One in 9 ships is likely to be wrecked when they are set on sail. When 6 ships are on sail, find the probability for (i) At least one will arrive safely (ii) Exactly three will arr... One in 9 ships is likely to be wrecked when they are set on sail. When 6 ships are on sail, find the probability for (i) At least one will arrive safely (ii) Exactly three will arrive safely. Read more

Steps to Solve

1. Define key variables

Let $p$ be the probability of a ship being wrecked, and $q$ be the probability of a ship arriving safely. Let $n$ be the total number of ships that set sail. Let $X$ be the random variable representing the number of ships arriving safely.

Given: $p = \frac{1}{9}$, so $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$. Also, $n = 6$.

2. Probability of at least one ship arriving safely

This is equivalent to 1 minus the probability that none of the ships arrive safely (all are wrecked). The probability that all ships are wrecked is $p^n = \left(\frac{1}{9}\right)^6$. Therefore, the probability of at least one ship arriving safely is $1 - \left(\frac{1}{9}\right)^6$.

3. Probability of exactly three ships arriving safely

We use the binomial probability formula: $P(X=k) = \binom{n}{k} q^k p^{n-k}$, where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

In our case, we want to find $P(X=3)$, so $k=3$. $P(X=3) = \binom{6}{3} \left(\frac{8}{9}\right)^3 \left(\frac{1}{9}\right)^{6-3} = \binom{6}{3} \left(\frac{8}{9}\right)^3 \left(\frac{1}{9}\right)^3$

4. Calculate the binomial coefficient

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

5. Calculate $P(X=3)$

$P(X=3) = 20 \times \left(\frac{8}{9}\right)^3 \left(\frac{1}{9}\right)^3 = 20 \times \frac{8^3}{9^3} \times \frac{1}{9^3} = 20 \times \frac{512}{9^6} = \frac{20 \times 512}{531441} = \frac{10240}{531441}$

6. Summarize the results

Probability of at least one ship arriving safely: $1 - \left(\frac{1}{9}\right)^6 = 1 - \frac{1}{531441} = \frac{531440}{531441}$. Probability of exactly three ships arriving safely: $\frac{10240}{531441}$.