Question

There are 18 hunters which are divided into four groups as,
$G_1$ : Consisting of 5 hunters each of whom can hit a target with the probability 0.8 .
$G_2$ : Consisting of 7 each hitting a target with probability 0.7 .
$G_3$ : Consisting of 4 each hitting a target with probability 0.6 , and
$G_4$ : Consisting of 2 each hitting a target with probability 0.5 .
A randomly selected hunter fires a shot and failed to hit the target. Most probable group of the hunter is

• A. $G_1$
• B. $G_2$
• C. $G_3$
• D. $G_4$

Answer 1

Answer: (B)

$G_1\to 5;G_2\to 7;G_3\to 4;G_4\to 2;$ let $P(G)\to$ Probability that a randomly selected the hunter belongs to the group $G$.
$\therefore P\left(G_1\right)=\frac{5}{18};P\left(G_2\right)=\frac{7}{18};P\left(G_3\right)=\frac{4}{18};P\left(G_4\right)=\frac{2}{18}$
$F$ : hunter failed to hit the target
$P(F)=P\left(G_1\cap F\right)+P\left(G_2\cap F\right)+P\left(G_3\cap F\right)+P\left(G_4\cap F\right)$
$P(F)=P\left(G_1\right)\cdot P\left(F/G_1\right)+P\left(G_2\right)\cdot P\left(F/G_2\right)+P\left(G_3\right)\cdot P\left(F/G_3\right)+P\left(G_4\right)P\left(F/G_4\right)$
$=\frac{5}{18}\cdot \left(\frac{2}{10}\right)+\frac{7}{18}\left(\frac{3}{10}\right)+\left(\frac{4}{18}\right)\left(\frac{4}{10}\right)+\frac{2}{18}\left(\frac{5}{10}\right)=\frac{10+21+16+10}{180}=\frac{57}{180}$
$P\left(G_1/F\right)=\frac{10}{57};P\left(G_2/F\right)=\frac{21}{57};P\left(G_3/F\right)=\frac{16}{57};P\left(G_4/F\right)=\frac{10}{57}$
Hence, most probably group of hunter is $G_2$.