Question

The probability that a missile hits a target successfully is $0.75.$ In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than $0.95$, is

Answer 1

Solve for the minimum number of missiles.

$P\left(H\right)=0.75=\frac{3}{4}$

$P\left(\overline{H}\right)=0.25=\frac{1}{4}$

Probability(target Hit) $\ge 0.95$

$1–P$(target not hit in n throws) $\ge 0.95$

$$\begin{gathered} {}^{n}C_3{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^{n-3}+{}^{n}C_4{\left(\frac{3}{4}\right)}^4{\left(\frac{1}{4}\right)}^{n-4}+...+{}^{n}C_n{\left(\frac{3}{4}\right)}^n{\left(\frac{1}{4}\right)}^0\ge \frac{95}{100} \\ 1-\left\{{}^{n}C_0{\left(\frac{3}{4}\right)}^0{\left(\frac{1}{4}\right)}^n+{}^{n}C_1{\left(\frac{3}{4}\right)}^1{\left(\frac{1}{4}\right)}^{n-1}+{}^{n}C_2{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^{n-2}\right\}\ge \frac{95}{100} \\ \Rightarrow 1-\frac{95}{100}\ge \frac{1}{4^n}+\frac{3n}{4^n}+\frac{9n(n-1)}{2\times 4^n} \\ \Rightarrow \frac{1}{20}\ge \frac{2+6n+9n^2-9n}{2\times 4^n} \\ \Rightarrow \frac{2^{2n+1}}{20}\ge 2-3n+9n^2 \\ \Rightarrow 2^{2n-1}\ge 10-15n+45n^2 \end{gathered}$$

$n=6$ satisfies the given condition.

Hence, the minimum number of missiles is $6.$