Let `A,B,C` be `3` events such that `P(A//B)=(1)/(5)`, `P(B)=(1)/(2)`, `P(A//C)=(2)/(7)` and `P(C )=(1)/(2)`, then `P(B//A)` is

Options

A. `(4)/(11)`
B. `(5)/(11)`
C. `(6)/(11)`
D. `(7)/(17)`

Explanation

To find \( P(B|A) \) using the given probabilities and Bayes' theorem, we can follow these steps:

### Step 1: Write down the known probabilities
We are given:
- \( P(A|B) = \frac{1}{5} \)
- \( P(B) = \frac{1}{2} \)
- \( P(A|C) = \frac{2}{7} \)
- \( P(C) = \frac{1}{2} \)

### Step 2: Apply Bayes' theorem
Bayes' theorem states that:
\[ P(B|A) = \frac{P(B) \cdot P(A|B)}{P(B) \cdot P(A|B) + P(C) \cdot P(A|C)} \]

### Step 3: Substitute the known values into the formula
Substituting the known values into the formula:
\[ P(B|A) = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{1}{5}\right)}{\left(\frac{1}{2} \cdot \frac{1}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{7}\right)} \]

### Step 4: Calculate the numerator
Calculating the numerator:
\[ \text{Numerator} = \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ \text{Denominator} = \left(\frac{1}{2} \cdot \frac{1}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{7}\right) \]
Calculating each term:
- First term: \( \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10} \)
- Second term: \( \frac{1}{2} \cdot \frac{2}{7} = \frac{1}{7} \)

Now, we need to add these two fractions:
\[ \text{Denominator} = \frac{1}{10} + \frac{1}{7} \]
To add these, we find a common denominator, which is 70:
\[ \frac{1}{10} = \frac{7}{70}, \quad \frac{1}{7} = \frac{10}{70} \]
Thus,
\[ \text{Denominator} = \frac{7}{70} + \frac{10}{70} = \frac{17}{70} \]

### Step 6: Putting it all together
Now we can substitute back into the formula for \( P(B|A) \):
\[ P(B|A) = \frac{\frac{1}{10}}{\frac{17}{70}} = \frac{1}{10} \cdot \frac{70}{17} = \frac{7}{17} \]

### Final Answer
Thus, the probability \( P(B|A) \) is:
\[ \boxed{\frac{7}{17}} \]