If on an average 1 vessel in every 10 is wrecked, the chance that out of 5 vessels expected 4 at least will arrive sefely is

Options

A. `3(9^(4)//10^(5))`
B. `4(9^(4)//10^(5))`
C. `(9^(4)//10^(5))`
D. `14(9^(4)//10^(5))`

Answer

To solve the problem, we need to find the probability that at least 4 out of 5 vessels arrive safely when the probability of a vessel being wrecked is \( \frac{1}{10} \).

### Step-by-Step Solution:

1. **Identify the probabilities**:
- The probability that a vessel is wrecked, \( P(W) = \frac{1}{10} \).
- The probability that a vessel is safe, \( P(S) = 1 - P(W) = 1 - \frac{1}{10} = \frac{9}{10} \).

2. **Define the scenario**:
- We want to find the probability that at least 4 vessels arrive safely out of 5. This means we need to calculate:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) \]
where \( X \) is the number of vessels that arrive safely.

3. **Calculate \( P(X = 4) \)**:
- The number of ways to choose 4 vessels out of 5 is given by \( \binom{5}{4} \).
- The probability that exactly 4 vessels are safe (and 1 is wrecked) is:
\[ P(X = 4) = \binom{5}{4} \left( \frac{9}{10} \right)^4 \left( \frac{1}{10} \right)^1 \]
- Calculating \( \binom{5}{4} = 5 \):
\[ P(X = 4) = 5 \left( \frac{9}{10} \right)^4 \left( \frac{1}{10} \right) \]

4. **Calculate \( P(X = 5) \)**:
- The probability that all 5 vessels are safe is:
\[ P(X = 5) = \left( \frac{9}{10} \right)^5 \]

5. **Combine the probabilities**:
- Now, we can combine both probabilities:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) \]
- Substituting the values we calculated:
\[ P(X \geq 4) = 5 \left( \frac{9}{10} \right)^4 \left( \frac{1}{10} \right) + \left( \frac{9}{10} \right)^5 \]

6. **Factor out common terms**:
- We can factor out \( \left( \frac{9}{10} \right)^4 \):
\[ P(X \geq 4) = \left( \frac{9}{10} \right)^4 \left( 5 \cdot \frac{1}{10} + \frac{9}{10} \right) \]
- Simplifying the expression inside the parentheses:
\[ P(X \geq 4) = \left( \frac{9}{10} \right)^4 \left( \frac{5}{10} + \frac{9}{10} \right) = \left( \frac{9}{10} \right)^4 \left( \frac{14}{10} \right) \]

7. **Final probability**:
- Thus, the final probability is:
\[ P(X \geq 4) = \frac{14}{10} \left( \frac{9}{10} \right)^4 = \frac{14 \cdot 9^4}{10^5} \]

### Final Answer:
\[ P(X \geq 4) = \frac{14 \cdot 9^4}{10^5} \]