The probability of any ship return safely to the port is 1/5 . Find the probability that the returning out of 5 ships, at least 3 ships returns safely. It is given that returning ships are independently.

Options

A. `256//625`
B. `369//625`
C. `248//625`
D. None of these

Explanation

To solve the problem, we need to find the probability that at least 3 out of 5 ships return safely to the port, given that the probability of any ship returning safely is \( \frac{1}{5} \).

### Step-by-Step Solution:

1. **Define the probabilities**:
- Let \( p = \frac{1}{5} \) (the probability that a ship returns safely).
- Therefore, the probability that a ship does not return safely is \( q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5} \).

2. **Identify the required event**:
- We need to find the probability that at least 3 ships return safely. This can be expressed as:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]
where \( X \) is the number of ships that return safely.

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

4. **Calculate \( P(X = 3) \)**:
- For \( k = 3 \):
\[ P(X = 3) = \binom{5}{3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^{5-3} \]
\[ = 10 \cdot \left(\frac{1}{5}\right)^3 \cdot \left(\frac{4}{5}\right)^2 \]
\[ = 10 \cdot \frac{1}{125} \cdot \frac{16}{25} = 10 \cdot \frac{16}{3125} = \frac{160}{3125} \]

5. **Calculate \( P(X = 4) \)**:
- For \( k = 4 \):
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{5-4} \]
\[ = 5 \cdot \left(\frac{1}{5}\right)^4 \cdot \left(\frac{4}{5}\right)^1 \]
\[ = 5 \cdot \frac{1}{625} \cdot \frac{4}{5} = 5 \cdot \frac{4}{3125} = \frac{20}{3125} \]

6. **Calculate \( P(X = 5) \)**:
- For \( k = 5 \):
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{5}\right)^5 \left(\frac{4}{5}\right)^{5-5} \]
\[ = 1 \cdot \left(\frac{1}{5}\right)^5 \cdot 1 = \frac{1}{3125} \]

7. **Combine the probabilities**:
- Now, add the probabilities:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]
\[ = \frac{160}{3125} + \frac{20}{3125} + \frac{1}{3125} = \frac{181}{3125} \]

### Final Answer:
The probability that at least 3 out of 5 ships return safely is \( \frac{181}{3125} \).