If on an average 1 vessel in every 10 is wrecked, the chance that out of 5 vessels expected 4 at least will arrive safely is

Probability of a vessel to arrive safely $=\frac{9}{10}$

Let $\mathrm{p}=\frac{9}{10}, \mathrm{q}=1-\frac{9}{10}=\frac{1}{10}, \mathrm{n}=5$

Now Probability of at least 4 arriving safely 

                $$\begin{gathered} =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\ ={}^5{\mathrm{c}}_4 {\left(\frac{9}{10}\right)}^4 {\left(\frac{1}{10}\right)}^1+{}^5{\mathrm{c}}_5 {\left(\frac{9}{10}\right)}^5 {\left(\frac{1}{10}\right)}^0 \\ =5\times \frac{9^4}{{10}^5}+1.\frac{9^5}{{10}^5} \\ =\frac{9^4}{{10}^5} (5+9) \\ =14 \left(\frac{9^4}{{10}^5}\right) \end{gathered}$$