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one in 9 ships is likely to be wrecked, when they are set on sail. When 6 ships set on sail the probability for exactly 3 will arrive safely is

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$1 + \frac{1}{9^{6}}$

$1 - \frac{1}{9^{6}}$

${6_{C_{3}}}_{} (\frac{8^{3}}{9^{6}})$

$6_{C_{3}} (\frac{8^{6}}{9^{3}})$

answer is C.

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Detailed Solution

$\begin{array}{l} q = \frac{1}{9}, p = \frac{8}{9}, n = 6, P (X = 3) = 6_{C_{3}} {(\frac{8}{9})}^{3} {(\frac{1}{9})}^{3} \\ = 6_{C_{3}} \frac{8^{3}}{9^{6}} \end{array}$

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