One out of 9 ships is likely to sink, when they are set on sail. When 6 ships are set on sail, the…
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
$1-\frac{1}{9^6}$
${ }^6 \mathrm{C}_3 \frac{8^3}{9^6}$
$\frac{25 \times 8^3}{9^5}$
${ }^6 \mathrm{C}_3 \frac{8}{9^6}$
Solution
Here, $p=\frac{1}{9} \Rightarrow q=1-\frac{1}{9}=\frac{8}{9}$
When out of 6 ships exactly 3 will not arrive safely then 3 will arrive safely.
So, the required probability is
$\begin{aligned} & p={ }^6 C_3 \cdot p^3 \cdot q^3={ }^6 C_3\left(\frac{1}{9}\right)^3\left(\frac{8}{9}\right)^3 \\ & ={ }^6 C_3 \frac{1}{9^3} \cdot \frac{8^3}{9^3} \Rightarrow p={ }^6 C_3 \cdot \frac{8^3}{9^6}\end{aligned}$