91Ó°ÊÓ

Referring to the exercise 6,

There are three men A, B, and C will shoot a target.

A shoots three times and the probability that he will hit a target is, \({\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 8}}\right.\ } \!\lower0.7ex\hbox{$8$}}\).

B shoots five times and the probability that he will hit a target is, \({\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 4}}\right.\ } \!\lower0.7ex\hbox{$4$}}\).

C shoots twice times and the probability that he will hit a target is, \({\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\ } \!\lower0.7ex\hbox{$2$}}\).

Let, \({N_A},\,\,{N_B},\,\,and\,\,{N_C}\) denote the number of times each man hits the target and they are independent.

Then,

\(\begin{align}Var\left( {{N_A} + {N_B} + {N_C}} \right) &= Var\left( {{N_A}} \right) + Var\left( {{N_B}} \right) + Var\left( {{N_C}} \right)\\ &= \left( {3 \times \frac{1}{8} \times \left( {1 - \frac{1}{8}} \right)} \right) + \left( {5 \times \frac{1}{4} \times \left( {1 - \frac{1}{4}} \right)} \right) + \left( {2 \times \frac{1}{2} \times \left( {1 - \frac{1}{2}} \right)} \right)\\ &= \left( {3 \times \frac{1}{8} \times \frac{7}{8}} \right) + \left( {5 \times \frac{1}{4} \times \frac{3}{4}} \right) + \left( {2 \times \frac{1}{2} \times \frac{1}{2}} \right)\\ &= \frac{{113}}{{64}}\\ &= 1.765625\end{align}\)

Since, all shots at the target are independent, covariance terms are zero.

Therefore, the variance of the number of times is 1.765625.