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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 2 · Q29
(a) 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…
Out of two bags, bag I contains $\displaystyle 3$ red and $\displaystyle 4$ white balls and bag II contains $\displaystyle 8$ red and $\displaystyle 6$ white balls. A die is thrown. If it shows a number less than $\displaystyle 3$ 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.
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 $\displaystyle \mathrm{P}\left(\mathrm{X}^{\prime}\right)+\mathrm{P}\left(\mathrm{Y}^{\prime}\right)=2-2 \mathrm{a}+\mathrm{b}$.
Marking-scheme solution
Let $\displaystyle E_1$ : The number appearing on the die $\displaystyle < 3$
$\displaystyle E_2$ : The number appearing on the die $\displaystyle \geq 3$
A: Red ball is drawn
$\displaystyle P(E_1) = \dfrac{2}{6}, P(A/E_1) = \dfrac{3}{7}, P(E_2) = \dfrac{4}{6}, P(A/E_2) = \dfrac{8}{14}$
The required probability $\displaystyle = P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)$
$\displaystyle = \dfrac{2}{6}\times\dfrac{3}{7} + \dfrac{4}{6}\times\dfrac{8}{14}$
$\displaystyle = \dfrac{11}{21}$
$\displaystyle P(X \cup Y) = a$
$\displaystyle P(X \cup Y) - P(X \cap Y) = b$
Hence, $\displaystyle P(X \cap Y) = a - b$
$\displaystyle P(X') + P(Y') = [1 - P(X)] + [1 - P(Y)]$
$\displaystyle = 2 - [P(X) + P(Y)]$
$\displaystyle = 2 - [P(X \cup Y) + P(X \cap Y)]$
$\displaystyle = 2 - [a + (a - b)]$
$\displaystyle = 2 - 2a + b$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.