Question

Three urns respectively contain 2 white and 3 black, 3 white and 2 black and 1 white and 4 black balls. If one ball is drawn from each um, then the probability that the selection contains 1 black and 2 white balls is

• A. $\frac{13}{125}$
• B. $\frac{37}{125}$
• C. $\frac{28}{125}$
• D. $\frac{33}{125}$

Hint:

List all combinations that satisfy the condition (e.g., BWW, WBW, WWB).

Answer 1

The Correct Option is C

Step 1: Individual Probabilities
Urn 1: $P(W_1) = 2/5, P(B_1) = 3/5$
Urn 2: $P(W_2) = 3/5, P(B_2) = 2/5$
Urn 3: $P(W_3) = 1/5, P(B_3) = 4/5$
Step 2: Selection Cases (1B, 2W)
- Case 1 ($B_1 W_2 W_3$): $\frac{3}{5} \times \frac{3}{5} \times \frac{1}{5} = \frac{9}{125}$
- Case 2 ($W_1 B_2 W_3$): $\frac{2}{5} \times \frac{2}{5} \times \frac{1}{5} = \frac{4}{125}$
- Case 3 ($W_1 W_2 B_3$): $\frac{2}{5} \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{125}$
Step 3: Total Probability
Sum $= \frac{9 + 4 + 24}{125} = \frac{37}{125}$.
*Recalculating sum:* $9+4+24 = 37$. (Option B).
Final Answer: (B)