CBSE Maths › Class 12 PYQs › Probability

Probability: 1 mark Questions (CBSE Class 12)

7 different 1 mark questions on Probability from CBSE Class 12 Maths board exams 2026, newest first.

Q191 markAssertion–ReasonProbability CBSE 2026 · 65/1/1

Assertion (A): In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is $\frac{2}{3}$ .
Reason (R): For any two events A and B, $P (A ∣ B) = \frac{P ( A \cup B )}{P ( B )}$

(A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true and Reason (R) is false.
(D)Assertion (A) is false and Reason (R) is true.

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Answer: (C) Assertion (A) is true and Reason (R) is false.

Odd outcomes: {1, 3, 5}. Prime among them: {3, 5}.
$P (prime ∣ odd) = \frac{2}{3}$ , so A is true.
The correct formula is $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , not with $A \cup B$ , so R is false.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q181 markMCQProbability CBSE 2026 · 65/2/1

For two events A and B such that $P (A) \neq = 0$ and $P (B) \neq = 1$ , $P (A^{'} / B^{'}) =$

(A) $1 - P (A / B)$
(B) $1 - P (A^{'} / B)$
(C) $\frac{1 - P ( A \cap B )}{P ( B ^{'} )}$
(D) $\frac{1 - P ( A \cup B )}{P ( B ^{'} )}$

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Answer: (D)

\frac{1 - P ( A \cup B )}{P ( B ^{'} )}

$P (A^{'} / B^{'}) = \frac{P ( A ^{'} \cap B ^{'} )}{P ( B ^{'} )}$ .
$A^{'} \cap B^{'} = (A \cup B)^{'}$ , so $P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$ .
$P (A^{'} / B^{'}) = \frac{1 - P ( A \cup B )}{P ( B ^{'} )}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q181 markMCQProbability CBSE 2026 · 65/3/1

If E and F are two independent events such that $P (E) = \frac{3}{10}$ , $P (E \cup F) = \frac{1}{2}$ , then $P (E ∣ F) - P (F ∣ E)$ is equal to :

(A) $\frac{2}{7}$
(B) $\frac{3}{35}$
(C) $\frac{1}{70}$
(D) $\frac{1}{7}$

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Answer: (C)

\frac{1}{70}

For independent events, $P (E \cup F) = P (E) + P (F) - P (E) P (F)$ .
$\frac{1}{2} = \frac{3}{10} + \frac{7}{10} P (F)$ , so $P (F) = \frac{2}{7}$ .
$P (E ∣ F) - P (F ∣ E) = P (E) - P (F) = \frac{3}{10} - \frac{2}{7} = \frac{1}{70}$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q181 markMCQProbability CBSE 2026 · 65/4/1

A box contains 4 red, 5 blue and 1 green marble. A child randomly takes out a marble from the box, notes down the colour and puts it back in the box. If the activity is repeated 3 times, what is the probability that at least one marble is red ?

(A) $\frac{27}{125}$
(B) $\frac{8}{125}$
(C) $\frac{2}{125}$
(D) $\frac{98}{125}$

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Answer: (D)

\frac{98}{125}

P(red in one draw) $= \frac{4}{10} = \frac{2}{5}$ , so P(not red) $= \frac{3}{5}$ .
Draws are independent (with replacement): P(no red in 3 draws) $= (\frac{3}{5})^{3} = \frac{27}{125}$ .
P(at least one red) $= 1 - \frac{27}{125} = \frac{98}{125}$ .

Q181 markMCQProbability CBSE 2026 · 65/4/2

The probability that it will rain tomorrow in cities A, B and C is 60%, 70% and 80% respectively. The probability that it will rain tomorrow in at least one of the cities is :

(A) $\frac{3}{250}$
(B) $\frac{244}{250}$
(C)1
(D) $\frac{9}{10}$

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Answer: (B)

\frac{244}{250}

Treating the cities as independent, P(no rain anywhere) $= 0.4 \times 0.3 \times 0.2 = 0.024 = \frac{6}{250}$ .
P(rain in at least one city) $= 1 - \frac{6}{250} = \frac{244}{250}$ .

Q181 markMCQProbability CBSE 2026 · 65/4/3

The probability that a particular item is available in three shops A, B and C is $\frac{4}{5}$ , $\frac{3}{4}$ and $\frac{2}{3}$ respectively. If a person visits all the three shops to buy the item, then what is the probability that it will be available in at least one of the shops ?

(A) $\frac{59}{60}$
(B)1
(C) $\frac{1}{60}$
(D) $\frac{9}{60}$

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Answer: (A)

\frac{59}{60}

Treating availability in the shops as independent, P(not available in any) $= \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} = \frac{1}{60}$ .
P(available in at least one) $= 1 - \frac{1}{60} = \frac{59}{60}$ .

Q181 markMCQProbability CBSE 2026 · 65/5/1

If $3 P (A) = P (B) = \frac{3}{5}$ and $P (A ∣ B) = \frac{2}{3}$ , then $P (A \cup B)$ is :

(A) $\frac{3}{5}$
(B) $\frac{1}{5}$
(C) $\frac{2}{15}$
(D) $\frac{2}{5}$

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Answer: (D)

\frac{2}{5}

$P (B) = \frac{3}{5}$ , $P (A) = \frac{1}{5}$ .
$P (A \cap B) = P (A ∣ B) P (B) = \frac{2}{3} \cdot \frac{3}{5} = \frac{2}{5}$ .
$P (A \cup B) = \frac{1}{5} + \frac{3}{5} - \frac{2}{5} = \frac{2}{5}$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

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