The probability that Ajay will go to office is `1/5` and probability that Ajay and Vijay will not go to office is `2/7` if their visit to office is independent of each other, then find the probability that Ajay will go to the office, but Vijay will not go, is?

Options

A. `1/14`
B. `1/17`
C. `1/20`
D. `1/18`

Answer

To solve the problem, we need to find the probability that Ajay goes to the office while Vijay does not go to the office, given their independent probabilities of going to the office.

### Step-by-Step Solution:

1. **Identify the given probabilities:**
- The probability that Ajay goes to the office, \( P(A) = \frac{1}{5} \).
- The probability that Ajay and Vijay do not go to the office, \( P(A') \cap P(V') = \frac{2}{7} \).

2. **Calculate the probability that Ajay does not go to the office:**
- The probability that Ajay does not go to the office, \( P(A') = 1 - P(A) = 1 - \frac{1}{5} = \frac{4}{5} \).

3. **Use the independence of events:**
- Since the events are independent, we can express the joint probability of Ajay and Vijay not going to the office as:
\[ P(A') \cap P(V') = P(A') \cdot P(V') \]
- We know \( P(A') = \frac{4}{5} \) and we can denote \( P(V') \) as \( x \). Therefore,
\[ P(A') \cdot P(V') = \frac{4}{5} \cdot x = \frac{2}{7} \]

4. **Solve for \( P(V') \):**
- Rearranging the equation gives:
\[ x = \frac{2}{7} \cdot \frac{5}{4} = \frac{10}{28} = \frac{5}{14} \]
- Thus, \( P(V') = \frac{5}{14} \).

5. **Calculate the probability that Vijay goes to the office:**
- The probability that Vijay goes to the office, \( P(V) = 1 - P(V') = 1 - \frac{5}{14} = \frac{9}{14} \).

6. **Find the probability that Ajay goes to the office and Vijay does not:**
- The probability that Ajay goes to the office and Vijay does not go to the office is given by:
\[ P(A) \cdot P(V') = \frac{1}{5} \cdot \frac{5}{14} = \frac{1 \cdot 5}{5 \cdot 14} = \frac{1}{14} \]

### Final Answer:
The probability that Ajay will go to the office but Vijay will not go is \( \frac{1}{14} \).