Question

Under the conditions of Exercise 6, assume also that all shots at the target are independent. What is the variance of the number of times that the target will be hit?

Short answer

The variance of the number of times that the target will be hit is 1.765625.

Step-by-step solution

Given information

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, \({\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\ } \!\lower0.7ex\hbox{$2$}}\).

Step 2:Calculation for the variance of the number of times that the target will be hit

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.