question_answer

A can hit target 4 times out of 5 times, B can hit target 3 times out of 4 times and C can hit target 2 times out of 3 times. They fire simultaneously. Find the probability that (i) Any two out of A, B and C will hit the target. (ii) None of them will hit the target.

Answer:

Here, P(A) = P(A hit the target) \[=\frac{4}{5},\] P(B)=P (B hit the target) \[=\frac{3}{4}\] and P(C) = (C hit the target) \[=\frac{2}{3}\] Then, \[P(\bar{A})=1-P(A)=1-\frac{4}{5}=\frac{1}{5}\] \[P(\bar{B})=1-P(B)=1-\frac{3}{4}=\frac{1}{4}\] and       \[P(\bar{C})=1-P(C)=1-\frac{2}{3}=\frac{1}{3}\] (i) P(any two of them hit the target) \[=P(A\cap B\cap \bar{C})+P(A\cap \bar{B}\cap C)+P(\bar{A}\cap B\cap C)\] \[=P(A)P(B)P(\bar{C})+P(A)P(\bar{B})P(C)+P(\bar{A})P(B)P(C)\]\[=\left( \frac{4}{5}\times \frac{3}{4}\times \frac{1}{3} \right)+\left( \frac{4}{5}\times \frac{1}{4}\times \frac{2}{3} \right)+\left( \frac{1}{5}\times \frac{3}{4}\times \frac{2}{3} \right)\] \[=\frac{12}{60}+\frac{8}{60}+\frac{6}{60}=\frac{26}{60}=\frac{13}{30}\] (ii) P(none Of them hit the target) \[=P(\bar{A})P(\bar{B})P(\bar{C})\] \[=\frac{1}{5}\times \frac{1}{4}\times \frac{1}{3}=\frac{1}{60}\]