The probability that a man aged x years will die in a year is p. The probability that out of n men`M_1, M_2, M_3, …… M_n`each aged years. `M_k`will die and be the first to die is-

Options

A. `1/n^2`
B. `1-(1-p)^n`
C. `1/(n [1-(1-p)^n])`
D. `1/n [1-(1-p)^n]`

Answer

To solve the problem, we need to determine the probability that a specific man, \( M_k \), out of \( n \) men, will die and be the first to die within a year given that the probability of any man aged \( x \) dying in a year is \( p \).

### Step-by-Step Solution:

1. **Understanding the Probability of Dying**:
- The probability that a man aged \( x \) will die in a year is given as \( p \).
- Therefore, the probability that he does not die in a year is \( 1 - p \).

2. **Probability that at least one man dies**:
- For \( n \) men, the probability that none of them die in a year is \( (1 - p)^n \).
- Thus, the probability that at least one man dies in a year is:
\[ P(\text{at least one dies}) = 1 - (1 - p)^n \]

3. **Probability that \( M_k \) dies and is the first to die**:
- For \( M_k \) to be the first to die, \( M_k \) must die, and all other \( n-1 \) men must survive the year.
- The probability that \( M_k \) dies is \( p \).
- The probability that each of the other \( n-1 \) men survives is \( (1 - p)^{n-1} \).

4. **Combining the Probabilities**:
- Since the events are independent, the probability that \( M_k \) dies and is the first to die is given by:
\[ P(M_k \text{ dies first}) = P(M_k \text{ dies}) \times P(\text{all others survive}) = p \times (1 - p)^{n-1} \]

5. **Final Result**:
- Therefore, the probability that \( M_k \) will die and be the first to die is:
\[ P(M_k \text{ is first to die}) = p \times (1 - p)^{n-1} \]

### Summary:
The probability that \( M_k \) will die and be the first to die among \( n \) men is given by:
\[ P(M_k \text{ is first to die}) = p \times (1 - p)^{n-1} \]