A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls. A third box `B_(3)` contains 3 white balls, 4 red balls, and 5 black balls.
If 2 balls are drawn (without replecement) from a randomly selected box and one of the balls is white and the other ball is red the probability that these 2 balls are drawn from box `B_(2)` is

Options

A. `116//182`
B. `126//181`
C. `65//181`
D. `55//181`

Explanation

To solve the problem step by step, we will use the concept of conditional probability and Bayes' theorem.

### Step 1: Define the events
Let:
- \( A \): Event that one ball is white and the other is red.
- \( E_1 \): Event that the balls are drawn from box \( B_1 \).
- \( E_2 \): Event that the balls are drawn from box \( B_2 \).
- \( E_3 \): Event that the balls are drawn from box \( B_3 \).

### Step 2: Calculate the total probability of drawing one white and one red ball from each box

#### From Box \( B_1 \):
- Box \( B_1 \) contains 1 white ball, 3 red balls, and 2 black balls (total = 6 balls).
- The number of ways to choose 1 white and 1 red ball:
\[ \text{Ways} = 1 \times 3 = 3 \]
- Total ways to choose 2 balls from 6:
\[ \binom{6}{2} = 15 \]
- Probability \( P(A | E_1) \):
\[ P(A | E_1) = \frac{3}{15} = \frac{1}{5} \]

#### From Box \( B_2 \):
- Box \( B_2 \) contains 2 white balls and 3 red balls (total = 5 balls).
- The number of ways to choose 1 white and 1 red ball:
\[ \text{Ways} = 2 \times 3 = 6 \]
- Total ways to choose 2 balls from 5:
\[ \binom{5}{2} = 10 \]
- Probability \( P(A | E_2) \):
\[ P(A | E_2) = \frac{6}{10} = \frac{3}{5} \]

#### From Box \( B_3 \):
- Box \( B_3 \) contains 3 white balls, 4 red balls, and 5 black balls (total = 12 balls).
- The number of ways to choose 1 white and 1 red ball:
\[ \text{Ways} = 3 \times 4 = 12 \]
- Total ways to choose 2 balls from 12:
\[ \binom{12}{2} = 66 \]
- Probability \( P(A | E_3) \):
\[ P(A | E_3) = \frac{12}{66} = \frac{2}{11} \]

### Step 3: Calculate the total probability of event \( A \)
Using the law of total probability:
\[ P(A) = P(A | E_1) P(E_1) + P(A | E_2) P(E_2) + P(A | E_3) P(E_3) \]
Since each box is equally likely to be chosen:
\[ P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \]
Thus,
\[ P(A) = \left(\frac{1}{5} \cdot \frac{1}{3}\right) + \left(\frac{3}{5} \cdot \frac{1}{3}\right) + \left(\frac{2}{11} \cdot \frac{1}{3}\right) \]
Calculating each term:
\[ P(A) = \frac{1}{15} + \frac{3}{15} + \frac{2}{33} \]
Finding a common denominator (which is 165):
\[ P(A) = \frac{11}{165} + \frac{33}{165} + \frac{10}{165} = \frac{54}{165} = \frac{18}{55} \]

### Step 4: Calculate the conditional probability \( P(E_2 | A) \) using Bayes' theorem
\[ P(E_2 | A) = \frac{P(A | E_2) P(E_2)}{P(A)} \]
Substituting the values:
\[ P(E_2 | A) = \frac{\left(\frac{3}{5}\right) \cdot \left(\frac{1}{3}\right)}{\frac{18}{55}} = \frac{\frac{3}{15}}{\frac{18}{55}} = \frac{3 \cdot 55}{15 \cdot 18} = \frac{11}{18} \]

### Final Answer
The probability that the two balls drawn are from box \( B_2 \) given that one is white and the other is red is:
\[ \boxed{\frac{11}{18}} \]