For `k=1,2,3` the box `B_k` contains k red balls and `(k+1)` white balls. Let `P(B_1)=1/2,P(B_2)=1/3 and P(B_3)=1/6`.A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box `B_2`, is

Options

A. `35/78`
B. `14/39`
C. `10/13`
D. `12/13`

Answer

To solve the problem, we will use Bayes' theorem to find the probability that a red ball drawn came from box \( B_2 \).

### Step 1: Define the Events
Let:
- \( B_1 \): Event that box 1 is chosen.
- \( B_2 \): Event that box 2 is chosen.
- \( B_3 \): Event that box 3 is chosen.
- \( R \): Event that a red ball is drawn.

### Step 2: Given Probabilities
From the problem, we have:
- \( P(B_1) = \frac{1}{2} \)
- \( P(B_2) = \frac{1}{3} \)
- \( P(B_3) = \frac{1}{6} \)

### Step 3: Calculate the Total Number of Balls in Each Box
- For box \( B_1 \): 1 red and 2 white → Total = 3 balls.
- For box \( B_2 \): 2 red and 3 white → Total = 5 balls.
- For box \( B_3 \): 3 red and 4 white → Total = 7 balls.

### Step 4: Calculate the Probability of Drawing a Red Ball from Each Box
- Probability of drawing a red ball from \( B_1 \):
\[ P(R | B_1) = \frac{1}{3} \]

- Probability of drawing a red ball from \( B_2 \):
\[ P(R | B_2) = \frac{2}{5} \]

- Probability of drawing a red ball from \( B_3 \):
\[ P(R | B_3) = \frac{3}{7} \]

### Step 5: Apply Bayes' Theorem
We want to find \( P(B_2 | R) \):
\[ P(B_2 | R) = \frac{P(R | B_2) \cdot P(B_2)}{P(R)} \]

### Step 6: Calculate \( P(R) \)
Using the law of total probability:
\[ P(R) = P(R | B_1) \cdot P(B_1) + P(R | B_2) \cdot P(B_2) + P(R | B_3) \cdot P(B_3) \]
Substituting the values:
\[ P(R) = \left(\frac{1}{3} \cdot \frac{1}{2}\right) + \left(\frac{2}{5} \cdot \frac{1}{3}\right) + \left(\frac{3}{7} \cdot \frac{1}{6}\right) \]
Calculating each term:
- From \( B_1 \): \( \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \)
- From \( B_2 \): \( \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15} \)
- From \( B_3 \): \( \frac{3}{7} \cdot \frac{1}{6} = \frac{1}{14} \)

Now, we need a common denominator to sum these fractions. The least common multiple of 6, 15, and 14 is 210.

Converting each fraction:
- \( \frac{1}{6} = \frac{35}{210} \)
- \( \frac{2}{15} = \frac{28}{210} \)
- \( \frac{1}{14} = \frac{15}{210} \)

Adding these:
\[ P(R) = \frac{35 + 28 + 15}{210} = \frac{78}{210} = \frac{39}{105} \]

### Step 7: Substitute Back into Bayes' Theorem
Now substituting back into Bayes' theorem:
\[ P(B_2 | R) = \frac{P(R | B_2) \cdot P(B_2)}{P(R)} = \frac{\left(\frac{2}{5}\right) \cdot \left(\frac{1}{3}\right)}{\frac{78}{210}} \]
Calculating the numerator:
\[ = \frac{2}{15} \]

Now, substituting:
\[ P(B_2 | R) = \frac{\frac{2}{15}}{\frac{78}{210}} = \frac{2}{15} \cdot \frac{210}{78} = \frac{2 \cdot 14}{78} = \frac{28}{78} = \frac{14}{39} \]

### Final Answer
Thus, the probability that a red ball drawn came from box \( B_2 \) is:
\[ \boxed{\frac{14}{39}} \]