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)

$G _1 \rightarrow 5 ; G _2 \rightarrow 7 ; G _3 \rightarrow 4 ; G _4 \rightarrow 2 ;$ let $P ( G ) \rightarrow$ 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 $G _2$.