A box 'A' contains 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is :

Options

A. `7//16`
B. `7//8`
C. `9//16`
D. `9//32`

Answer

To solve the problem, we need to find the probability that both balls drawn are from box B given that one ball is white and the other ball is red. We will use Bayes' theorem for this.

### Step 1: Define the events
Let:
- \( A \): The event that both balls are drawn from box A.
- \( B \): The event that both balls are drawn from box B.
- \( W \): The event that one ball is white and the other is red.

We need to find \( P(B | W) \).

### Step 2: Apply Bayes' theorem
Using Bayes' theorem, we have:
\[ P(B | W) = \frac{P(W | B) \cdot P(B)}{P(W)} \]

### Step 3: Calculate \( P(B) \) and \( P(A) \)
Since the box is chosen at random:
\[ P(A) = P(B) = \frac{1}{2} \]

### Step 4: Calculate \( P(W | A) \)
From box A, we have:
- 2 white balls
- 3 red balls
- 2 black balls

The total number of balls in box A is \( 2 + 3 + 2 = 7 \).

To find \( P(W | A) \):
- The number of ways to choose 1 white and 1 red ball from box A is:
\[ \binom{2}{1} \cdot \binom{3}{1} = 2 \cdot 3 = 6 \]
- The total ways to choose 2 balls from box A is:
\[ \binom{7}{2} = 21 \]
Thus,
\[ P(W | A) = \frac{6}{21} = \frac{2}{7} \]

### Step 5: Calculate \( P(W | B) \)
From box B, we have:
- 4 white balls
- 2 red balls
- 3 black balls

The total number of balls in box B is \( 4 + 2 + 3 = 9 \).

To find \( P(W | B) \):
- The number of ways to choose 1 white and 1 red ball from box B is:
\[ \binom{4}{1} \cdot \binom{2}{1} = 4 \cdot 2 = 8 \]
- The total ways to choose 2 balls from box B is:
\[ \binom{9}{2} = 36 \]
Thus,
\[ P(W | B) = \frac{8}{36} = \frac{2}{9} \]

### Step 6: Calculate \( P(W) \)
Using the law of total probability:
\[ P(W) = P(W | A) \cdot P(A) + P(W | B) \cdot P(B) \]
Substituting the values:
\[ P(W) = \left(\frac{2}{7} \cdot \frac{1}{2}\right) + \left(\frac{2}{9} \cdot \frac{1}{2}\right) = \frac{1}{7} + \frac{1}{9} \]
To add these fractions, find a common denominator (63):
\[ P(W) = \frac{9}{63} + \frac{7}{63} = \frac{16}{63} \]

### Step 7: Substitute into Bayes' theorem
Now substituting back into Bayes' theorem:
\[ P(B | W) = \frac{P(W | B) \cdot P(B)}{P(W)} = \frac{\left(\frac{2}{9}\right) \cdot \left(\frac{1}{2}\right)}{\frac{16}{63}} = \frac{\frac{2}{18}}{\frac{16}{63}} = \frac{2}{18} \cdot \frac{63}{16} \]
This simplifies to:
\[ P(B | W) = \frac{63}{144} = \frac{7}{16} \]

### Final Answer
Thus, the probability that both balls are drawn from box B given that one ball is white and the other is red is:
\[ \boxed{\frac{7}{16}} \]