A) 0.06
B) 0.14
C) 0.2
D) 0.7
Correct Answer: D
Solution :
Firstly, find the probability of incorrect target. Let the events A = 1st aeroplane hit the target, B = 2nd aeroplane hit the target and their corresponding probabilities are \[p(A)=0.3\]and\[p(B)=0.2\] \[p(\overline{A})=0.7\]and\[p(\overline{B})=0.8\] The desired probability \[=P(\overline{A})P(B)+P(\overline{A})P(\overline{B})P(\overline{A})P(B)+....\] \[=(0.7)(0.2)+(0.7)(0.8)(0.7)(0.2)\] \[+(0.7)(0.8)(0.7)(0.8)(0.7)(0.2)+...\] \[=0.14[1+(0.56)+{{(0.56)}^{2}}+...]\] \[=0.14\left[ \frac{1}{1-0.56} \right]=\frac{0.14}{0.44}=\frac{7}{22}=0.32\] From above, it is clear that no option is correct.You need to login to perform this action.
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