A husband and wife appear in an interview for two vacancies in the same post .The probability of husband 's selection is 1/7 and that of wife's 1/5
What is the probability that only one of them will be selected ?

Options

A. `(2)/(7)`
B. `(1)/(35)`
C. `(24)/(35)`
D. `(11)/(35)`

Answer

To find the probability that only one of the husband or wife will be selected for the vacancies, we can break the problem down into two cases:

### Step 1: Define the probabilities
- Let \( P(H) \) be the probability that the husband is selected. Given \( P(H) = \frac{1}{7} \).
- Let \( P(W) \) be the probability that the wife is selected. Given \( P(W) = \frac{1}{5} \).

### Step 2: Calculate the probabilities of not being selected
- The probability that the husband is not selected, \( P(H') \), is:
\[ P(H') = 1 - P(H) = 1 - \frac{1}{7} = \frac{6}{7} \]
- The probability that the wife is not selected, \( P(W') \), is:
\[ P(W') = 1 - P(W) = 1 - \frac{1}{5} = \frac{4}{5} \]

### Step 3: Calculate the probability of only one being selected
We need to consider two scenarios:
1. The husband is selected and the wife is not selected.
2. The wife is selected and the husband is not selected.

#### Case 1: Husband selected, wife not selected
The probability for this case is:
\[ P(H \text{ and } W') = P(H) \times P(W') = \frac{1}{7} \times \frac{4}{5} = \frac{4}{35} \]

#### Case 2: Wife selected, husband not selected
The probability for this case is:
\[ P(W \text{ and } H') = P(W) \times P(H') = \frac{1}{5} \times \frac{6}{7} = \frac{6}{35} \]

### Step 4: Combine the probabilities
Now, we add the probabilities of both cases to find the total probability that only one of them is selected:
\[ P(\text{Only one selected}) = P(H \text{ and } W') + P(W \text{ and } H') = \frac{4}{35} + \frac{6}{35} = \frac{10}{35} \]

### Step 5: Simplify the result
Now, we simplify \( \frac{10}{35} \):
\[ \frac{10}{35} = \frac{2}{7} \]

### Final Answer
Thus, the probability that only one of them will be selected is:
\[ \boxed{\frac{2}{7}} \]

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