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Step-by-Step Solution
Step 1: Understand the Problem
We have two boxes:
• Box A with 2 white, 3 red, and 2 black balls (total 7 balls).
• Box B with 4 white, 2 red, and 3 black balls (total 9 balls).
A box is selected at random (so each box has probability 1/2 of being chosen). Then two balls are drawn from the chosen box, without replacement. We know that the outcome is one white ball and one red ball (in any order). We need to find the probability that these two balls came from box B, given this observed outcome.
Step 2: Probability of Drawing One White and One Red from Each Box
Let us first calculate the probability of drawing one white and one red from each box (in a specific order: first white then red). Because the question only states “one white and one red,” we will handle the order carefully, or notice that these computations can be adapted for any order if needed. However, we will follow the given short method that effectively handles one specific order in both cases and uses ratio arguments.
Box B:
Total balls: 9.
Number of ways to choose 2 balls from 9: $^9C_2$.
Number of ways to choose 1 white (from 4) and then 1 red (from 2): $^4C_1 \times {}^2C_1$.
Hence, selecting one white and one red (in that order) from Box B is:
$$
\frac{^4C_1 \times {}^2C_1}{^9C_2}
= \frac{4 \times 2}{\frac{9 \times 8}{2}}
= \frac{8}{36}
= \frac{2}{9}.
$$
Box A:
Total balls: 7.
Number of ways to choose 2 balls from 7: $^7C_2$.
Number of ways to choose 1 white (from 2) and then 1 red (from 3): $^2C_1 \times {}^3C_1$.
Hence, selecting one white and one red (in that order) from Box A is:
$$
\frac{^2C_1 \times {}^3C_1}{^7C_2}
= \frac{2 \times 3}{\frac{7 \times 6}{2}}
= \frac{6}{21}
= \frac{2}{7}.
$$
Step 3: Use the Conditional Probability to Find the Required Probability
We are given that the drawn balls (from whichever box was chosen) turned out to be one white and one red. We want the probability that those two balls came from box B given this event. For a random selection of the box, the relevant ratio for conditional probability becomes:
$$
P(\text{chosen box B} \mid \text{1 white and 1 red}) \;=\; \frac{P(\text{1 white, 1 red} \mid \text{box B})}{P(\text{1 white, 1 red} \mid \text{box A}) + P(\text{1 white, 1 red} \mid \text{box B})}.
$$
From our computations:
$P(\text{1 white, 1 red} \mid \text{box B}) = \frac{2}{9}.$
$P(\text{1 white, 1 red} \mid \text{box A}) = \frac{2}{7}.$
The fraction then becomes:
$$
\frac{\frac{2}{9}}
{\frac{2}{7} + \frac{2}{9}}
= \frac{\frac{2}{9}}
{\frac{2 \times 9 + 2 \times 7}{63}}
= \frac{\frac{2}{9}}
{\frac{18 + 14}{63}}
= \frac{\frac{2}{9}}
{\frac{32}{63}}
= \frac{2}{9} \times \frac{63}{32}
= \frac{2 \times 63}{9 \times 32}
= \frac{126}{288}
= \frac{7}{16}.
$$
Step 4: Conclude the Answer
Thus, the probability that the two drawn balls (one white and one red) are from Box B is
$$\frac{7}{16}.$$
