Question

An urn A contains 4 white and 1 black ball; urn B contains 3 white and 2 black balls; urn C contains 2 white and 3 black balls. One ball is transferred randomly from A to B; then one ball is transferred randomly from B to C. Finally, a ball is drawn randomly from C. Find the probability that it is black.

• A. \( \frac{7}{12} \)
• B. \( \frac{89}{180} \)
• C. \( \frac{101}{180} \)
• D. \( \frac{17}{36} \)

Hint:

Use the law of total probability to handle multi-stage random transfers. Calculate probabilities step-by-step for each sequence and sum weighted outcomes.

Answer 1

The Correct Option is C

Initial compositions: \[ A: 4W, 1B; B: 3W, 2B; C: 2W, 3B \] Step 1: Transfer one ball from A to B.
\[ P(W \text{ from A}) = \frac{4}{5}, P(B \text{ from A}) = \frac{1}{5} \] After transfer: \[ B' = \begin{cases} 4W, 2B & \text{if W transferred}
3W, 3B & \text{if B transferred} \end{cases} \] Step 2: Transfer one ball from B' to C.
\[ \begin{cases} P(W \text{ from } B') = \frac{4}{6}, P(B \text{ from } B') = \frac{2}{6} & \text{if } B' = 4W,2B
P(W \text{ from } B'') = \frac{3}{6}, P(B \text{ from } B'') = \frac{3}{6} & \text{if } B'' = 3W,3B \end{cases} \] Step 3: Final composition of C and probability of drawing black: \[ \begin{aligned} & P(\text{Black} | W \to B, W \to C) = \frac{3}{6} = \frac{1}{2}, P(\text{Black} | W \to B, B \to C) = \frac{4}{6} = \frac{2}{3},
& P(\text{Black} | B \to B, W \to C) = \frac{3}{6} = \frac{1}{2}, P(\text{Black} | B \to B, B \to C) = \frac{4}{6} = \frac{2}{3}.
\end{aligned} \] Step 4: Total probability using law of total probability: \[ \begin{aligned} P(\text{Black}) &= \frac{4}{5} \times \frac{4}{6} \times \frac{1}{2} + \frac{4}{5} \times \frac{2}{6} \times \frac{2}{3}
& + \frac{1}{5} \times \frac{3}{6} \times \frac{1}{2} + \frac{1}{5} \times \frac{3}{6} \times \frac{2}{3}
&= \frac{4}{15} + \frac{8}{45} + \frac{1}{20} + \frac{1}{15} = \frac{101}{180}.
\end{aligned} \] This matches option (3).