Question

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 $\frac{45920+k}{50000}$. Find the value of $k$.

Answer 1

Let the probability of a vessel wrecking be $q$ and of safe arrival be $p$ so that $q=\frac{1}{10}$and $p=1-\frac{1}{10}=\frac{9}{10}$.
The probabilities of no vessel, one vessel, two vessel, etc., arriving safely are the first, second, third terms, etc., in the binomial expansion
$(q+p{)}^5=q^5+{}^5{C}_1{q}^{4}p+{}^5{C}_2{q}^3{p}^2{+}^5{C}_3{q}^2{p}^3{+}^5{C}_{4}q{p}^4+p^5.$
$\therefore$ The probability of at least $4$ vessels arriving safely is the sum of last two terms.
Hence the required probability $={}^5{C}_{4}Q{P}^4+P^5=5\frac{1}{10}{\left(\frac{9}{10}\right)}^4+{\left(\frac{9}{10}\right)}^5=\frac{45927}{50000}.$