Question

The probability of a man hitting a target is 1/4. If he fires 7 times, what is the probability of his hitting the target at least twice?

Answer 1

Let X be number of times the target is hit. Then, X follows a binomial distribution with n =7, $p\mathit{}=\frac{1}{4}\mathrm{and}q=\frac{3}{4}$

$$\begin{gathered} P(X=r)={}^{\mathit{7}}C_r{\left(\frac{1}{4}\right)}^{\mathrm{r}}{\left(\frac{3}{4}\right)}^{7-r} \\ \mathrm{P}(\mathrm{hitting}\mathrm{the}\mathrm{target}\mathrm{at}\mathrm{least}\mathrm{twice}) \\ =P(X\ge 2) \\ =1-\left\{P(X=0)+P(X=1)\right\} \\ =1-{}^{7}C_0{\left(\frac{1}{4}\right)}^0{\left(\frac{3}{4}\right)}^{7-0}-{}^{7}C_1{\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^{7-1} \\ =1-{\left(\frac{3}{4}\right)}^7-7\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^6 \\ =1-\frac{1}{16384}(2187+5103) \\ =1-\frac{3645}{8192} \\ =\frac{4547}{8192} \end{gathered}$$