One in 9 ships is likely to be wrecked when they are set on a sail. When 6 ships are on sail, find the probability for
i) atleast one will arrive safely
ii) exactly three will arrive safely.

Answer

To solve the problem, we need to determine the probabilities based on the information given about the ships.

### Given:
- Probability of a ship being wrecked (failure) \( q = \frac{1}{9} \)
- Probability of a ship arriving safely (success) \( p = 1 - q = 1 - \frac{1}{9} = \frac{8}{9} \)
- Number of ships \( n = 6 \)

### Part (i): Probability that at least one ship will arrive safely

To find the probability that at least one ship arrives safely, we can use the complement rule. The complement of "at least one ship arrives safely" is "no ships arrive safely."

1. **Calculate the probability that no ships arrive safely (all ships are wrecked)**:
\[ P(X = 0) = \binom{n}{0} p^0 q^n = \binom{6}{0} \left(\frac{8}{9}\right)^0 \left(\frac{1}{9}\right)^6 \]
\[ P(X = 0) = 1 \cdot 1 \cdot \left(\frac{1}{9}\right)^6 = \frac{1}{9^6} \]

2. **Use the complement to find the probability that at least one ship arrives safely**:
\[ P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{1}{9^6} \]

### Part (ii): Probability that exactly three ships will arrive safely

To find the probability that exactly three ships arrive safely, we can use the binomial probability formula:

1. **Calculate the probability for exactly three ships arriving safely**:
\[ P(X = 3) = \binom{n}{3} p^3 q^{n-3} = \binom{6}{3} \left(\frac{8}{9}\right)^3 \left(\frac{1}{9}\right)^{6-3} \]
\[ P(X = 3) = \binom{6}{3} \left(\frac{8}{9}\right)^3 \left(\frac{1}{9}\right)^3 \]

2. **Calculate the binomial coefficient and substitute values**:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
\[ P(X = 3) = 20 \cdot \left(\frac{8}{9}\right)^3 \cdot \left(\frac{1}{9}\right)^3 \]
\[ P(X = 3) = 20 \cdot \frac{512}{729} \cdot \frac{1}{729} = 20 \cdot \frac{512}{729^2} \]

### Final Answers:
- **Part (i)**: The probability that at least one ship will arrive safely is:
\[ P(X \geq 1) = 1 - \frac{1}{9^6} \]

- **Part (ii)**: The probability that exactly three ships will arrive safely is:
\[ P(X = 3) = 20 \cdot \frac{512}{729^2} \]