Question

Three men A, B, and C shoot at a target. Suppose that A shoots three times and the probability that he will hit the target on any given shot is\({\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{8}}}}\right.\ } \!\lower0.7ex\hbox{\({\bf{8}}\)}}\), B shoots five times and the probability that he will hit the target on any given shot is\({\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{4}}}}\right.\ } \!\lower0.7ex\hbox{\({\bf{4}}\)}}\), and C shoots twice and the probability that he will hit the target on any given shot is\({\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{2}}}}\right.\ } \!\lower0.7ex\hbox{\({\bf{2}}\)}}\). What is the expected number of times that the target will be hit?

Short answer

The expected number of times the target will be hit is 2.625.

Step-by-step solution

Given information

There are three men A, B, and C who shoot the target.

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

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

Calculation for the expected 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.

Then,

\(\begin{align}E\left( {{N_A} + {N_B} + {N_C}} \right) &= E\left( {{N_A}} \right) + E\left( {{N_B}} \right) + E\left( {{N_C}} \right)\\ &= \frac{3}{8} + \frac{5}{4} + \frac{2}{2}\\ &= \frac{{21}}{8}\\ &= 2.625\end{align}\)

Therefore, the expected number of times the target will hit is 2.625.