UPSC NDA I 2023 Mathematics Solved Paper

Concept:This problem uses conditional probability and combinations to find the probability after transferring balls.Explanation:Bag 1 has 3 white (W) and 2 black (B) balls. Bag 2 has 2 W and 3 B balls.Two balls are drawn from Bag 1 and placed into Bag 2. Then one ball is drawn from Bag 2.We need the probability that the final ball is white.Three mutually exclusive cases exist for the two balls drawn from Bag 1:Case 1: Two black balls (BB). Bag 2 then has 2W and 5B. Probability of drawing white from Bag 2 is $\frac{2}{7}$.Case 2: Two white balls (WW). Bag 2 then has 4W and 3B. Probability of drawing white is $\frac{4}{7}$.Case 3: One white and one black (WB or BW). Bag 2 then has 3W and 4B. Probability of drawing white is $\frac{3}{7}$.Compute probabilities of each case:Total ways to choose 2 balls from Bag 1: $\binom{5}{2}=10$.Case 1: $\binom{2}{2}=1$ way. Probability = $\frac{1}{10}$.Case 2: $\binom{3}{2}=3$ ways. Probability = $\frac{3}{10}$.Case 3: $\binom{3}{1}\binom{2}{1}=6$ ways. Probability = $\frac{6}{10}$.Total probability: $\frac{1}{10}\cdot\frac{2}{7}+\frac{3}{10}\cdot\frac{4}{7}+\frac{6}{10}\cdot\frac{3}{7} = \frac{2}{70}+\frac{12}{70}+\frac{18}{70} = \frac{32}{70}$.Answer:$\frac{32}{70}$ (Option B).