The probability that a man who is 85 years old will die before attaining the age of 90 is 1/3. Four persons `A_1,A_2,A_3` and `A_4` are 85 years old. The probability that `A_1` will die before attaining the age of 90 and will be the first to die is

Options

A. `65/81`
B. `13/81`
C. `65/324`
D. `13/108`

Answer

To solve the problem, we need to find the probability that person \( A_1 \) will die before attaining the age of 90 and will be the first to die among the four persons \( A_1, A_2, A_3, \) and \( A_4 \).

### Step-by-Step Solution:

1. **Understanding the Given Probability**:
- The probability that a man who is 85 years old will die before reaching the age of 90 is given as \( P(E) = \frac{1}{3} \).
- Therefore, the probability that he will live beyond 90 years is \( P(E') = 1 - P(E) = 1 - \frac{1}{3} = \frac{2}{3} \).

2. **Probability of Each Person Dying**:
- Since there are four persons \( A_1, A_2, A_3, \) and \( A_4 \) who are equally likely to die first, the probability that any one of them dies first is \( P(\text{any one dies first}) = \frac{1}{4} \).

3. **Calculating the Probability of \( A_1 \) Dying First**:
- We need to find the probability that \( A_1 \) dies first and that all four persons die after the age of 90.
- The probability that all four persons die after the age of 90 is \( P(E')^4 = \left(\frac{2}{3}\right)^4 = \frac{16}{81} \).

4. **Combining the Probabilities**:
- The events of \( A_1 \) dying first and all four dying after 90 are independent events. Therefore, we can multiply their probabilities:
\[ P(A_1 \text{ dies first and all die after 90}) = P(A_1 \text{ dies first}) \times P(\text{all die after 90}) = \frac{1}{4} \times \frac{16}{81} = \frac{16}{324} = \frac{4}{81}. \]

5. **Considering the Two Cases**:
- Now we consider the two mutually exclusive cases for \( A_1 \):
- Case 1: \( A_1 \) dies before 90.
- Case 2: \( A_1 \) dies after 90.
- The probability that \( A_1 \) dies before 90 is \( P(E) = \frac{1}{3} \).
- The probability that \( A_1 \) dies after 90 is \( P(E') = \frac{2}{3} \).

6. **Final Calculation**:
- We can use the law of total probability:
\[ P(A_1 \text{ dies first}) = P(A_1 \text{ dies before 90}) + P(A_1 \text{ dies after 90}). \]
- We already calculated \( P(A_1 \text{ dies first and all die after 90}) = \frac{4}{81} \).
- The probability that \( A_1 \) dies before 90 and is the first to die is:
\[ P(A_1 \text{ dies first and before 90}) = P(A_1 \text{ dies first}) - P(A_1 \text{ dies first and all die after 90}) = \frac{1}{4} - \frac{4}{81}. \]
- To perform this subtraction, we need a common denominator:
\[ \frac{1}{4} = \frac{81}{324}, \quad \text{and} \quad \frac{4}{81} = \frac{16}{324}. \]
- Thus,
\[ P(A_1 \text{ dies first and before 90}) = \frac{81}{324} - \frac{16}{324} = \frac{65}{324}. \]

### Final Answer:
The probability that \( A_1 \) will die before attaining the age of 90 and will be the first to die is \( \frac{65}{324} \).