Important Questions for Class 10 Maths Chapter 15 Probability
Updated on June 15, 2025 | By Learnzy Academy
Q1. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?
Total number of outcomes = Prizes + Blanks = 10 + 25 = 35
Number of favorable outcomes (getting a prize) = 10
So, the probability of getting a prize =
Number of prizes / Total number of outcomes
= 10 / 35
= 2 / 7
Answer: 2/7
Q2. The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times. (i) What is the probability that on a given day it was correct? (ii) What is the probability that it was not correct on a given day?
Total number of days = 250
Number of days forecast was correct = 175
Number of days forecast was not correct = 250 − 175 = 75
(i) Probability that the forecast was correct:
= 175 / 250
= 7 / 10
(ii) Probability that the forecast was not correct:
= 75 / 250
= 3 / 10
Q3. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.
When two dice are thrown at the same time, the total number of possible outcomes is:
6 × 6 = 36
We need to find the number of outcomes where the product of the two numbers is less than 9.
Let’s list all such outcomes where product is less than 9:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (4,1), (4,2), (5,1), (6,1)
There are 16 such outcomes where the product is less than 9.
So, the required probability = 16 / 36 = 4 / 9
Hence answer is 4/9
Q4. 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.
We are given two dice:
Die A has numbers: 1, 2, 3, 4, 5, 6
Die B has numbers: 1, 1, 2, 2, 3, 3
Total possible outcomes = 6 × 6 = 36
Now we will count how many times each sum from 2 to 9 appears.
Sum = 2
(1,1), (1,1) → 2 outcomes
Probability = 2 / 36 = 1 / 18
Sum = 3
(1,2), (1,2), (2,1), (2,1) → 4 outcomes
Probability = 4 / 36 = 1 / 9
Sum = 4
(1,3), (1,3), (2,2), (2,2), (3,1), (3,1) → 6 outcomes
Probability = 6 / 36 = 1 / 6
Sum = 5
(2,3), (2,3), (3,2), (3,2), (4,1), (4,1) → 6 outcomes
Probability = 6 / 36 = 1 / 6
Sum = 6
(3,3), (3,3), (4,2), (4,2), (5,1), (5,1) → 6 outcomes
Probability = 6 / 36 = 1 / 6
Sum = 7
(4,3), (4,3), (5,2), (5,2), (6,1), (6,1) → 6 outcomes
Probability = 6 / 36 = 1 / 6
Sum = 8
(5,3), (5,3), (6,2), (6,2) → 4 outcomes
Probability = 4 / 36 = 1 / 9
Sum = 9
(6,3), (6,3) → 2 outcomes
Probability = 2 / 36 = 1 / 18
Q5. 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
(i) A driver attempts to start a car. The car starts or does not start.
Not equally likely. The chances depend on the condition of the car. If the car is in good condition, it is more likely to start than not.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Not equally likely.The outcome depends on the player's skill. A skilled player is more likely to score than miss.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Equally likely. If the person is guessing, both outcomes have an equal chance of occurring.
(iv) A baby is born. It is a boy or a girl.
Equally likely. The chances of a baby being a boy or a girl are almost equal, so we consider these outcomes equally likely
Q6. Two dice are thrown at the same time. Determine the probabiity that the difference of the numbers on the two dice is 2.
When two dice are thrown at the same time, the total number of possible outcomes is:
6 × 6 = 36
We are asked to find the probability that the difference of the numbers on the two dice is 2.
Let’s list all such pairs:
(1, 3) → difference = 2
(2, 4) → difference = 2
(3, 5) → difference = 2
(4, 6) → difference = 2
(3, 1) → difference = 2
(4, 2) → difference = 2
(5, 3) → difference = 2
(6, 4) → difference = 2
There are 8 favorable outcomes.
So, the required probability = 8 / 36 = 2 / 9
Q7. 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?
A bag contains 3 red balls and 5 black balls.
Total number of balls = 3 + 5 = 8
Formula:
Probability of an event = Number of favourable outcomes / Total number of outcomes
(i) Probability that the ball drawn is red:
Number of red balls = 3
P(red) = 3 / 8
(ii) Probability that the ball drawn is not red:
Number of balls that are not red (i.e., black balls) = 5
P(not red) = 5 / 8
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 taken out is a good one.
Total number of pens = 12 defective + 132 good = 144
Number of good pens = 132
Probability of taking out a good pen = Number of good pens / Total number of pens
= 132 / 144
=> 11 / 12
Hence answer is : 11/12
Q9. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is (i) 6 (ii) 12 (iii) 7
When two dice are thrown together, there are a total of 36 possible outcomes.
Now let's find the required probabilities.
(i) Product = 6
The pairs of numbers whose product is 6 are:
(1, 6), (2, 3), (3, 2), (6, 1)
So, there are 4 favorable outcomes.
Probability = 4 / 36 = 1 / 9
(ii) Product = 12
The pairs of numbers whose product is 12 are:
(2, 6), (3, 4), (4, 3), (6, 2)
So, there are 4 favorable outcomes.
Probability = 4 / 36 = 1 / 9
(iii) Product = 7
There are no pairs of numbers between 1 and 6 whose product is 7.
So, there are 0 favorable outcomes.
Probability = 0 / 36 = 0
Q10. What is the probability that an ordinary year has 53 Sundays?
An ordinary year has 365 days.
365 ÷ 7 = 52 weeks and 1 extra day
So, an ordinary year always has 52 Sundays.
The extra day can be either Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or Saturday — 7 possibilities in total.
Only 1 of these is a Sunday.
Therefore, the probability that an ordinary year has 53 Sundays is: 1/7
Q11. Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken out at random from the box. Find the probability that the drawn bulb is not defective.
Total number of bulbs = 400
Number of defective bulbs = 15
So, number of non-defective bulbs = 400 − 15 = 385
The probability that the drawn bulb is not defective =
Number of non-defective bulbs / Total number of bulbs
= 385 / 400
= 77 / 80
Answer: 77/80
Q12. 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 50p coin? (ii) will not be a ₹5 coin?
Total number of coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180 coins
Formula:
Probability of an event = (Number of favourable outcomes) / (Total number of outcomes)
(i) Probability that the coin will be a 50p coin:
Number of 50p coins = 100
P(50p coin) = 100 / 180 = 5/9
(ii) Probability that the coin will not be a ₹5 coin:
Number of coins that are not ₹5 = 100 + 50 + 20 = 170
P(not ₹5 coin) = 170 / 180 = 17/18
Q13. A coin is tossed two times. Find the probability of getting at most one head.
When a coin is tossed two times, the total number of possible outcomes is:
HH, HT, TH, TT
So, total outcomes = 4
We are asked to find the probability of getting at most one head, which means either no head or only one head.
The outcomes with at most one head are:
TT → 0 heads
HT → 1 head
TH → 1 head
So, favorable outcomes = 3
Therefore, the required probability is = 3 / 4
Q14. 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?
The box contains:
Red marbles = 5
White marbles = 8
Green marbles = 4
So total marbles = 5 + 8 + 4 = 17
Formula:
Probability of an event = Number of favourable outcomes / Total number of outcomes
(i) Probability that the marble is red:
Number of red marbles = 5
P(red) = 5 / 17
(ii) Probability that the marble is white:
Number of white marbles = 8
P(white) = 8 / 17
(iii) Probability that the marble is not green:
Number of marbles that are not green = 5 (red) + 8 (white) = 13
P(not green) = 13 / 17
Q15. 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?
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 + 8 = 13
Probability of an event = (Number of favourable outcomes) / (Total number of outcomes)
Probability that the fish is male = 5 / 13
Q16. 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
(i) Probability of getting a king of red colour:
There are 2 red kings — King of hearts and King of diamonds.
Favorable outcomes = 2
So,
Probability = 2/52 = 1/26
(ii) Probability of getting a face card:
Face cards = Jacks, Queens, Kings
Each suit has 3 face cards
Total face cards = 4 suits × 3 = 12
So,
Probability = 12/52 = 3/13
(iii) Probability of getting a red face card:
Red suits = Hearts and Diamonds
Each red suit has 3 face cards
Total red face cards = 2 × 3 = 6
So,
Probability = 6/52 = 3/26
(iv) Probability of getting the jack of hearts:
Favorable outcomes = 1
So,
Probability = 1/52
(v) Probability of getting a spade:
Spade suit has 13 cards
So,
Probability = 13/52 = 1/4
(vi) Probability of getting the queen of diamonds:
Only one card Queen of diamonds
Favorable outcomes = 1
So,
Probability = 1/52
Q17. 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?
(i) What is the probability that the card is the queen?
There is only 1 queen in the 5 cards.
So,
Probability = 1/5
(ii) If the queen is drawn and put aside,
The remaining cards are: 10, J, K, A i.e 4 cards remaining
Let’s now find:
(a) Probability that the second card picked is an ace
Only 1 ace left in 4 cards.
So,
Probability = 1/4
(b) Probability that the second card picked is a queen
Since the queen has already been drawn and put aside, there is no queen left.
So,
Probability = 0
Q18. A coin is tossed 3 times. List the possible outcomes. Find the probability of getting (i) all heads (ii) at least 2 heads
When a coin is tossed 3 times, the possible outcomes are:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
So, total number of outcomes = 8
(i) All heads:
Only one outcome has all heads → HHH
Favorable outcomes = 1
Probability = 1 / 8
(ii) At least 2 heads:
Outcomes with at least 2 heads are → HHH, HHT, HTH, THH
Favorable outcomes = 4
Probability = 4 / 8 = 1 / 2
Q19. 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 _________.
(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.
Q20. 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
A standard die has 6 faces numbered: 1, 2, 3, 4, 5, 6
Total number of outcomes = 6
(i) Probability of getting a prime number:
Prime numbers between 1 and 6 are: 2, 3, 5
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2
(ii) Probability of getting a number lying between 2 and 6:
"Between 2 and 6" means: 3, 4, 5 (not including 2 and 6)
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2
(iii) Probability of getting an odd number:
Odd numbers between 1 and 6 are: 1, 3, 5
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2
Q21. 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?
Tossing a coin is considered a fair way of deciding which team should get the ball at the beginning of a football game because:
- A coin has two sides: heads and tails.
- When tossed properly, each side has an equal chance (50%) of landing face up.
- The outcome is random and unbiased, meaning it does not favor either team.
- Both teams agree to the rules before the toss and have an equal opportunity to call heads or tails.
Therefore, since each team has an equal probability of winning the toss, it is a fair method for making the decision.
Q22. 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 , 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?
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1 to 8. All outcomes are equally likely. So, the total number of outcomes = 8.
(i) Probability that it will point at 8:
Favorable outcome = 1 (only the number 8)
Probability = 1/8
(ii) Probability that it points at an odd number:
Odd numbers = 1, 3, 5, 7
So favorable outcomes is 4
Probability = 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
So favorable outcomes is 6
Probability = 6/8 = 3/4
(iv) Probability that it points at a number less than 9:
All numbers (1 to 8) are less than 9
So favorable outcomes is 8
Probability = 8/8 = 1
Q23. Which of the following cannot be the probability of an event? (A) 2/3 (B) –1.5 (C) 15% (D) 0.7
Answer: (B) –1.5
Explanation:
The probability of any event must be a number between 0 and 1.
–1.5 is less than 0, so it is not a valid probability.
All other options (2/3, 15%, and 0.7) are between 0 and 1, so they are valid probabilities.
Q24. 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?
(i) Probability of taking out an orange flavoured candy:
Since there are no orange flavoured candies in the bag,
So Probability of orange flavoured candy is = 0
(ii) Probability of taking out a lemon flavoured candy:
All candies are lemon flavoured, so the probability is
P(lemon flavoured candy) = 1
Q25. 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?
Probability of an event happening = 1 – Probability of the event not happening
So,
Probability that 2 students have the same birthday = 1 – Probability that they do not have the same birthday
=> 1 – 0.992
=> 0.008
Q26. 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.
A box contains 90 discs numbered from 1 to 90. One disc is drawn at random.
Total number of outcomes = 90
(i) Probability of getting a two-digit number:
Two-digit numbers are from 10 to 90
Total two-digit numbers = 81
Probability = 81 / 90 = 9 / 10
(ii) Probability of getting a perfect square number:
Perfect square numbers between 1 and 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81
Total perfect squares = 9
Probability = 9 / 90 = 1 / 10
(iii) Probability of getting a number divisible by 5:
Numbers divisible by 5 from 1 to 90 are: 5, 10, 15, ..., 90
Total such numbers = 18
Probability = 18 / 90 = 1 / 5
Q27. If P(E) = 0.05, what is the probability of ‘not E’?
If P(E) = 0.05, then the probability of ‘not E’ is:
P(not E) = 1 – P(E)
= 1 – 0.05
= 0.95
Hence ‘not E’ is: 0.95
Q28. (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 ?
(i) A lot of 20 bulbs contains 4 defective bulbs.
So, total bulbs = 20
Number of defective bulbs = 4
Probability that the bulb drawn is defective = 4 / 20 = 1 / 5
(ii) The first bulb drawn is not defective and is not replaced.
That means:
One good bulb is removed.
Remaining bulbs = 19
Remaining good bulbs = 16 − 1 = 15 (since total good bulbs were 16 = 20 − 4)
Probability that the second bulb drawn is not defective = 15 / 19
Q29. 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 ?
Total number of pens = 144
Number of defective pens = 20
Number of good pens = 144 − 20 = 124
(i) Probability that Nuri will buy the pen (i.e., it is good):
Probability = Number of good pens / Total number of pens
=> 124 / 144
=> 31 / 36
(ii) Probability that Nuri will not buy the pen (i.e., it is defective):
Probability = Number of defective pens / Total number of pens
=> 20 / 144
=> 5 / 36
Q30. 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.
Total number of possible outcomes when a coin is tossed 3 times:
Each toss has 2 outcomes: Head (H) or Tail (T)
The outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
So, total outcomes are 8
Favorable outcomes (Hanif wins):
Hanif wins if the result is either HHH or TTT
Number of winning outcomes = 2
Unfavorable outcomes (Hanif loses) = Total outcomes − Winning outcomes
= 8 − 2 = 6
Probability of an event = (Number of favorable outcomes) ÷ (Total number of outcomes)
So, The probability that Hanif loses = 6 ÷ 8 = 3/4
Q31. 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?
Total possible outcomes when a die is thrown twice:
= 6 × 6 = 36
i) 5 will not come up either time
Let’s count how many outcomes include 5:
5 appears in the first throw: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes
5 appears in the second throw: (1,5), (2,5), (3,5), (4,5), (6,5) → 5 outcomes
(We already counted (5,5) once, so don’t count it again.)
So, total outcomes where 5 appears at least once = 6 + 5 = 11
Therefore, outcomes where 5 does not appear at all = 36 − 11 = 25
Answer (i): Probability = 25 / 36
(ii) 5 will come up at least once
From above, favorable outcomes = 11
(Where 5 appears in at least one position)
Answer (ii): Probability = 11 / 36
Q32. 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.
(i) Not correct.
Reason:
When two coins are tossed, the actual outcomes are: HH, HT, TH, TT — a total of 4 outcomes.
Two heads = HH → 1 outcome
Two tails = TT → 1 outcome
One of each = HT or TH → 2 outcomes
So the probability of:
Two heads = 1/4
Two tails = 1/4
One of each = 2/4 = 1/2
Since these grouped outcomes are not equally likely, assigning a probability of 1/3 to each is incorrect.
(ii) Correct.
Reason:
When a die is thrown, possible outcomes are: 1, 2, 3, 4, 5, 6
Odd numbers = 1, 3, 5 → 3 outcomes
Even numbers = 2, 4, 6 → 3 outcomes
So, the probability of getting an odd number = 3/6 = 1/2.
This reasoning is correct because the outcomes are equally likely.