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(B) Let $p$ be the probability that a ship sinks,so $p = \frac{1}{9}$.Let $q$ be the probability that a ship arrives safely,so $q = 1 - \frac{1}{9} =

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${}^6 C_3 \frac{8^3}{9^6}$

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(B) Let $p$ be the probability that a ship sinks,so $p = \frac{1}{9}$.Let $q$ be the probability that a ship arrives safely,so $q = 1 - \frac{1}{9} = \frac{8}{9}$.We are looking for the probability that exactly $3$ ships sink out of $6$ ships.Using the binomial distribution formula $P(X=k) = {}^n C_k \cdot p^k \cdot q^{n-k}$,where $n=6$,$k=3$,$p=\frac{1}{9}$,and $q=\frac{8}{9}$:$P(X=3) = {}^6 C_3 \cdot \left(\frac{1}{9}\right)^3 \cdot \left(\frac{8}{9}\right)^3$$P(X=3) = {}^6 C_3 \cdot \frac{1}{9^3} \cdot \frac{8^3}{9^3}$$P(X=3) = {}^6 C_3 \cdot \frac{8^3}{9^6}$

One out of $9$ ships is likely to sink when they are set on sail. When $6$ ships are set on sail,the probability that exactly $3$ of them will not arrive safely is

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