Step 1: Understanding the Concept:
We draw one ball from each of the three urns. We want exactly 1 black and 2 white balls in total. There are three mutually exclusive ways this can happen based on which urn provides the black ball.
Step 2: Key Formula or Approach:
Let $W_i$ and $B_i$ denote drawing a white or black ball from the $i^{th}$ urn.
$P(W_1) = 2/5, P(B_1) = 3/5$
$P(W_2) = 3/5, P(B_2) = 2/5$
$P(W_3) = 1/5, P(B_3) = 4/5$
Step 3: Detailed Explanation:
The required probability $P$ is the sum of probabilities of these three scenarios:
1. (Black from Urn 1, White from 2, White from 3):
$P_1 = P(B_1) \cdot P(W_2) \cdot P(W_3) = \frac{3}{5} \times \frac{3}{5} \times \frac{1}{5} = \frac{9}{125}$
2. (White from Urn 1, Black from 2, White from 3):
$P_2 = P(W_1) \cdot P(B_2) \cdot P(W_3) = \frac{2}{5} \times \frac{2}{5} \times \frac{1}{5} = \frac{4}{125}$
3. (White from Urn 1, White from 2, Black from 3):
$P_3 = P(W_1) \cdot P(W_2) \cdot P(B_3) = \frac{2}{5} \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{125}$
Total probability $= P_1 + P_2 + P_3 = \frac{9 + 4 + 24}{125} = \frac{37}{125}$
Step 4: Final Answer:
The probability of selecting 1 black and 2 white balls is $\frac{37}{125}$.