Question

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

• A. 0.024
• B. 0.188
• C. 0.336
• D. 0.452

Answer 1

Answer: (B)

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