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 step by step, we will use the concept of conditional probability and the law of total probability.

### 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 \( P(B | W) \), the probability that both balls are drawn from box B given that one ball is white and the other is red.

### Step 2: Use 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 boxes are chosen at random:
\[ P(A) = P(B) = \frac{1}{2} \]

### Step 4: Calculate \( P(W | B) \)
To find \( P(W | B) \), we calculate the probability of drawing one white and one red ball from box B.

Box B contains:
- 4 white balls
- 2 red balls
- 3 black balls

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

The number of ways to choose 1 white and 1 red ball from box B is:
\[ \text{Ways to choose 1 white} = \binom{4}{1} = 4 \]
\[ \text{Ways to choose 1 red} = \binom{2}{1} = 2 \]
Thus, the total ways to choose 1 white and 1 red from box B is:
\[ 4 \times 2 = 8 \]

The total ways to choose any 2 balls from box B is:
\[ \binom{9}{2} = \frac{9 \times 8}{2} = 36 \]

Therefore,
\[ P(W | B) = \frac{8}{36} = \frac{2}{9} \]

### Step 5: Calculate \( P(W | A) \)
Now, we calculate \( P(W | A) \) for box A.

Box A contains:
- 2 white balls
- 3 red balls
- 2 black balls

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

The number of ways to choose 1 white and 1 red ball from box A is:
\[ \text{Ways to choose 1 white} = \binom{2}{1} = 2 \]
\[ \text{Ways to choose 1 red} = \binom{3}{1} = 3 \]
Thus, the total ways to choose 1 white and 1 red from box A is:
\[ 2 \times 3 = 6 \]

The total ways to choose any 2 balls from box A is:
\[ \binom{7}{2} = \frac{7 \times 6}{2} = 21 \]

Therefore,
\[ P(W | A) = \frac{6}{21} = \frac{2}{7} \]

### Step 6: Calculate \( P(W) \)
Now we can find \( P(W) \) using the law of total probability:
\[ P(W) = P(W | A) \cdot P(A) + P(W | B) \cdot P(B) \]
Substituting the values:
\[ P(W) = \left(\frac{2}{7} \cdot \frac{1}{2}\right) + \left(\frac{2}{9} \cdot \frac{1}{2}\right) \]
\[ = \frac{1}{7} + \frac{1}{9} \]
Finding a common denominator (63):
\[ = \frac{9}{63} + \frac{7}{63} = \frac{16}{63} \]

### Step 7: Substitute Values into Bayes' Theorem
Now substituting back into Bayes' theorem:
\[ P(B | W) = \frac{P(W | B) \cdot P(B)}{P(W)} = \frac{\left(\frac{2}{9}\right) \cdot \left(\frac{1}{2}\right)}{\frac{16}{63}} \]
\[ = \frac{\frac{2}{18}}{\frac{16}{63}} = \frac{2 \cdot 63}{18 \cdot 16} = \frac{126}{288} = \frac{21}{48} = \frac{7}{16} \]

### 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:
\[ \boxed{\frac{7}{16}} \]