UPSC NDA II 2024 Mathematics Paper

Concept:The problem involves conditional probability where one ball is transferred from urn A to urn B before drawing from urn B.Explanation:Let $E_1$ be the event of transferring a white ball from urn A to urn B.Probability of $E_1$ = $\frac{2}{4} = \frac{1}{2}$.After $E_1$, urn B has 4 white and 2 black balls, so drawing a white ball from B has probability $\frac{4}{6} = \frac{2}{3}$.Let $E_2$ be the event of transferring a black ball from urn A to urn B.Probability of $E_2$ = $\frac{2}{4} = \frac{1}{2}$.After $E_2$, urn B has 3 white and 3 black balls, so drawing a white ball from B has probability $\frac{3}{6} = \frac{1}{2}$.Let $F$ be the event that the ball drawn from urn B is white.By the law of total probability:$P(F) = P(E_1) \cdot P(F|E_1) + P(E_2) \cdot P(F|E_2)$$= \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{2}$$= \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$.Answer:$\frac{7}{12}$ (Option B).