CBSE Class 10 · Maths
CBSE Class 10 Maths Probability Previous Year Questions
Help your child score full marks in CBSE Class 10 Maths Probability with this curated collection of previous year board exam questions spanning 2016–2025. Every question comes with a detailed step-by-step solution — covering cards, dice, coins, and real-life experiments — so your child builds the confidence to tackle any probability problem on exam day.
CBSE Class 10 Maths Probability — Questions with Solutions
A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the card drawn is not an ace is:
Solution
Answer: Option (D) is correct.
Explanation: Number of ace cards in a pack of 52 cards $= 4$.
Therefore, number of non-ace cards $= 52 – 4 = 48$.
$$\text{Required probability} = \frac{48}{52} = \frac{12}{13}$$
Explanation: Number of ace cards in a pack of 52 cards $= 4$.
Therefore, number of non-ace cards $= 52 – 4 = 48$.
$$\text{Required probability} = \frac{48}{52} = \frac{12}{13}$$
Two dice are thrown together. The probability of getting the difference of numbers on their upper faces equals to 3 is:
Solution
Answer: Option (C) is correct.
Explanation: When two dice are thrown, the total number of outcomes $= 36$.
Pairs where $|a – b| = 3$: $(4,1),\ (5,2),\ (6,3),\ (1,4),\ (2,5),\ (3,6)$ — that is 6 favourable outcomes.
$$\text{Probability} = \frac{6}{36} = \frac{1}{6}$$
Explanation: When two dice are thrown, the total number of outcomes $= 36$.
Pairs where $|a – b| = 3$: $(4,1),\ (5,2),\ (6,3),\ (1,4),\ (2,5),\ (3,6)$ — that is 6 favourable outcomes.
$$\text{Probability} = \frac{6}{36} = \frac{1}{6}$$
A bag contains 5 red balls and $n$ green balls. If the probability of drawing a green ball is three times that of a red ball, then the value of $n$ is:
Solution
Answer: Option (B) is correct.
Explanation: Total balls $= 5 + n$.
$P(R) = \dfrac{5}{5+n}$, $\quad P(G) = \dfrac{n}{5+n}$
Given $P(G) = 3 \cdot P(R)$: $$\frac{n}{5+n} = 3 \times \frac{5}{5+n} \Rightarrow n = 15$$
Explanation: Total balls $= 5 + n$.
$P(R) = \dfrac{5}{5+n}$, $\quad P(G) = \dfrac{n}{5+n}$
Given $P(G) = 3 \cdot P(R)$: $$\frac{n}{5+n} = 3 \times \frac{5}{5+n} \Rightarrow n = 15$$
A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball?
Solution
Answer: Option (C) is correct.
Explanation: Total balls $= 5 + 8 + 7 = 20$.
Getting neither blue nor pink means getting a yellow ball. There are 7 yellow balls.
$$\text{Probability} = \frac{7}{20}$$
Explanation: Total balls $= 5 + 8 + 7 = 20$.
Getting neither blue nor pink means getting a yellow ball. There are 7 yellow balls.
$$\text{Probability} = \frac{7}{20}$$
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability of getting a face card is:
Solution
Answer: Option (B) is correct.
Explanation: Total number of cards $= 52$. Total number of face cards (Jack, Queen, King across 4 suits) $= 12$.
$$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$$
Explanation: Total number of cards $= 52$. Total number of face cards (Jack, Queen, King across 4 suits) $= 12$.
$$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$$
A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 23 is:
Solution
Answer: Option (C) is correct.
Explanation: Prime numbers less than 23 are: 2, 3, 5, 7, 11, 13, 17, 19 — that is 8 primes.
Total number of discs $= 90$.
$$P = \frac{8}{90} = \frac{4}{45}$$
Explanation: Prime numbers less than 23 are: 2, 3, 5, 7, 11, 13, 17, 19 — that is 8 primes.
Total number of discs $= 90$.
$$P = \frac{8}{90} = \frac{4}{45}$$
Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out of the bag at random. What is the probability that the number on the card taken out is an even number?
Solution
Answer: Option (B) is correct.
Explanation: Total cards (3 to 20) $= 20 – 3 + 1 = 18$.
Even numbers from 3 to 20: 4, 6, 8, 10, 12, 14, 16, 18, 20 — that is 9 even numbers.
$$\text{Probability} = \frac{9}{18} = \frac{1}{2}$$
Explanation: Total cards (3 to 20) $= 20 – 3 + 1 = 18$.
Even numbers from 3 to 20: 4, 6, 8, 10, 12, 14, 16, 18, 20 — that is 9 even numbers.
$$\text{Probability} = \frac{9}{18} = \frac{1}{2}$$
2 cards of hearts and 4 cards of spades are missing from a pack of 52 cards. What is the probability of getting a black card from the remaining pack?
Solution
Answer: Option (B) is correct.
Explanation: Remaining total cards $= 52 – 6 = 46$.
Total black cards $= 26$. Remaining black cards (spades removed) $= 26 – 4 = 22$.
$$P(\text{black card}) = \frac{22}{46}$$
Explanation: Remaining total cards $= 52 – 6 = 46$.
Total black cards $= 26$. Remaining black cards (spades removed) $= 26 – 4 = 22$.
$$P(\text{black card}) = \frac{22}{46}$$
Ginny flipped a fair coin three times and tails came up each time. Ginny wants to flip the coin again. What is the probability of getting heads in the next coin flip?
Solution
Answer: Option (C) is correct.
Explanation: Each coin flip is an independent event — past outcomes do not affect future flips. Total outcomes $= 2$ (Head or Tail). Favourable outcomes $= 1$ (Head).
$$P(\text{Head}) = \frac{1}{2} = 0.5$$
Explanation: Each coin flip is an independent event — past outcomes do not affect future flips. Total outcomes $= 2$ (Head or Tail). Favourable outcomes $= 1$ (Head).
$$P(\text{Head}) = \frac{1}{2} = 0.5$$
For an event $E$, $P(E) + P(\bar{E}) = x$, then the value of $x^3 – 3$ is
Solution
Answer: Option (A) is correct.
Explanation: By the law of probability, $P(E) + P(\bar{E}) = 1$, so $x = 1$.
$$x^3 – 3 = (1)^3 – 3 = 1 – 3 = -2$$
Explanation: By the law of probability, $P(E) + P(\bar{E}) = 1$, so $x = 1$.
$$x^3 – 3 = (1)^3 – 3 = 1 – 3 = -2$$
In a random experiment of throwing a die, which of the following is a sure event?
Solution
Answer: Option (D) is correct.
Explanation: The outcomes of a die are 1, 2, 3, 4, 5, 6 — all natural numbers less than 7. Since every possible outcome satisfies this condition, it is a sure event (probability = 1). Options (A), (B), and (C) do not cover all six outcomes.
Explanation: The outcomes of a die are 1, 2, 3, 4, 5, 6 — all natural numbers less than 7. Since every possible outcome satisfies this condition, it is a sure event (probability = 1). Options (A), (B), and (C) do not cover all six outcomes.
A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is:
Solution
Answer: Option (D) is correct.
Explanation: Red suits are hearts and diamonds. Each has 3 face cards (Jack, Queen, King), giving $3 + 3 = 6$ red face cards in total.
$$P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$$
Explanation: Red suits are hearts and diamonds. Each has 3 face cards (Jack, Queen, King), giving $3 + 3 = 6$ red face cards in total.
$$P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$$
Three coins are tossed together. The probability of getting exactly one tail is:
Solution
Answer: Option (D) is correct.
Explanation: Total outcomes when tossing 3 coins $= 2^3 = 8$.
Outcomes with exactly one tail: HHT, HTH, THH — that is 3 favourable outcomes.
$$P(\text{exactly one tail}) = \frac{3}{8}$$
Explanation: Total outcomes when tossing 3 coins $= 2^3 = 8$.
Outcomes with exactly one tail: HHT, HTH, THH — that is 3 favourable outcomes.
$$P(\text{exactly one tail}) = \frac{3}{8}$$
On a throw of a die, if getting 6 is considered success then probability of losing the game is:
Solution
Answer: Option (D) is correct.
Explanation: $P(\text{success}) = P(6) = \dfrac{1}{6}$. Losing means rolling any number other than 6 — there are 5 such outcomes.
$$P(\text{losing}) = \frac{5}{6}$$
Explanation: $P(\text{success}) = P(6) = \dfrac{1}{6}$. Losing means rolling any number other than 6 — there are 5 such outcomes.
$$P(\text{losing}) = \frac{5}{6}$$
If the probability of a player winning a game is 0.79, then the probability of his losing the same game is:
Solution
Answer: Option (D) is correct.
Explanation: The sum of probabilities of all outcomes of an event equals 1.
$$P(\text{losing}) = 1 – P(\text{winning}) = 1 – 0.79 = 0.21$$
Explanation: The sum of probabilities of all outcomes of an event equals 1.
$$P(\text{losing}) = 1 – P(\text{winning}) = 1 – 0.79 = 0.21$$
In a medical centre, 780 randomly selected people were observed to find if there is a relationship between age and the likelihood of getting a heart attack. The following results were observed.
(i) Based on this table, what is the probability that a randomly chosen person from the same sample is younger than or equal to 55 years and has had a heart attack?
(ii) Looking at the data in the table, Giri says “If a person is randomly chosen, then the probability that the person have had a heart attack is about $12.5\%$.” Is the statement true or false? Justify your reason.
| Younger than or equal to 55 | Older than 55 | Total | |
|---|---|---|---|
| Number of persons who have had a heart attack | 29 | 75 | 104 |
| Number of persons who have never had a heart attack | 401 | 275 | 676 |
| Total | 430 | 350 | 780 |
(i) Based on this table, what is the probability that a randomly chosen person from the same sample is younger than or equal to 55 years and has had a heart attack?
(ii) Looking at the data in the table, Giri says “If a person is randomly chosen, then the probability that the person have had a heart attack is about $12.5\%$.” Is the statement true or false? Justify your reason.
Answer
(i) Number of persons who are younger than or equal to 55 years and had a heart attack $= 29$.
Total number of people observed $= 780$.
$$\therefore \quad \text{Probability} = \frac{29}{780}$$
(ii) The statement is false.
Total number of persons who have had a heart attack $= 104$.
$$\text{Probability} = \frac{104}{780} \approx 13.33\%$$ Since $13.33\% \neq 12.5\%$, Giri’s statement is incorrect.
Total number of people observed $= 780$.
$$\therefore \quad \text{Probability} = \frac{29}{780}$$
(ii) The statement is false.
Total number of persons who have had a heart attack $= 104$.
$$\text{Probability} = \frac{104}{780} \approx 13.33\%$$ Since $13.33\% \neq 12.5\%$, Giri’s statement is incorrect.
Shown below is a square dart board with circular rings inside.
(Note: The figure is not to scale)
Find the probability that a dart thrown at random lands on the shaded area. Show your steps.
(Note: The figure is not to scale)
Find the probability that a dart thrown at random lands on the shaded area. Show your steps.
Answer
Area of the square dartboard $= 4^2 = 16$ sq. units.
Area of circular ring (radius 2 units) $= \pi(2)^2 = 4\pi$ sq. units.
Probability that dart lands on shaded area: $$= \frac{16 – 4\pi}{16} = \frac{4(4-\pi)}{16} = \frac{4-\pi}{4}$$
Area of circular ring (radius 2 units) $= \pi(2)^2 = 4\pi$ sq. units.
Probability that dart lands on shaded area: $$= \frac{16 – 4\pi}{16} = \frac{4(4-\pi)}{16} = \frac{4-\pi}{4}$$
Shown below are two baskets with grey and black balls.
Abhishek is playing a game with his friend where he has to close his eyes and pick a black ball from one of the baskets in one trial. He said “I will try with basket 2 as it has a higher number of black balls than basket 1 and hence the probability of picking a black ball from basket 2 is higher.” Is Abhishek’s statement correct? Justify your answer.
Abhishek is playing a game with his friend where he has to close his eyes and pick a black ball from one of the baskets in one trial. He said “I will try with basket 2 as it has a higher number of black balls than basket 1 and hence the probability of picking a black ball from basket 2 is higher.” Is Abhishek’s statement correct? Justify your answer.
Answer
No, Abhishek’s statement is not correct.
Justification:
Probability of picking a black ball from Basket 1 $= \dfrac{4}{7}$
Probability of picking a black ball from Basket 2 $= \dfrac{8}{14} = \dfrac{4}{7}$
Both baskets have the same probability of picking a black ball. The higher number of black balls in Basket 2 is offset by the higher total number of balls.
Justification:
Probability of picking a black ball from Basket 1 $= \dfrac{4}{7}$
Probability of picking a black ball from Basket 2 $= \dfrac{8}{14} = \dfrac{4}{7}$
Both baskets have the same probability of picking a black ball. The higher number of black balls in Basket 2 is offset by the higher total number of balls.
On a particular day, Vidhi and Unnati couldn’t decide on who would get to drive the car. They had one coin each and flipped their coins exactly three times. The following was agreed upon:
• If Vidhi gets two heads in a row, she would drive the car.
• If Unnati gets a head immediately followed by a tail, she would drive the car.
Who has more probability to drive the car that day? List all outcomes and show your steps.
• If Vidhi gets two heads in a row, she would drive the car.
• If Unnati gets a head immediately followed by a tail, she would drive the car.
Who has more probability to drive the car that day? List all outcomes and show your steps.
Answer
All possible outcomes: {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}.
Total number of possible outcomes $= 8$.
Favourable outcomes for Vidhi (two heads in a row): HHH, HHT, THH — that is 3 outcomes.
$$P(\text{Vidhi drives}) = \frac{3}{8}$$
Favourable outcomes for Unnati (head immediately followed by tail: HT appears): HHT, HTH, THT, HTT — that is 4 outcomes.
$$P(\text{Unnati drives}) = \frac{4}{8} = \frac{1}{2}$$
Since $\dfrac{1}{2} > \dfrac{3}{8}$, Unnati is more likely to drive the car that day.
Total number of possible outcomes $= 8$.
Favourable outcomes for Vidhi (two heads in a row): HHH, HHT, THH — that is 3 outcomes.
$$P(\text{Vidhi drives}) = \frac{3}{8}$$
Favourable outcomes for Unnati (head immediately followed by tail: HT appears): HHT, HTH, THT, HTT — that is 4 outcomes.
$$P(\text{Unnati drives}) = \frac{4}{8} = \frac{1}{2}$$
Since $\dfrac{1}{2} > \dfrac{3}{8}$, Unnati is more likely to drive the car that day.
Shown below is a game with 3 spinners attached to 3 rooms. Each player is given a treasure which has to be kept in one of the rooms. All the spinners are rotated and if any spinner lands on the grey region, that room is opened. If the player kept his treasure in that room, he loses the treasure. If he kept it in any other room, he wins the treasure.
Zaira kept her treasure in the room which is least likely to be opened. In which room did she keep her treasure? Show your work.
Zaira kept her treasure in the room which is least likely to be opened. In which room did she keep her treasure? Show your work.
Answer
Probability of opening Room 1 $= \dfrac{3}{6} = \dfrac{1}{2}$
Probability of opening Room 2 $= \dfrac{4}{16} = \dfrac{1}{4}$
(The white region has 8 parts, so total regions $= 16$)
Probability of opening Room 3 $= \dfrac{4}{8} = \dfrac{1}{2}$
Since Room 2 has the lowest probability of being opened $\left(\dfrac{1}{4}\right)$, Zaira kept her treasure in Room 2.
Probability of opening Room 2 $= \dfrac{4}{16} = \dfrac{1}{4}$
(The white region has 8 parts, so total regions $= 16$)
Probability of opening Room 3 $= \dfrac{4}{8} = \dfrac{1}{2}$
Since Room 2 has the lowest probability of being opened $\left(\dfrac{1}{4}\right)$, Zaira kept her treasure in Room 2.
Farah and Sheena are playing a game with number tokens. Each of them has four number tokens, 2, 3, 4 and 5. A token is randomly picked by each of them from their stack simultaneously. If the sum of the numbers picked by each of them is a prime number, Farah wins the game and if it is a composite number, then Sheena wins the game.
Find the probability of each of them winning the game and state who has a higher probability of winning the game. Show your work.
Find the probability of each of them winning the game and state who has a higher probability of winning the game. Show your work.
Answer
Total number of outcomes $= 4 \times 4 = 16$.
Favourable outcomes for Farah (sum is prime): (2,3), (3,2), (2,5), (5,2), (4,3), (3,4) — that is 6 outcomes.
$$P(\text{Farah wins}) = \frac{6}{16} = \frac{3}{8}$$
Probability that Sheena wins: $$P(\text{Sheena wins}) = 1 – \frac{3}{8} = \frac{5}{8}$$
Since $\dfrac{5}{8} > \dfrac{3}{8}$, Sheena has a higher probability of winning the game.
Favourable outcomes for Farah (sum is prime): (2,3), (3,2), (2,5), (5,2), (4,3), (3,4) — that is 6 outcomes.
$$P(\text{Farah wins}) = \frac{6}{16} = \frac{3}{8}$$
Probability that Sheena wins: $$P(\text{Sheena wins}) = 1 – \frac{3}{8} = \frac{5}{8}$$
Since $\dfrac{5}{8} > \dfrac{3}{8}$, Sheena has a higher probability of winning the game.
Naima is playing a game and has two identical 6-sided dice. The faces of the dice have 3 even numbers and 3 odd numbers. She has to roll the two dice simultaneously and has two options to choose from before rolling the dice. She wins a prize if:
Option 1: The sum of the two numbers appearing on the top of the two dice is odd.
Option 2: The product of the two numbers appearing on top of the two dice is odd.
Which option should Naima choose so that her chances of winning a prize is higher? Show your work.
Option 1: The sum of the two numbers appearing on the top of the two dice is odd.
Option 2: The product of the two numbers appearing on top of the two dice is odd.
Which option should Naima choose so that her chances of winning a prize is higher? Show your work.
Answer
Option 1 (sum is odd):
Total possible parity combinations $= 4$: {even+even, even+odd, odd+even, odd+odd}.
Sum is odd only when: even + odd = odd, or odd + even = odd → 2 favourable outcomes.
$$P(\text{Option 1}) = \frac{2}{4} = \frac{1}{2}$$
Option 2 (product is odd):
Product is odd only when: odd × odd = odd → 1 favourable outcome.
$$P(\text{Option 2}) = \frac{1}{4}$$
Since $\dfrac{1}{2} > \dfrac{1}{4}$, Naima should choose Option 1 for a higher chance of winning.
Total possible parity combinations $= 4$: {even+even, even+odd, odd+even, odd+odd}.
Sum is odd only when: even + odd = odd, or odd + even = odd → 2 favourable outcomes.
$$P(\text{Option 1}) = \frac{2}{4} = \frac{1}{2}$$
Option 2 (product is odd):
Product is odd only when: odd × odd = odd → 1 favourable outcome.
$$P(\text{Option 2}) = \frac{1}{4}$$
Since $\dfrac{1}{2} > \dfrac{1}{4}$, Naima should choose Option 1 for a higher chance of winning.
An integer is chosen between 70 and 100. Find the probability that it is:
(i) a prime number
(ii) divisible by 7.
(i) a prime number
(ii) divisible by 7.
Answer
Total integers between 70 and 100 (exclusive): 71, 72, …, 99 → $n(S) = 29$.
(i) Prime numbers between 70 and 100: 71, 73, 79, 83, 89, 97 → 6 primes.
$$P(\text{prime}) = \frac{6}{29}$$
(ii) Numbers divisible by 7 between 70 and 100: 77, 84, 91, 98 → 4 numbers.
$$P(\text{divisible by 7}) = \frac{4}{29}$$
(i) Prime numbers between 70 and 100: 71, 73, 79, 83, 89, 97 → 6 primes.
$$P(\text{prime}) = \frac{6}{29}$$
(ii) Numbers divisible by 7 between 70 and 100: 77, 84, 91, 98 → 4 numbers.
$$P(\text{divisible by 7}) = \frac{4}{29}$$
Read the following passage and answer the questions given at the end:
Diwali Fair
A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. Prizes are given when a black marble is picked. Shweta plays the game once.
(i) What is the probability that she will be allowed to pick a marble from the bag?
(ii) Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains 20 marbles out of which 6 are black?
Diwali Fair
A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. Prizes are given when a black marble is picked. Shweta plays the game once.
(i) What is the probability that she will be allowed to pick a marble from the bag?
(ii) Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains 20 marbles out of which 6 are black?
Answer
(i) She is allowed to pick a marble only if the spinner lands on an even number. Based on the spinner shown, the probability depends on the proportion of even-numbered sections. (Refer to the figure for exact sections; the probability equals the fraction of even-numbered sections on the spinner.)
(ii) If she is allowed to pick a marble:
Total marbles $= 20$, Black marbles $= 6$.
$$P(\text{prize}) = \frac{6}{20} = \frac{3}{10}$$
(ii) If she is allowed to pick a marble:
Total marbles $= 20$, Black marbles $= 6$.
$$P(\text{prize}) = \frac{6}{20} = \frac{3}{10}$$
Two coins are tossed simultaneously. What is the probability of getting:
(i) At least one head?
(ii) At most one tail?
(iii) A head and a tail?
(i) At least one head?
(ii) At most one tail?
(iii) A head and a tail?
Answer
All possible outcomes: {HH, HT, TH, TT} → Total $= 4$.
(i) At least one head — {HH, HT, TH} → 3 outcomes. $$P(\text{at least one head}) = \frac{3}{4}$$
(ii) At most one tail — {HH, HT, TH} → 3 outcomes. $$P(\text{at most one tail}) = \frac{3}{4}$$
(iii) A head and a tail — {HT, TH} → 2 outcomes. $$P(\text{one head and one tail}) = \frac{2}{4} = \frac{1}{2}$$
(i) At least one head — {HH, HT, TH} → 3 outcomes. $$P(\text{at least one head}) = \frac{3}{4}$$
(ii) At most one tail — {HH, HT, TH} → 3 outcomes. $$P(\text{at most one tail}) = \frac{3}{4}$$
(iii) A head and a tail — {HT, TH} → 2 outcomes. $$P(\text{one head and one tail}) = \frac{2}{4} = \frac{1}{2}$$
A child has a die whose six faces show the letters as shown below:
The die is thrown once. What is the probability of getting:
(i) A
(ii) D
The die is thrown once. What is the probability of getting:
(i) A
(ii) D
Answer
Total possible outcomes $= 6$.
(i) Getting letter A: There are 2 faces showing A.
$$P(A) = \frac{2}{6} = \frac{1}{3}$$
(ii) Getting letter D: There is 1 face showing D.
$$P(D) = \frac{1}{6}$$
(i) Getting letter A: There are 2 faces showing A.
$$P(A) = \frac{2}{6} = \frac{1}{3}$$
(ii) Getting letter D: There is 1 face showing D.
$$P(D) = \frac{1}{6}$$
A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.
Answer
Total number of outcomes $= 8$.
Favourable outcomes for success (all same): {HHH, TTT} → 2 outcomes.
$$P(\text{success}) = \frac{2}{8} = \frac{1}{4}$$ $$P(\text{losing}) = 1 – \frac{1}{4} = \frac{3}{4}$$
Favourable outcomes for success (all same): {HHH, TTT} → 2 outcomes.
$$P(\text{success}) = \frac{2}{8} = \frac{1}{4}$$ $$P(\text{losing}) = 1 – \frac{1}{4} = \frac{3}{4}$$
A die is thrown once. Find the probability of getting:
(i) a composite number
(ii) a prime number.
(i) a composite number
(ii) a prime number.
Answer
Possible outcomes on a die: {1, 2, 3, 4, 5, 6} → Total $= 6$.
(i) Composite numbers (more than one factor, greater than 1): {4, 6} → 2 outcomes.
$$P(\text{composite}) = \frac{2}{6} = \frac{1}{3}$$
(ii) Prime numbers: {2, 3, 5} → 3 outcomes.
$$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$$
(i) Composite numbers (more than one factor, greater than 1): {4, 6} → 2 outcomes.
$$P(\text{composite}) = \frac{2}{6} = \frac{1}{3}$$
(ii) Prime numbers: {2, 3, 5} → 3 outcomes.
$$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$$
Find the probability of getting a black queen when a card is drawn at random from a well-shuffled pack of 52 cards.
Answer
Total number of cards $= 52$.
Number of black queens (Queen of Spades + Queen of Clubs) $= 2$.
$$P(\text{black queen}) = \frac{2}{52} = \frac{1}{26}$$
Number of black queens (Queen of Spades + Queen of Clubs) $= 2$.
$$P(\text{black queen}) = \frac{2}{52} = \frac{1}{26}$$
Three different coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least two tails.
(i) exactly two heads
(ii) at least two heads
(iii) at least two tails.
Answer
Total possible outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} → Total $= 8$.
(i) Exactly two heads: {HHT, HTH, THH} → 3 outcomes. $$P = \frac{3}{8}$$
(ii) At least two heads: {HHH, HHT, HTH, THH} → 4 outcomes. $$P = \frac{4}{8} = \frac{1}{2}$$
(iii) At least two tails: {HTT, THT, TTH, TTT} → 4 outcomes. $$P = \frac{4}{8} = \frac{1}{2}$$
(i) Exactly two heads: {HHT, HTH, THH} → 3 outcomes. $$P = \frac{3}{8}$$
(ii) At least two heads: {HHH, HHT, HTH, THH} → 4 outcomes. $$P = \frac{4}{8} = \frac{1}{2}$$
(iii) At least two tails: {HTT, THT, TTH, TTT} → 4 outcomes. $$P = \frac{4}{8} = \frac{1}{2}$$
The King, Queen and Jack of clubs are removed from a pack of 52 cards and then the remaining cards are well shuffled. A card is selected from the remaining cards. Find the probability of getting a card:
(i) of spade
(ii) of black king
(iii) of club
(iv) of jacks
(i) of spade
(ii) of black king
(iii) of club
(iv) of jacks
Answer
Total number of remaining cards $= 52 – 3 = 49$.
(i) Spade cards (none removed) $= 13$.
$$P(\text{spade}) = \frac{13}{49}$$
(ii) Black king — King of Clubs was removed; King of Spades remains → 1 black king.
$$P(\text{black king}) = \frac{1}{49}$$
(iii) Club cards — 3 removed, so $13 – 3 = 10$ club cards remain.
$$P(\text{club}) = \frac{10}{49}$$
(iv) Jacks — Jack of Clubs removed; 3 jacks remain.
$$P(\text{jack}) = \frac{3}{49}$$
(i) Spade cards (none removed) $= 13$.
$$P(\text{spade}) = \frac{13}{49}$$
(ii) Black king — King of Clubs was removed; King of Spades remains → 1 black king.
$$P(\text{black king}) = \frac{1}{49}$$
(iii) Club cards — 3 removed, so $13 – 3 = 10$ club cards remain.
$$P(\text{club}) = \frac{10}{49}$$
(iv) Jacks — Jack of Clubs removed; 3 jacks remain.
$$P(\text{jack}) = \frac{3}{49}$$
A box contains 125 shirts out of which 110 are good, 12 have minor defects and 3 have major defects. Ram Lal will buy only those shirts which are good while Naveen will reject only those which have major defects. A shirt is taken out at random from the box. Find the probability that:
(i) Ram Lal will buy it.
(ii) Naveen will buy it.
(i) Ram Lal will buy it.
(ii) Naveen will buy it.
Answer
Total number of shirts $= 125$.
(i) Ram Lal buys only good shirts. Number of good shirts $= 110$.
$$P(\text{Ram Lal buys}) = \frac{110}{125} = \frac{22}{25}$$
(ii) Naveen rejects only shirts with major defects. He accepts the rest: $125 – 3 = 122$ shirts.
$$P(\text{Naveen buys}) = \frac{122}{125}$$
(i) Ram Lal buys only good shirts. Number of good shirts $= 110$.
$$P(\text{Ram Lal buys}) = \frac{110}{125} = \frac{22}{25}$$
(ii) Naveen rejects only shirts with major defects. He accepts the rest: $125 – 3 = 122$ shirts.
$$P(\text{Naveen buys}) = \frac{122}{125}$$
Two different dice are thrown together. Find the probability that the numbers obtained have:
(i) even sum
(ii) even product.
(i) even sum
(ii) even product.
Answer
Total outcomes $= 36$.
(i) Even sum: Sum is even when both numbers are even or both are odd.
Even+Even: $3 \times 3 = 9$ outcomes. Odd+Odd: $3 \times 3 = 9$ outcomes. Total $= 18$.
$$P(\text{even sum}) = \frac{18}{36} = \frac{1}{2}$$
(ii) Even product: Product is odd only when both numbers are odd ($9$ outcomes). So product is even in $36 – 9 = 27$ outcomes.
$$P(\text{even product}) = \frac{27}{36} = \frac{3}{4}$$
(i) Even sum: Sum is even when both numbers are even or both are odd.
Even+Even: $3 \times 3 = 9$ outcomes. Odd+Odd: $3 \times 3 = 9$ outcomes. Total $= 18$.
$$P(\text{even sum}) = \frac{18}{36} = \frac{1}{2}$$
(ii) Even product: Product is odd only when both numbers are odd ($9$ outcomes). So product is even in $36 – 9 = 27$ outcomes.
$$P(\text{even product}) = \frac{27}{36} = \frac{3}{4}$$
A number $x$ is selected at random from the numbers 1, 2, 3, and 4. Another number $y$ is selected at random from the numbers 1, 4, 9, and 16. Find the probability that the product of $x$ and $y$ is less than 16.
Answer
Total possible outcomes $= 4 \times 4 = 16$.
Favourable outcomes where $xy < 16$:
$(1,1),\ (1,4),\ (1,9),\ (2,1),\ (2,4),\ (3,1),\ (3,4),\ (4,1)$ → 8 outcomes.
$$P(xy < 16) = \frac{8}{16} = \frac{1}{2}$$
Favourable outcomes where $xy < 16$:
$(1,1),\ (1,4),\ (1,9),\ (2,1),\ (2,4),\ (3,1),\ (3,4),\ (4,1)$ → 8 outcomes.
$$P(xy < 16) = \frac{8}{16} = \frac{1}{2}$$
A number $x$ is selected at random from the numbers 1, 4, 9, 16 and another number $y$ is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of $xy$ is more than 16.
Answer
Total possible outcomes $= 4 \times 4 = 16$.
Favourable outcomes where $xy > 16$:
$(9,2),\ (9,3),\ (9,4),\ (16,2),\ (16,3),\ (16,4)$ → 6 outcomes.
$$P(xy > 16) = \frac{6}{16} = \frac{3}{8}$$
Favourable outcomes where $xy > 16$:
$(9,2),\ (9,3),\ (9,4),\ (16,2),\ (16,3),\ (16,4)$ → 6 outcomes.
$$P(xy > 16) = \frac{6}{16} = \frac{3}{8}$$
Read the following and answer any four questions from Q.1 to Q.5.
On a weekend Rani was playing cards with her family. The deck has 52 cards. If her brother drew one card.
1. Find the probability of getting a king of red colour.
(A) $\frac{1}{26}$ (B) $\frac{1}{13}$ (C) $\frac{1}{52}$ (D) $\frac{1}{4}$
2. Find the probability of getting a face card.
(A) $\frac{1}{26}$ (B) $\frac{1}{13}$ (C) $\frac{2}{13}$ (D) $\frac{3}{13}$
3. Find the probability of getting a jack of hearts.
(A) $\frac{1}{26}$ (B) $\frac{1}{52}$ (C) $\frac{3}{52}$ (D) $\frac{3}{26}$
4. Find the probability of getting a red face card.
(A) $\frac{3}{26}$ (B) $\frac{1}{13}$ (C) $\frac{1}{52}$ (D) $\frac{1}{4}$
5. Find the probability of getting a spade.
(A) $\frac{1}{26}$ (B) $\frac{1}{13}$ (C) $\frac{1}{52}$ (D) $\frac{1}{4}$
On a weekend Rani was playing cards with her family. The deck has 52 cards. If her brother drew one card.
1. Find the probability of getting a king of red colour.
(A) $\frac{1}{26}$ (B) $\frac{1}{13}$ (C) $\frac{1}{52}$ (D) $\frac{1}{4}$
2. Find the probability of getting a face card.
(A) $\frac{1}{26}$ (B) $\frac{1}{13}$ (C) $\frac{2}{13}$ (D) $\frac{3}{13}$
3. Find the probability of getting a jack of hearts.
(A) $\frac{1}{26}$ (B) $\frac{1}{52}$ (C) $\frac{3}{52}$ (D) $\frac{3}{26}$
4. Find the probability of getting a red face card.
(A) $\frac{3}{26}$ (B) $\frac{1}{13}$ (C) $\frac{1}{52}$ (D) $\frac{1}{4}$
5. Find the probability of getting a spade.
(A) $\frac{1}{26}$ (B) $\frac{1}{13}$ (C) $\frac{1}{52}$ (D) $\frac{1}{4}$
Answer
1. Ans. Option (A) is correct.
Number of red kings (King of Hearts + King of Diamonds) $= 2$. Total cards $= 52$.
$$P = \frac{2}{52} = \frac{1}{26}$$
2. Ans. Option (D) is correct.
Total face cards $= 12$. $$P = \frac{12}{52} = \frac{3}{13}$$
3. Ans. Option (B) is correct.
There is only 1 Jack of Hearts. $$P = \frac{1}{52}$$
4. Ans. Option (A) is correct.
Red face cards $= 6$ (3 from hearts + 3 from diamonds). $$P = \frac{6}{52} = \frac{3}{26}$$
5. Ans. Option (D) is correct.
Spade cards $= 13$. $$P = \frac{13}{52} = \frac{1}{4}$$
Number of red kings (King of Hearts + King of Diamonds) $= 2$. Total cards $= 52$.
$$P = \frac{2}{52} = \frac{1}{26}$$
2. Ans. Option (D) is correct.
Total face cards $= 12$. $$P = \frac{12}{52} = \frac{3}{13}$$
3. Ans. Option (B) is correct.
There is only 1 Jack of Hearts. $$P = \frac{1}{52}$$
4. Ans. Option (A) is correct.
Red face cards $= 6$ (3 from hearts + 3 from diamonds). $$P = \frac{6}{52} = \frac{3}{26}$$
5. Ans. Option (D) is correct.
Spade cards $= 13$. $$P = \frac{13}{52} = \frac{1}{4}$$
Read the following passage and answer the following questions:
“Eight Ball” is a game played on a pool table with 15 balls numbered 1 to 15 and a “cue ball” that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8 and seven are striped balls numbered 9 to 15.
The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random. Based on the above information, answer the following questions:
1. What is the probability that the drawn ball bears number 8?
2. What is the probability that the drawn ball bears an even number?
OR
What is the probability that the drawn ball bears a number which is a multiple of 3?
3. What is the probability that the drawn ball is a solid coloured and bears an even number?
“Eight Ball” is a game played on a pool table with 15 balls numbered 1 to 15 and a “cue ball” that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8 and seven are striped balls numbered 9 to 15.
The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random. Based on the above information, answer the following questions:
1. What is the probability that the drawn ball bears number 8?
2. What is the probability that the drawn ball bears an even number?
OR
What is the probability that the drawn ball bears a number which is a multiple of 3?
3. What is the probability that the drawn ball is a solid coloured and bears an even number?
Answer
Total number of balls $= 15$.
1. Only one ball bears number 8.
$$P(\text{ball number 8}) = \frac{1}{15}$$
2. Even numbers from 1 to 15: 2, 4, 6, 8, 10, 12, 14 → 7 balls.
$$P(\text{even number}) = \frac{7}{15}$$
OR
Multiples of 3 from 1 to 15: 3, 6, 9, 12, 15 → 5 balls.
$$P(\text{multiple of 3}) = \frac{5}{15} = \frac{1}{3}$$
3. Solid coloured balls are numbered 1 to 8. Even numbers among them: 2, 4, 6, 8 → 4 balls.
$$P(\text{solid coloured and even}) = \frac{4}{15}$$
1. Only one ball bears number 8.
$$P(\text{ball number 8}) = \frac{1}{15}$$
2. Even numbers from 1 to 15: 2, 4, 6, 8, 10, 12, 14 → 7 balls.
$$P(\text{even number}) = \frac{7}{15}$$
OR
Multiples of 3 from 1 to 15: 3, 6, 9, 12, 15 → 5 balls.
$$P(\text{multiple of 3}) = \frac{5}{15} = \frac{1}{3}$$
3. Solid coloured balls are numbered 1 to 8. Even numbers among them: 2, 4, 6, 8 → 4 balls.
$$P(\text{solid coloured and even}) = \frac{4}{15}$$
Frequently Asked Questions
What does the Probability chapter cover in CBSE Class 10 Maths? ⌄
The Probability chapter in CBSE Class 10 covers the theoretical approach to probability, including basic concepts such as experiments, outcomes, events, and the probability formula. Students learn to calculate probabilities for everyday scenarios — tossing coins, throwing dice, drawing cards — and apply the complementary event rule. These concepts are directly tested across multiple question types in the board exam.
How many marks does the Probability chapter carry in the CBSE Class 10 board exam? ⌄
Probability is part of the Statistics and Probability unit, which together carry approximately 11 marks in the CBSE Class 10 Mathematics board paper. Questions from Probability typically appear as 1-mark MCQs, 2-mark short answers, and occasionally as 3-mark or case-study questions, making consistent practice across all formats essential for your child’s score.
What are the most important topics in Class 10 Probability for board exams? ⌄
The most frequently tested topics are: probability of drawing cards from a deck (aces, face cards, red/black cards), outcomes when tossing one or more coins, outcomes when throwing one or two dice, and applying the complementary probability rule $P(\bar{E}) = 1 – P(E)$. Case study questions involving real-life data tables have also become common in recent board papers.
What are the most common mistakes students make in Probability questions? ⌄
A very common error is miscounting the sample space — for example, forgetting that a standard deck has 52 cards with exactly 12 face cards and 4 aces, or miscounting outcomes when two dice are thrown (there are 36, not 12). Another frequent mistake is confusing “at least” and “at most” conditions when listing favourable outcomes. Practising a wide variety of previous year questions helps your child avoid these errors confidently.
How does Angle Belearn help students score well in Probability? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn directly from real board papers (2016–2025), each paired with clear, step-by-step solutions. Regular practice with these verified questions helps your child build a systematic approach to listing outcomes, identifying favourable cases, and applying the probability formula — earning full marks through accurate and well-structured working in the board exam.
