CBSE Maths › Class 12 PYQs › Probability

Probability: 3 marks Questions (CBSE Class 12)

8 different 3 marks questions on Probability from CBSE Class 12 Maths board exams 2026, newest first.

Q313 marksShort AnswerProbability CBSE 2026 · 65/1/1

Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.

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Answer:

\frac{11}{21}

Let $E_{1}$ : die shows 1 or 2 (bag I), $E_{2}$ : die shows 3, 4, 5 or 6 (bag II). $P (E_{1}) = \frac{1}{3}$ , $P (E_{2}) = \frac{2}{3}$ .
Let R: red ball drawn. $P (R ∣ E_{1}) = \frac{3}{7}$ , $P (R ∣ E_{2}) = \frac{8}{14} = \frac{4}{7}$ .
$P (R) = \frac{1}{3} \cdot \frac{3}{7} + \frac{2}{3} \cdot \frac{4}{7} = \frac{3}{21} + \frac{8}{21} = \frac{11}{21}$ .

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

Q31 (OR) (OR)3 marksShort AnswerProbability CBSE 2026 · 65/1/1

The probability of simultaneous occurrence of atleast one of the two events X and Y is a. If the probability that exactly one of the events X, Y occurs is b, prove that $P (X^{'}) + P (Y^{'}) = 2 - 2 a + b$ .

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Answer: Proved.

Given $P (X \cup Y) = P (X) + P (Y) - P (X \cap Y) = a$ .
Exactly one occurs: $P (X) + P (Y) - 2 P (X \cap Y) = b$ .
Subtracting: $P (X \cap Y) = a - b$ , so $P (X) + P (Y) = a + (a - b) = 2 a - b$ .
$P (X^{'}) + P (Y^{'}) = 2 - [P (X) + P (Y)] = 2 - 2 a + b$ . Hence proved.

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

Q313 marksShort AnswerProbability CBSE 2026 · 65/2/1

The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed.
The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that
(i) target is hit
(ii) atleast one shot misses the target.

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Answer: (i)

\frac{15}{16}

(ii)

\frac{7}{16}

Let $p$ = P(hit). $p = 3 (1 - p) \Rightarrow p = \frac{3}{4}$ , P(miss) $= \frac{1}{4}$ . Shots are independent.
(i) Target is hit (at least once) $= 1 - P (both miss) = 1 - \frac{1}{16} = \frac{15}{16}$ .
(ii) At least one shot misses $= 1 - P (both hit) = 1 - \frac{9}{16} = \frac{7}{16}$ .

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

Q31 (OR) (OR)3 marksShort AnswerProbability CBSE 2026 · 65/2/1

Mother, Father and Son line up at random for a family picture. Let events E : Son on one end and F : Father in the middle. Find P(E/F).

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Answer:

P (E / F) = 1

Sample space: $3! = 6$ equally likely arrangements.
F = {MFS, SFM}, so $P (F) = \frac{2}{6} = \frac{1}{3}$ .
In both arrangements of F the son is at an end, so $E \cap F = F$ and $P (E \cap F) = \frac{1}{3}$ .
$P (E / F) = \frac{P ( E \cap F )}{P ( F )} = 1$ .

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

Q313 marksShort AnswerProbability CBSE 2026 · 65/3/1

A survey was conducted on the patients who have undergone knee replacement surgeries.
It was found that, Robotic Knee replacement surgeries have 90% success rate.
On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that :
(i) exactly one surgery is successful,
(ii) at most two surgeries are successful.

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Answer: (i) 0.027 (ii) 0.271

P(success) = 0.9, P(failure) = 0.1 for each surgery, independently.
(i) Exactly one success: $3 \times 0.9 \times 0.1 \times 0.1 = 0.027$ .
(ii) At most two successes = 1 − P(all three succeed) $= 1 - (0.9)^{3} = 1 - 0.729 = 0.271$ .

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

Q313 marksShort AnswerProbability CBSE 2026 · 65/4/1

In a school, the probability of holding a debate competition is $\frac{1}{3}$ and that of a quiz competition is $\frac{2}{3}$ . In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.

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Answer:

\frac{22}{45}

Let $E_{1}$ : debate (team A), $E_{2}$ : quiz (team B); $P (E_{1}) = \frac{1}{3}$ , $P (E_{2}) = \frac{2}{3}$ . Let G: one girl and one boy chosen.
$P (G ∣ E_{1}) = \frac{4 \times 6}{^{10} C _{2}} = \frac{24}{45}$ , $P (G ∣ E_{2}) = \frac{7 \times 3}{^{10} C _{2}} = \frac{21}{45}$ .
$P (G) = \frac{1}{3} \cdot \frac{24}{45} + \frac{2}{3} \cdot \frac{21}{45} = \frac{24 + 42}{135} = \frac{66}{135} = \frac{22}{45}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q313 marksShort AnswerProbability CBSE 2026 · 65/5/1

A die is rolled. Consider events :
A = {1, 2, 5}, B = {3, 5}, C = {2, 3, 4, 5}
and hence find :
(i) $P (A ∣ C)$ and $P (C ∣ A)$
(ii) $P (A \cap B ∣ C)$ and $P (A \cup B ∣ C)$

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Answer: (i)

P (A ∣ C) = \frac{1}{2}

P (C ∣ A) = \frac{2}{3}

(ii)

P (A \cap B ∣ C) = \frac{1}{4}

P (A \cup B ∣ C) = \frac{3}{4}

$P (A) = \frac{3}{6}$ , $P (C) = \frac{4}{6}$ , $A \cap C = {2, 5}$ so $P (A \cap C) = \frac{2}{6}$ .
$P (A ∣ C) = \frac{2/6}{4/6} = \frac{1}{2}$ ; $P (C ∣ A) = \frac{2/6}{3/6} = \frac{2}{3}$ .
$A \cap B = {5}$ , $(A \cap B) \cap C = {5}$ : $P (A \cap B ∣ C) = \frac{1/6}{4/6} = \frac{1}{4}$ .
$A \cup B = {1, 2, 3, 5}$ , $(A \cup B) \cap C = {2, 3, 5}$ : $P (A \cup B ∣ C) = \frac{3/6}{4/6} = \frac{3}{4}$ .

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

Q31 (OR) (OR)3 marksShort AnswerProbability CBSE 2026 · 65/5/1

A box contains 6 cards numbered 1 to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as 10, and B, the event of a number other than 4 on the first card selected.
Find P(A and B) and find whether the events A and B are independent events or not.

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Answer:

P (A and B) = \frac{1}{18}

; A and B are not independent.

There are 36 equally likely outcomes.
A = {(4, 6), (5, 5), (6, 4)}, so $P (A) = \frac{3}{36} = \frac{1}{12}$ .
$P (B) = \frac{5}{6}$ (first card is not 4).
$A \cap B = {(5, 5), (6, 4)}$ , so $P (A \cap B) = \frac{2}{36} = \frac{1}{18}$ .
$P (A) P (B) = \frac{1}{12} \cdot \frac{5}{6} = \frac{5}{72} \neq = \frac{1}{18}$ , so A and B are not independent.

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

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