃

 (ii) Total number of faces having D on it = 1 

P (getting D)= ¹/₆

Question 20
Suppose you drop a die at random on the rectangular region shown in Fig. . What is the probability that it will land inside the circle with diameter 1 m?

Solution:

Question 21:
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?

Solution:

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 = ¹²⁴/₁₄₄ = ³¹/₃₆

P(Nuri buys a pen) =  ³¹/₃₆

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

Question 22:
Two dice, one blue and one grey, are thrown at the same time.Write down all the possible outcomes and complete the following table:

Event: Sum of two dice	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36						5/36				1/36

 
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?

Solution:

(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)

Question 23:
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.

Solution:

The possible outcomes are: {HHH, TTT, HHT, HTH, THH, TTH, THT, and HTT}

Number of total possible outcomes = 8

Number of favorable outcomes = 2 

P (Hanif will win the game)= ²/₈ = ¹/₄

P (Hanif will lose the game)= 1-¹/₄= ³/₄

Question 24:
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].

Solution:

Total number of outcomes = 6 × 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 favorable cases = 11

P (5 will come up either time)=¹¹/₃₆

P (5 will not come up either time)=1-¹¹/₃₆= ²⁵/₃₆

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

P (5 will come at least once)=¹¹/₃₆

Question 25:
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.

Solution:

(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 ¹/₄, the probability of getting two tails is ¹/₄, and the probability of getting one of each is ¹/₂.