NCERT Solutions for Class 10 Maths Chapter 14 – Probability Ex 14.1

NCERT Solutions for Class 10 Maths Chapter 14 – Probability Ex 14.1 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. Probability tells us how likely an event is to happen. It is a number between 0 and 1. A probability of 0 means the event is impossible, and a probability of 1 means it is certain. Exercise 14.1 is the main exercise of the chapter and uses theoretical probability.

This exercise has 25 questions. They are based on everyday objects of chance, such as coins, dice, playing cards, marbles of different colours, and defective and non-defective items. In every question you count the number of favourable outcomes and divide by the total number of outcomes. Every question below is solved in short steps, with the final answer in bold. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.

Exercise 14.1 – Questions and Answers

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) Probability of an event E + Probability of the event 'not E' = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 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 does not have equally likely outcomes. The car may not start because of some fault, so the two outcomes are not equally probable.
(ii) It does not have equally likely outcomes. Whether the player shoots or misses depends on the player's skill, so the two outcomes are not equally probable.
(iii) It has equally likely outcomes. The answer to a true-false question is either right or wrong, and both are equally probable.
(iv) It has equally likely outcomes. The baby is either a boy or a girl, and both are equally probable.

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:
The outcomes of a coin toss are equally likely. So the result of a coin toss is completely unpredictable, and neither team gets an unfair advantage.

Q4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) −1.5 (C) 15% (D) 0.7

Answer:
The probability of an event always lies between 0 and 1, so it can never be negative.
Here, −1.5 is negative.
The correct answer is (B).

Q5. 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) a lemon flavoured candy?

Answer:
(i) There is no orange flavoured candy in the bag. So, the probability of taking out an orange flavoured candy is zero. It is an impossible event.
(ii) The probability of taking out a lemon flavoured candy is 1, as there are only lemon flavoured candies in the bag. It is a certain event.

Q6. If P(E) = 0.05, what is the probability of 'not E'?

Answer:
Probability of 'not E' = 1 − P(E)
= 1 − 0.05
= 0.95

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:
The probability of 2 students not having the same birthday = P(E') = 0.992
The probability of 2 students having the same birthday = 1 − P(E')
= 1 − 0.992
= 0.008

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:
Number of red balls = 3
Number of black balls = 5
Total number of outcomes = 3 + 5 = 8

(i) Probability of getting a red ball
= Number of favourable outcomes / Total number of outcomes
= 3/8

(ii) Probability of not getting a red ball
= 1 − Probability of getting a red ball
= 1 − 3/8
= 5/8

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:
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of outcomes = 5 + 8 + 4 = 17

(i) Probability of getting a red marble
= Number of favourable outcomes / Total number of outcomes
= 5/17

(ii) Probability of getting a white marble
= Number of favourable outcomes / Total number of outcomes
= 8/17

(iii) Probability of getting a marble that is not green
= 1 − Probability of getting a green marble
= 1 − 4/17
= 13/17

Q10. A piggy bank contains hundred 50p 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 a ₹5 coin?

Answer:
Number of 50p coins = 100
Number of ₹1 coins = 50
Number of ₹2 coins = 20
Number of ₹5 coins = 10
Total number of outcomes = 100 + 50 + 20 + 10 = 180

(i) Probability of getting a 50p coin
= Number of favourable outcomes / Total number of outcomes
= 100/180
= 5/9

(ii) Probability of not getting a ₹5 coin
= 1 − Probability of getting a ₹5 coin
= 1 − (Number of favourable outcomes / Total number of outcomes)
= 1 − 10/180
= 17/18

Q11. 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 (see the following figure). What is the probability that the fish taken out is a male fish?

Answer:
Number of male fish = 5
Number of female fish = 8
Total number of outcomes = 5 + 8 = 13

Probability of getting a male fish
= Number of favourable outcomes / Total number of outcomes
= 5/13

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 the following figure), 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 outcomes = 8

(i) Probability of the arrow pointing at 8
= Number of favourable outcomes / Total number of outcomes
= 1/8

(ii) Probability of the arrow pointing at an odd number
The odd numbers are 1, 3, 5 and 7, so there are 4 favourable outcomes.
= 4/8
= 1/2

(iii) Probability of the arrow pointing at a number greater than 2
The numbers greater than 2 are 3, 4, 5, 6, 7 and 8, so there are 6 favourable outcomes.
= 6/8
= 3/4

(iv) Probability of the arrow pointing at a number less than 9
All 8 numbers are less than 9, so there are 8 favourable outcomes.
= 8/8
= 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:
Total number of outcomes = 6

(i) Prime numbers on the die are 2, 3 and 5.
∴ Number of prime numbers on the die = 3
Probability of getting a prime number
= Number of favourable outcomes / Total number of outcomes
= 3/6
= 1/2

(ii) Numbers lying between 2 and 6 on the die are 3, 4 and 5.
∴ Number of such numbers on the die = 3
Probability of getting a number lying between 2 and 6
= 3/6
= 1/2

(iii) Odd numbers on the die are 1, 3 and 5.
∴ Number of odd numbers on the die = 3
Probability of getting an odd number
= 3/6
= 1/2

Q14. One card is drawn from a well-shuffled 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:
Total number of outcomes = 52

(i) Number of kings of red colour = 2
Probability of getting a king of red colour
= 2/52
= 1/26

(ii) Number of face cards = 12
Probability of getting a face card
= 12/52
= 3/13

(iii) Number of red face cards = 6
Probability of getting a red face card
= 6/52
= 3/26

(iv) Number of the jack of hearts = 1
Probability of getting the jack of hearts
= 1/52

(v) Number of spades = 13
Probability of getting a spade
= 13/52
= 1/4

(vi) Number of the queen of diamonds = 1
Probability of getting the queen of diamonds
= 1/52

Q15. 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?

Answer:

(i) Total number of outcomes = 5
Probability of getting the queen
= Number of favourable outcomes / Total number of outcomes
= 1/5

(ii) The queen is put aside.
∴ Total number of outcomes = 4

(a) Probability of getting an ace
= Number of favourable outcomes / Total number of outcomes
= 1/4

(b) Probability of getting a queen
= 0/4
= 0
There is no queen left, so it is an impossible event.

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:
Number of defective pens = 12
Number of good pens = 132
Total number of outcomes = 132 + 12 = 144

Probability of getting a good pen
= Number of favourable outcomes / Total number of outcomes
= 132/144
= 11/12

Q17. (i) A lot of 20 bulbs contains 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?

Answer:

(i) Number of defective bulbs = 4
Total number of outcomes = 20
Probability of getting a defective bulb
= Number of favourable outcomes / Total number of outcomes
= 4/20
= 1/5

(ii) Number of defective bulbs = 4
Number of good bulbs = 16 − 1 = 15
Total number of outcomes = 15 + 4 = 19
Probability of not getting a defective bulb
= 1 − Probability of getting a defective bulb
= 1 − 4/19
= 15/19

Q18. 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.

Answer:
Number of two-digit numbers among 1 to 90 = 81
Number of perfect square numbers among 1 to 90 = 9
Number of multiples of 5 which are less than or equal to 90 = 18
Total number of outcomes = 90

(i) Probability of getting a disc bearing a two-digit number
= Number of favourable outcomes / Total number of outcomes
= 81/90
= 9/10

(ii) Probability of getting a perfect square number
= 9/90
= 1/10

(iii) Probability of getting a number divisible by 5
= 18/90
= 1/5

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 outcomes = 6

(i) The letter A appears on 2 faces.
Probability of getting A
= Number of favourable outcomes / Total number of outcomes
= 2/6
= 1/3

(ii) The letter D appears on 1 face.
Probability of getting D
= 1/6

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

Answer:

Favourable outcomes = Area of the circle
Radius of the circle = 1/2 = 0.5 m
Area of circle = πr² = π(0.5)² = 0.25π m²

Total outcomes = Area of the rectangle
Area of rectangle = 2 m × 3 m = 6 m²

Therefore,
P(landing inside the circle) = Area of the circle / Area of the rectangle
= 0.25π / 6
= π/24

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 it 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:
Number of defective pens = 20
Number of good pens = 144 − 20 = 124
Total number of outcomes = 144

(i) Probability of buying a pen = Probability of drawing a good pen
= Number of favourable outcomes / Total number of outcomes
= 124/144
= 31/36

(ii) Probability of not buying a pen = 1 − Probability of buying a pen
= 1 − 31/36
= 5/36

Q22. Two dice, one blue and one grey, are thrown at the same time.

(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) Total possible outcomes = 36

Possible outcomes for getting the sum 2 are (1, 1).
Possible outcomes for getting the sum 3 are (1, 2) and (2, 1).
Possible outcomes for getting the sum 4 are (1, 3); (3, 1) and (2, 2).
Possible outcomes for getting the sum 5 are (1, 4); (2, 3); (3, 2) and (4, 1).
Possible outcomes for getting the sum 6 are (1, 5); (2, 4); (3, 3); (4, 2) and (5, 1).
Possible outcomes for getting the sum 7 are (1, 6); (2, 5); (3, 4); (4, 3); (5, 2) and (6, 1).
Possible outcomes for getting the sum 8 are (2, 6); (3, 5); (4, 4); (5, 3) and (6, 2).
Possible outcomes for getting the sum 9 are (3, 6); (4, 5); (5, 4) and (6, 3).
Possible outcomes for getting the sum 10 are (4, 6); (5, 5) and (6, 4).
Possible outcomes for getting the sum 11 are (5, 6) and (6, 5).
Possible outcomes for getting the sum 12 are (6, 6).

Now the complete table is as follows:

(ii) No. The eleven sums are not equally likely. For example, the sum 2 can happen in only one way (1, 1), while the sum 7 can happen in six different ways. So they cannot all have the probability 1/11.

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:
Possible outcomes are
{HHH, TTT, HTT, THH, HTH, THT, HHT, TTH}
Number of possible outcomes = 8
Number of favourable outcomes = 2 (HHH and TTT)

P(Hanif will win the game)
= Number of favourable outcomes / Total number of outcomes
= 2/8
= 1/4

∴ P(Hanif will lose the game)
= 1 − P(Hanif will win 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 × 6 = 36

(i) Possible outcomes when 5 comes up on either throw are
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)
Number of favourable outcomes = 11

P(5 will come up either time)
= Number of favourable outcomes / Total number of outcomes
= 11/36

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

(ii) Number of cases when 5 will come up at least once = 11
P(5 will come up at least once)
= Number of favourable outcomes / Total number of outcomes
= 11/36

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) The given statement is incorrect. There are 4 possible outcomes, which are (H, H), (T, T), (H, T) and (T, H).

P(getting two heads)
= Number of favourable outcomes / Total number of outcomes
= 1/4

P(getting two tails)
= 1/4

P(getting one head and one tail)
= 2/4
= 1/2

Since the three outcomes do not have the same probability, the argument is not correct.

(ii) Correct, because the two outcomes are equally likely. There are 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6) on a die, so P(odd number) = 3/6 = 1/2.

Related Links

• NCERT Solutions Class 10 Maths Chapter 14 – Probability