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  4. Three persons, A , B and C fire at a target in turn, starting with A. Their probability of hitting the target are 0.4,0.3 and 0.2 respectively. The probability of two hits is

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Q. Three persons, $A , B$ and $C$ fire at a target in turn, starting with $A$. Their probability of hitting the target are $0.4,0.3$ and $0.2$ respectively. The probability of two hits is

Solution:

$P(A)=0.4 \,\,\, P(B)=0.3 \,\, P(C)=0.2 $
$P(\bar{A})=0.6 \,\, P(\bar{B})=0.7 \,\,\,P(\bar{C})=0.8$
Probability of two hits $=P(A) P(B) P(\bar{C})+P(A) P(\bar{B}) P(C)+P(\bar{A}) P(B) P(C)$
$=(0.4)(0.3)(0.8)+(0.4)(0.7)(0.8)+(0.6)(0.3)(0.2)$
$=0.096+0.056+0.036=0.188$