Class 10 Mathematics
Chapter 14: Probability
Exercise 14.1 – Stepwise Solutions
1. Complete the following statements
(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 sure (or 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.
2. Which of the following experiments have equally likely outcomes?
(i) A driver attempts to start a car. The car starts or does not start.
These outcomes are not equally likely, because the car may be in good condition and is more likely to start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
These outcomes are not equally likely, because success depends on the player’s skill.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
These outcomes are equally likely if the answer is guessed randomly, because there are only two possible outcomes.
(iv) A baby is born. It is a boy or a girl.
These outcomes are considered equally likely.
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?
A coin has two outcomes: head and tail.
Both outcomes are equally likely.
So, each team gets an equal chance.
Therefore, tossing a coin is a fair method.
4. Which of the following cannot be the probability of an event?
Options are:
(A) 2/3 (B) −1.5 (C) 15
We know probability always lies between 0 and 1.
2/3 = 0.666… → possible
−1.5 → negative, not possible
15
0.7 → possible
Answer: (B) −1.5
5. If P(E) = 0.05, what is the probability of ‘not E’?
We know:
P(not E) = 1 − P(E)
= 1 − 0.05
= 0.95
Answer: 0.95
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?
Since there are no orange candies in the bag,
Probability = 0
(ii) a lemon flavoured candy?
All candies are lemon flavoured, so
Probability = 1
7. 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?
We know:
P(same birthday) = 1 − P(not same birthday)
= 1 − 0.992
= 0.008
Answer: 0.008
8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random. Find the probability that the ball drawn is:
Total balls = 3 + 5 = 8
(i) red
Number of red balls = 3
Probability = 3/8
(ii) not red
Not red means black.
Number of black balls = 5
Probability = 5/8
9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out at random. Find the probability that the marble will be:
Total marbles = 5 + 8 + 4 = 17
(i) red
Probability = 5/17
(ii) white
Probability = 8/17
(iii) not green
Not green means red or white.
Number of not green marbles = 5 + 8 = 13
Probability = 13/17
Final Answers Summary
- 1(i) 1
- 1(ii) 0, impossible event
- 1(iii) 1, sure event
- 1(iv) 1
- 1(v) 0 and 1
- 2(i) Not equally likely
- 2(ii) Not equally likely
- 2(iii) Equally likely
- 2(iv) Equally likely
- 3) Because both outcomes are equally likely
- 4) −1.5
- 5) 0.95
- 6(i) 0, 6(ii) 1
- 7) 0.008
- 8(i) 3/8, 8(ii) 5/8
- 9(i) 5/17, 9(ii) 8/17, 9(iii) 13/17
Class 10 Mathematics
Chapter 14: Probability
Exercise 14.1 – Stepwise Solutions (Q10 to Q18)
10. Piggy Bank Coins
Given:
50p coins = 100
₹1 coins = 50
₹2 coins = 20
₹5 coins = 10
Total coins = 100 + 50 + 20 + 10 = 180
(i) Probability that the coin is a 50p coin
P(50p coin) = Number of 50p coins / Total coins
= 100/180
= 5/9
Answer: 5/9
(ii) Probability that the coin will not be a ₹5 coin
Number of coins that are not ₹5 coins = 180 − 10 = 170
P(not a ₹5 coin) = 170/180
= 17/18
Answer: 17/18
11. Fish Tank Problem
Given:
Male fish = 5
Female fish = 8
Total fish = 5 + 8 = 13
Probability that the fish taken out is male
P(male fish) = 5/13
Answer: 5/13
12. Spinner Game
Numbers on spinner = 1, 2, 3, 4, 5, 6, 7, 8
Total equally likely outcomes = 8
(i) Probability that it points at 8
Favourable outcome = {8}
P(8) = 1/8
(ii) Probability that it points at an odd number
Odd numbers = {1, 3, 5, 7}
Number of favourable outcomes = 4
P(odd number) = 4/8 = 1/2
(iii) Probability that it points at a number greater than 2
Numbers greater than 2 = {3, 4, 5, 6, 7, 8}
Favourable outcomes = 6
P(number greater than 2) = 6/8 = 3/4
(iv) Probability that it points at a number less than 9
All numbers 1 to 8 are less than 9
Favourable outcomes = 8
P(number less than 9) = 8/8 = 1
Answers: (i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1
13. A Die is Thrown Once
Possible outcomes = {1, 2, 3, 4, 5, 6}
Total outcomes = 6
(i) Probability of getting a prime number
Prime numbers on die = {2, 3, 5}
Favourable outcomes = 3
P(prime) = 3/6 = 1/2
(ii) Probability of getting a number lying between 2 and 6
Numbers between 2 and 6 = {3, 4, 5}
Favourable outcomes = 3
P(number between 2 and 6) = 3/6 = 1/2
(iii) Probability of getting an odd number
Odd numbers = {1, 3, 5}
Favourable outcomes = 3
P(odd) = 3/6 = 1/2
Answers: (i) 1/2 (ii) 1/2 (iii) 1/2
14. One Card Drawn from a Deck of 52 Cards
Total cards = 52
(i) Probability of getting a king of red colour
Red kings = King of hearts, King of diamonds
Favourable outcomes = 2
P = 2/52 = 1/26
(ii) Probability of getting a face card
Face cards = Jack, Queen, King in each suit
Total face cards = 3 × 4 = 12
P = 12/52 = 3/13
(iii) Probability of getting a red face card
Red suits = hearts and diamonds
Red face cards = 3 × 2 = 6
P = 6/52 = 3/26
(iv) Probability of getting the jack of hearts
Only one such card
P = 1/52
(v) Probability of getting a spade
Spades = 13
P = 13/52 = 1/4
(vi) Probability of getting the queen of diamonds
Only one such card
P = 1/52
Answers:
(i) 1/26
(ii) 3/13
(iii) 3/26
(iv) 1/52
(v) 1/4
(vi) 1/52
15. Five Cards: 10, J, Q, K, A of Diamonds
Total cards = 5
(i) Probability that the card is the queen
Only one queen
P = 1/5
(ii) If the queen is drawn and put aside
Now remaining cards = 4
(a) Probability that the second card is an ace
Only one ace remains
P = 1/4
(b) Probability that the second card is a queen
Queen has already been removed
P = 0
Answers: (i) 1/5 (ii)(a) 1/4 (ii)(b) 0
16. Defective and Good Pens
Defective pens = 12
Good pens = 132
Total pens = 12 + 132 = 144
Probability that the pen taken out is good
P(good pen) = 132/144
= 11/12
Answer: 11/12
17. Bulbs Problem
Total bulbs = 20
Defective bulbs = 4
Good bulbs = 20 − 4 = 16
(i) Probability that the bulb is defective
P(defective) = 4/20 = 1/5
(ii) First bulb drawn is not defective and is not replaced
So one good bulb is removed.
Remaining bulbs = 19
Remaining good bulbs = 15
P(second bulb is not defective) = 15/19
Answers: (i) 1/5 (ii) 15/19
18. Box of 90 Discs Numbered 1 to 90
Total outcomes = 90
(i) Probability of getting a two-digit number
Two-digit numbers from 10 to 90
Count = 90 − 10 + 1 = 81
P(two-digit) = 81/90 = 9/10
(ii) Probability of getting a perfect square number
Perfect squares from 1 to 90 are:
1, 4, 9, 16, 25, 36, 49, 64, 81
Count = 9
P(perfect square) = 9/90 = 1/10
(iii) Probability of getting a number divisible by 5
Numbers divisible by 5 from 1 to 90:
5, 10, 15, …, 90
Count = 90/5 = 18
P(divisible by 5) = 18/90 = 1/5
Answers: (i) 9/10 (ii) 1/10 (iii) 1/5
Final Answers Summary
- Q10: (i) 5/9 (ii) 17/18
- Q11: 5/13
- Q12: (i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1
- Q13: (i) 1/2 (ii) 1/2 (iii) 1/2
- Q14: (i) 1/26 (ii) 3/13 (iii) 3/26 (iv) 1/52 (v) 1/4 (vi) 1/52
- Q15: (i) 1/5 (ii)(a) 1/4 (ii)(b) 0
- Q16: 11/12
- Q17: (i) 1/5 (ii) 15/19
- Q18: (i) 9/10 (ii) 1/10 (iii) 1/5
Class 10 Mathematics
Chapter 14 – Probability
Exercise 14.1 (Questions 19–24) Solutions
19. Die with letters A, B, C, D, E, A
Total faces on die = 6
Faces = A, B, C, D, E, A
(i) Probability of getting A
Number of A faces = 2
P(A) = 2 / 6
= 1 / 3
Answer: 1/3
(ii) Probability of getting D
Number of D faces = 1
P(D) = 1 / 6
Answer: 1/6
20. Die dropped on rectangular region
Rectangle dimensions = 3 m × 2 m
Area of rectangle = 3 × 2
= 6 m²
Diameter of circle = 1 m
Radius r = 0.5 m
Area of circle = πr²
= π(0.5)²
= π × 0.25
= 0.25π
Probability = Area of circle / Area of rectangle
= (0.25π) / 6
= π / 24
If π = 22/7
Probability ≈ 0.131
Answer: π/24
21. Ball Pens Problem
Total pens = 144
Defective pens = 20
Good pens = 144 − 20
= 124
(i) Probability that she will buy it
She buys only good pens.
P(good pen) = 124 / 144
= 31 / 36
Answer: 31/36
(ii) Probability that she will not buy it
P(defective pen) = 20 / 144
= 5 / 36
Answer: 5/36
22. Sum on two dice
Total outcomes when two dice are thrown = 36
| Sum | Probability |
|---|---|
| 2 | 1/36 |
| 3 | 2/36 |
| 4 | 3/36 |
| 5 | 4/36 |
| 6 | 5/36 |
| 7 | 6/36 |
| 8 | 5/36 |
| 9 | 4/36 |
| 10 | 3/36 |
| 11 | 2/36 |
| 12 | 1/36 |
(ii) Student’s argument
The student says each sum has probability 1/11.
This is incorrect.
Because sums do not occur equally likely.
Example:
Sum 7 has 6 outcomes while sum 2 has only 1 outcome.
Therefore the argument is wrong.
23. Tossing a coin 3 times
Total outcomes = 2³ = 8
Possible outcomes:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Hanif wins if all results are same.
Winning outcomes:
HHH, TTT
Total winning outcomes = 2
P(win) = 2 / 8
= 1 / 4
P(lose) = 1 − 1/4
= 3/4
Answer: 3/4
24. A die is thrown twice
Total outcomes = 6 × 6 = 36
(i) Probability that 5 will not come either time
Probability of not getting 5 in one throw = 5/6
For two throws:
P(no 5 both times) = (5/6)²
= 25 / 36
Answer: 25/36
(ii) Probability that 5 will come at least once
P(at least one 5)
= 1 − P(no 5)
= 1 − 25/36
= 11/36
Answer: 11/36
Final Answers Summary
- Q19 → (i) 1/3 (ii) 1/6
- Q20 → π/24
- Q21 → (i) 31/36 (ii) 5/36
- Q22 → Probabilities table above
- Q23 → 3/4
- Q24 → (i) 25/36 (ii) 11/36
Exercise 14.1 – Question 25
Question: Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) Argument:
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.
Explanation:
When two coins are tossed, the actual possible outcomes are:
S = {HH, HT, TH, TT}
Total outcomes = 4
These outcomes are equally likely.
The event “one head and one tail” includes two outcomes:
{HT, TH}
So the probabilities are:
- P(HH) = 1/4
- P(TT) = 1/4
- P(one head and one tail) = 2/4 = 1/2
Thus the three outcomes mentioned in the argument are not equally likely.
Conclusion: The argument is incorrect.
(ii) Argument:
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.
Explanation:
The possible outcomes when a die is thrown are:
S = {1, 2, 3, 4, 5, 6}
Odd numbers = {1, 3, 5}
Even numbers = {2, 4, 6}
Total outcomes = 6
Favourable outcomes for odd numbers = 3
P(odd number) = 3/6 = 1/2
Similarly,
P(even number) = 3/6 = 1/2
Conclusion: The argument is correct.
Final Answer
- (i) Incorrect argument
- (ii) Correct argument