CBSE Maths › Class 10 PYQs › Probability

Probability: 2 marks Questions (CBSE Class 10)

32 different 2 marks questions on Probability from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (133)2 marks (32)3 marks (26)4 marks (9)

Q242 marksVery Short AnswerProbability CBSE 2026 · Basic 430/5/1

Find the probability that a number selected at random from the numbers 30, 31, 32, 33, ....., 60 is (i) a prime number (ii) a multiple of 6.

Show answer & solution

Answer: (i)

\frac{7}{31}

(ii)

\frac{6}{31}

Total numbers from 30 to 60 = 31.
(i) Primes: 31, 37, 41, 43, 47, 53, 59, i.e. 7. P = 7/31.
(ii) Multiples of 6: 30, 36, 42, 48, 54, 60, i.e. 6. P = 6/31.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q24 (OR) (OR)2 marksVery Short AnswerProbability CBSE 2026 · Basic 430/5/1

Slips of letters of the word ‘BACKGROUND’ are put in a bowl and thoroughly mixed. One slip is picked up at random. Find the probability that picked up slip’s letter is (i) a vowel (ii) present in the word ‘BALL’.

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Answer: (i)

\frac{3}{10}

(ii)

\frac{1}{5}

BACKGROUND has 10 letters: B, A, C, K, G, R, O, U, N, D.
(i) Vowels: A, O, U, i.e. 3. P = 3/10.
(ii) Letters also in 'BALL': B and A, i.e. 2 slips. P = 2/10 = 1/5.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q252 marksVery Short AnswerProbability CBSE 2026 · Standard 30/5/1

A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is 3/5, then find the number of yellow balls.

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Answer: 10 yellow balls

Number of green balls $= \frac{3}{5} \times 25 = 15$
Number of yellow balls $= 25 - 15 = 10$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q232 marksVery Short AnswerProbability CBSE 2025 · Basic 430/4/1

A coin is dropped at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with radius 0.7 m ?

Diagram for CBSE 2025 Class 10 Maths question 23

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Answer:

\frac{77}{300}

(about 0.257)

Area of rectangle $= 3 \times 2 = 6$ m $^{2}$ .
Area of circle $= \frac{22}{7} \times 0.7 \times 0.7 = 1.54$ m $^{2}$ .
P(coin lands inside the circle) $= \frac{1.54}{6} = \frac{154}{600} = \frac{77}{300}$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q23 (OR) (OR)2 marksVery Short AnswerProbability CBSE 2025 · Basic 430/4/1

A die is thrown twice. What is the probability that (i) difference between two numbers obtained is 3 ? (ii) sum of the numbers obtained is 8 ?

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Answer: (i)

\frac{1}{6}

(ii)

\frac{5}{36}

Total outcomes $= 6 \times 6 = 36$ .
(i) Difference 3: (1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3), i.e. 6 outcomes. P $= \frac{6}{36} = \frac{1}{6}$ .
(ii) Sum 8: (2, 6), (6, 2), (3, 5), (5, 3), (4, 4), i.e. 5 outcomes. P $= \frac{5}{36}$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q242 marksVery Short AnswerProbability CBSE 2025 · Basic 430/5/1

A bag contains 40 marbles out of which some are white and others are black. If the probability of drawing a black marble is $\frac{3}{5}$ , then find the number of white marbles.

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Answer: 16

Number of black marbles $= \frac{3}{5} \times 40 = 24$ .
Number of white marbles $= 40 - 24 = 16$ .

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q24 (OR) (OR)2 marksVery Short AnswerProbability CBSE 2025 · Basic 430/5/1

In a pre-primary class, a teacher put cards numbered 20 to 59 in a bowl. A student picked up a card at random and read the number. Find the probability that the number read was (i) a prime number (ii) a perfect square.

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Answer: (i)

\frac{9}{40}

(ii)

\frac{3}{40}

Total cards $= 59 - 20 + 1 = 40$ .
(i) Primes: 23, 29, 31, 37, 41, 43, 47, 53, 59, i.e. 9 cards. P $= \frac{9}{40}$ .
(ii) Perfect squares: 25, 36, 49, i.e. 3 cards. P $= \frac{3}{40}$ .

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q232 marksVery Short AnswerProbability CBSE 2025 · Basic 430/6/1

A box contains 120 discs, which are numbered from 1 to 120. If one disc is drawn at random from the box, find the probability that
(i) it bears a 2– digit number
(ii) the number is a perfect square.

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Answer: (i)

\frac{3}{4}

(ii)

\frac{1}{12}

Total outcomes = 120.
(i) 2-digit numbers are 10 to 99, i.e. 90 numbers. P = $\frac{90}{120} = \frac{3}{4}$ .
(ii) Perfect squares up to 120: 1, 4, 9, ..., 100, i.e. 10 numbers ( $1 1^{2} = 121 > 120$ ). P $= \frac{10}{120} = \frac{1}{12}$ .

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q222 marksVery Short AnswerProbability CBSE 2025 · Standard 30/3/1

The probability of guessing the correct answer of a certain test question is $\frac{x}{12}$ . If the probability of not guessing the correct answer is $\frac{5}{6}$ , then find the value of $x$ .

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Answer:

x = 2

$P (E) + P (not E) = 1$ .
$\frac{x}{12} + \frac{5}{6} = 1$ , so $\frac{x}{12} = \frac{1}{6}$ .
$x = 2$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q252 marksVery Short AnswerProbability CBSE 2025 · Standard 30/4/1

Two friends Anil and Ashraf were born in the December month in the year 2010. Find the probability that :
(i) they share same date of birth.
(ii) they have different dates of birth.

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Answer: (i)

\frac{1}{31}

(ii)

\frac{30}{31}

December has 31 days. Whatever Anil's date is, Ashraf's date can be any of 31 equally likely days.
(i) P(same date) $= \frac{1}{31}$
(ii) P(different dates) $= 1 - \frac{1}{31} = \frac{30}{31}$

Q222 marksVery Short AnswerProbability CBSE 2025 · Standard 30/4/2

Saima and Aryaa were born in the month of June in the year 2012. Find the probability that :
(i) they have different dates of birth.
(ii) they have same date of birth.

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Answer: (i)

\frac{29}{30}

(ii)

\frac{1}{30}

June has 30 days. Whatever Saima's date is, Aryaa's date can be any of 30 equally likely days.
(ii) P(same date) $= \frac{1}{30}$
(i) P(different dates) $= 1 - \frac{1}{30} = \frac{29}{30}$

Q242 marksVery Short AnswerProbability CBSE 2025 · Standard 30/4/3

Renu and Simran were born in the year 2000 which is a leap year. Find the probability that :
(i) both have same birthday.
(ii) both have different birthdays.

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Answer: (i)

\frac{1}{366}

(ii)

\frac{365}{366}

A leap year has 366 days. Whatever Renu's birthday is, Simran's can be any of 366 equally likely days.
(i) P(same birthday) $= \frac{1}{366}$
(ii) P(different birthdays) $= 1 - \frac{1}{366} = \frac{365}{366}$

Q252 marksVery Short AnswerProbability CBSE 2025 · Standard 30/5/1

While shuffling a pack of 52 cards, one card was accidently dropped. Find the probability that the dropped card
(i) is not a face card.
(ii) is a black king.

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Answer: (i)

\frac{10}{13}

(ii)

\frac{1}{26}

Total outcomes = 52.
(i) Face cards = 12, so non-face cards = 40. $P = \frac{40}{52} = \frac{10}{13}$
(ii) Black kings = 2. $P = \frac{2}{52} = \frac{1}{26}$

Q222 marksVery Short AnswerProbability CBSE 2025 · Standard 30/5/2

All the face cards are removed from the pack of 52 cards and a card is drawn at random from the remaining cards. Find the probability that the card so drawn is
(i) a spade.
(ii) not an ace.

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Answer: (i)

\frac{1}{4}

(ii)

\frac{9}{10}

Remaining cards $= 52 - 12 = 40$ .
(i) Spades left $= 13 - 3 = 10$ . $P = \frac{10}{40} = \frac{1}{4}$
(ii) Aces $= 4$ , so non-aces $= 36$ . $P = \frac{36}{40} = \frac{9}{10}$

Q242 marksVery Short AnswerProbability CBSE 2025 · Standard 30/5/3

From a pack of 52 cards, all aces and all kings are removed. A card is drawn at random from the remaining cards. Find the probability that the card so drawn is
(i) a face card.
(ii) a card of red colour.

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Answer: (i)

\frac{2}{11}

(ii)

\frac{1}{2}

Remaining cards $= 52 - 8 = 44$ .
(i) Face cards left (jacks and queens) $= 8$ . $P = \frac{8}{44} = \frac{2}{11}$
(ii) Red cards left $= 26 - 4 = 22$ . $P = \frac{22}{44} = \frac{1}{2}$

Q252 marksVery Short AnswerProbability CBSE 2025 · Standard 30/6/1

The number of red balls in a bag is three more than the number of black balls. If the probability of drawing a red ball at random from the given bag is $\frac{12}{23}$ , find the total number of balls in the given bag.

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Answer: 69

Let black balls $= x$ ; red balls $= x + 3$ ; total $= 2 x + 3$
$\frac{x + 3}{2 x + 3} = \frac{12}{23}$
$23 x + 69 = 24 x + 36 \Rightarrow x = 33$
Total $= 2 (33) + 3 = 69$ balls

Q222 marksVery Short AnswerProbability CBSE 2025 · Standard 30/6/2

A bag contains cards which are numbered from 5 to 100 such that each card bears a different number. A card is drawn at random. Find the probability that number on the card is
(i) a perfect square
(ii) a 2-digit number

Show answer & solution

Answer: (i)

\frac{1}{12}

(ii)

\frac{15}{16}

Total cards $= 100 - 5 + 1 = 96$
(i) Perfect squares: 9, 16, 25, 36, 49, 64, 81, 100 (8 cards). $P = \frac{8}{96} = \frac{1}{12}$
(ii) 2-digit numbers: 10 to 99 (90 cards). $P = \frac{90}{96} = \frac{15}{16}$

Q242 marksVery Short AnswerProbability CBSE 2025 · Standard 30/6/3

A bag contains balls numbered 2 to 91 such that each ball bears a different number. A ball is drawn at random from the bag. Find the probability that
(i) it bears a 2– digit number
(ii) it bears a multiple of 1.

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Answer: (i)

\frac{41}{45}

(ii) 1

Total balls $= 91 - 2 + 1 = 90$
(i) 2-digit numbers: 10 to 91, i.e. 82 balls. $P = \frac{82}{90} = \frac{41}{45}$
(ii) Every number is a multiple of 1, so all 90 balls qualify. $P = \frac{90}{90} = 1$

Q222 marksVery Short AnswerProbability CBSE 2024 · Basic 430/1/2

A bag contains 4 red, 5 white and some yellow balls. If probability of drawing a red ball at random is $\frac{1}{5}$ , then find the probability of drawing a yellow ball at random.

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Answer:

\frac{11}{20}

Let the number of yellow balls be $y$ . Total $= 9 + y$ .
$\frac{4}{9 + y} = \frac{1}{5}$ , so $9 + y = 20$ and $y = 11$ .
P(yellow) $= \frac{11}{20}$ .

Also asked in: 2024 Basic 430/1/3

Q232 marksVery Short AnswerProbability CBSE 2024 · Basic 430/2/1

15 defective pens are accidentally mixed with 145 good ones. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

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Answer:

\frac{29}{32}

Total pens $= 15 + 145 = 160$
Favourable outcomes (good pens) $= 145$
P(good) $= \frac{145}{160} = \frac{29}{32}$

Also asked in: 2024 Basic 430/2/3

Q222 marksVery Short AnswerProbability CBSE 2024 · Basic 430/2/2

A lot consists of 165 ball pens of which 30 are defective and the others are good. Rakshita will buy a pen if it is good. The shopkeeper draws one pen at random and gives it to Rakshita. What is the probability that she will buy it ?

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Answer:

\frac{9}{11}

Total pens = 165; good pens $= 165 - 30 = 135$
P(she buys) = P(good) $= \frac{135}{165} = \frac{9}{11}$

Q252 marksVery Short AnswerProbability CBSE 2024 · Basic 430/3/1

There are 80 cards numbered from 1 to 80. One card is drawn at random from them. Find the probability that the number on the selected card is not divisible by 8.

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Answer:

\frac{7}{8}

Numbers from 1 to 80 divisible by 8: 8, 16, ..., 80, i.e. 10 numbers.
Numbers not divisible by 8 $= 80 - 10 = 70$ .
Required probability $= \frac{70}{80} = \frac{7}{8}$ .

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q222 marksVery Short AnswerProbability CBSE 2024 · Standard 30/1/1

In a pack of 52 playing cards one card is lost. From the remaining cards, a card is drawn at random. Find the probability that the drawn card is queen of heart, if the lost card is a black card.

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Answer:

\frac{1}{51}

After a black card is lost, 51 cards remain.
The queen of hearts is red, so it is still in the pack: 1 favourable outcome.
Required probability $= \frac{1}{51}$ .

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q212 marksVery Short AnswerProbability CBSE 2024 · Standard 30/3/1

One card is drawn at random from a well shuffled deck of 52 cards. Find the probability that the card drawn
(i) is queen of hearts;
(ii) is not a jack.

Show answer & solution

Answer: (i)

\frac{1}{52}

(ii)

\frac{12}{13}

Total outcomes = 52.
(i) There is 1 queen of hearts, so P = $\frac{1}{52}$ .
(ii) There are 4 jacks, so 48 cards are not jacks. P = $\frac{48}{52} = \frac{12}{13}$ .

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q252 marksVery Short AnswerProbability CBSE 2024 · Standard 30/5/1

A carton consists of 60 shirts of which 48 are good, 8 have major defects and 4 have minor defects. Nigam, a trader, will accept the shirts which are good but Anmol, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. Find the probability that it is acceptable to Anmol.

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Answer:

\frac{13}{15}

Anmol accepts all shirts except those with major defects: $60 - 8 = 52$ shirts.
P(acceptable to Anmol) $= \frac{52}{60} = \frac{13}{15}$ .

Also asked in: 2024 Standard 30/5/2

Q212 marksVery Short AnswerProbability CBSE 2024 · Standard 30/5/3

The king, queen and ace of clubs and diamonds are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is randomly drawn from the remaining cards. Find the probability of getting
(i) a card of clubs.
(ii) a red coloured card.

Show answer & solution

Answer: (i)

\frac{5}{23}

(ii)

\frac{1}{2}

Cards removed: 3 clubs and 3 diamonds, so $52 - 6 = 46$ cards remain.
(i) Clubs left $= 13 - 3 = 10$ ; P(club) $= \frac{10}{46} = \frac{5}{23}$ .
(ii) Red cards left $= 13$ hearts $+ 10$ diamonds $= 23$ ; P(red) $= \frac{23}{46} = \frac{1}{2}$ .

Q252 marksVery Short AnswerProbability CBSE 2023 · Basic 430/2/1

From a well-shuffled deck of 52 playing cards, all diamond cards are removed. Now, a card is drawn from the remaining pack at random. Find the probability that the selected card is a king.

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Answer:

\frac{1}{13}

Cards left after removing 13 diamonds = 52 − 13 = 39
Kings left (king of diamonds removed) = 3
P(king) $= \frac{3}{39} = \frac{1}{13}$

Also asked in: 2023 Basic 430/2/2

Q222 marksVery Short AnswerProbability CBSE 2023 · Basic 430/2/3

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random, then find the probability that it bears a multiple of 15.

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Answer:

\frac{1}{15}

Multiples of 15 from 1 to 90: 15, 30, 45, 60, 75, 90, i.e. 6 discs.
P(multiple of 15) $= \frac{6}{90} = \frac{1}{15}$

Q242 marksVery Short AnswerProbability CBSE 2023 · Basic 430/6/1

A box contains 20 discs which are numbered from 1 to 20. If one disc is drawn at random from the box, then find the probability that the number on the drawn disc is a
(i) 2-digit number
(ii) number less than 10

Show answer & solution

Answer: (i)

\frac{11}{20}

(ii)

\frac{9}{20}

Total outcomes = 20.
(i) 2-digit numbers: 10 to 20, i.e. 11 numbers. P $= \frac{11}{20}$ .
(ii) Numbers less than 10: 1 to 9, i.e. 9 numbers. P $= \frac{9}{20}$ .

Also asked in: 2023 Basic 430/6/2

Q212 marksVery Short AnswerProbability CBSE 2023 · Basic 430/6/3

A bag contains 30 discs numbered from 1 to 30. One disc is drawn at random from the bag. Find the probability that it bears a number
(a) divisible by 6.
(b) greater than 25.

Show answer & solution

Answer: (a)

\frac{1}{6}

(b)

\frac{1}{6}

Total outcomes = 30.
(a) Multiples of 6: 6, 12, 18, 24, 30, i.e. 5 numbers. P $= \frac{5}{30} = \frac{1}{6}$ .
(b) Numbers greater than 25: 26, 27, 28, 29, 30, i.e. 5 numbers. P $= \frac{5}{30} = \frac{1}{6}$ .

Q252 marksVery Short AnswerProbability CBSE 2023 · Standard 30/1/1

A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random from the bag. Find the probability that drawn ball is (i) red (ii) yellow.

Show answer & solution

Answer: (i)

\frac{4}{9}

(ii)

\frac{2}{9}

Total balls = 4 + 3 + 2 = 9.
(i) P(red) $= \frac{4}{9}$
(ii) P(yellow) $= \frac{2}{9}$

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q242 marksVery Short AnswerProbability CBSE 2023 · Standard 30/4/1

If a fair coin is tossed twice, find the probability of getting ‘atmost one head’.

Show answer & solution

Answer:

\frac{3}{4}

Outcomes: HH, HT, TH, TT (4 outcomes).
At most one head: HT, TH, TT (3 outcomes).
P $= \frac{3}{4}$

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

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