The probability that a ship safety reaches a port is `(1)/(3)`. The probability that out of 5 ships, at least 4 ships would arrive safely is:

Options

A. `(1)/(243)`
B. `(10)/(243)`
C. `(11)/(243)`
D. `(13)/(243)`

Answer

To solve the problem, we need to calculate the probability that at least 4 out of 5 ships arrive safely at the port, given that the probability of a ship arriving safely is \( \frac{1}{3} \).

### Step-by-Step Solution:

1. **Identify the probabilities**:
- The probability that a ship arrives safely, \( P(S) = \frac{1}{3} \).
- The probability that a ship does not arrive safely, \( P(N) = 1 - P(S) = 1 - \frac{1}{3} = \frac{2}{3} \).

2. **Define the event**:
- We need to find the probability of at least 4 ships arriving safely. This can be expressed as the sum of the probabilities of exactly 4 ships arriving safely and exactly 5 ships arriving safely.

3. **Use the binomial probability formula**:
The probability of exactly \( k \) successes (ships arriving safely) in \( n \) trials (total ships) is given by:
\[ P(X = k) = \binom{n}{k} (P(S))^k (P(N))^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

4. **Calculate for exactly 4 ships**:
- Here, \( n = 5 \) and \( k = 4 \):
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{5-4} \]
- Calculate \( \binom{5}{4} = 5 \):
\[ P(X = 4) = 5 \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = 5 \cdot \frac{2}{243} = \frac{10}{243} \]

5. **Calculate for exactly 5 ships**:
- Here, \( n = 5 \) and \( k = 5 \):
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^{5-5} \]
- Calculate \( \binom{5}{5} = 1 \):
\[ P(X = 5) = 1 \cdot \left(\frac{1}{3}\right)^5 \cdot 1 = \frac{1}{243} \]

6. **Sum the probabilities**:
- Now, we add the probabilities of exactly 4 and exactly 5 ships arriving safely:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243} \]

### Final Answer:
The probability that at least 4 out of 5 ships arrive safely is \( \frac{11}{243} \).

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