An insurance agent insures lives of 5 men, all of the same age and in good health. The probability that a man of this age will survive tha next 30 years is known to be `(2)/(3)`. The probability that in the next 30 years at most three men will survive is

Options

A. `131/243`
B. `80/243`
C. `32/243`
D. `163/243`

Answer

To solve the problem of finding the probability that at most three out of five men will survive the next 30 years, we can use the binomial distribution formula. Here’s a step-by-step solution:

### Step 1: Identify the parameters of the binomial distribution
- Let \( n = 5 \) (the number of men).
- The probability of survival for each man is \( p = \frac{2}{3} \).
- The probability of not surviving is \( q = 1 - p = \frac{1}{3} \).

### Step 2: Define the event we are interested in
We want to find the probability that at most three men survive, which can be represented as:
\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]

### Step 3: Use the binomial probability formula
The binomial probability formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

### Step 4: Calculate each probability
1. **For \( P(X = 0) \)**:
\[ P(X = 0) = \binom{5}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^{5} = 1 \cdot 1 \cdot \frac{1}{243} = \frac{1}{243} \]

2. **For \( P(X = 1) \)**:
\[ P(X = 1) = \binom{5}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^{4} = 5 \cdot \frac{2}{3} \cdot \frac{1}{81} = \frac{10}{243} \]

3. **For \( P(X = 2) \)**:
\[ P(X = 2) = \binom{5}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^{3} = 10 \cdot \frac{4}{9} \cdot \frac{1}{27} = \frac{40}{243} \]

4. **For \( P(X = 3) \)**:
\[ P(X = 3) = \binom{5}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^{2} = 10 \cdot \frac{8}{27} \cdot \frac{1}{9} = \frac{80}{243} \]

### Step 5: Sum the probabilities
Now, we sum the probabilities calculated:
\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]
\[ = \frac{1}{243} + \frac{10}{243} + \frac{40}{243} + \frac{80}{243} = \frac{131}{243} \]

### Final Answer
Thus, the probability that at most three men will survive the next 30 years is:
\[ \boxed{\frac{131}{243}} \]