Question

Bag A contains 5 white and 2 black balls. Bag B contains 2 white and 5 black balls. Two balls are randomly chosen from bag A and placed in bag B. Now a ball is drawn randomly from bag B and found that it is white. The probability that the two balls drawn from bag A are of different colour is:

• A. \( \frac{10}{21} \)
• B. \( \frac{10}{63} \)
• C. \( \frac{5}{12} \)
• D. \( \frac{2}{189} \)

Hint:

Always split such problems into conditioning cases before applying Bayes’ theorem.

Answer 1

The Correct Option is B

Concept: We use conditional probability and Bayes’ theorem. First we classify all possible ways of transferring balls from Bag A to Bag B, then compute probability of drawing a white ball from Bag B.

Step 1: Find probabilities of transfer cases from Bag A.
Total ways: \[ \binom{7}{2} = 21 \] Cases: \[ P(WW)=\frac{\binom{5}{2}}{21}=\frac{10}{21}, \quad P(WB)=\frac{\binom{5}{1}\binom{2}{1}}{21}=\frac{10}{21}, \quad P(BB)=\frac{1}{21} \]

Step 2: Update Bag B composition and compute \(P(W)\).
Initial Bag B: 2W, 5B (7 balls) After transfer: \[ WW \Rightarrow 4W,5B,\quad P(W|WW)=\frac{4}{9} \] \[ WB \Rightarrow 3W,6B,\quad P(W|WB)=\frac{3}{9} \] \[ BB \Rightarrow 2W,7B,\quad P(W|BB)=\frac{2}{9} \] \[ P(W)=\frac{10}{21}\cdot\frac{4}{9}+\frac{10}{21}\cdot\frac{3}{9}+\frac{1}{21}\cdot\frac{2}{9} = \frac{72}{189} \]

Step 3: Compute required probability (WB case).
\[ P(WB|W)=\frac{\frac{10}{21}\cdot\frac{3}{9}}{\frac{72}{189}} = \frac{10}{63} \]