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 Select the correct answer from above options

Answer 1

Here, P(A)=0.4 P(B)=0.3 and P(C)=0.2 And, P(A’)=1-P(A)=[1-0.4] = 0.6 P(B’)=1-P(B)=[1-0.3] = 0.7 P(C’)=1-P(C)=[1-0.2] = 0.8 P(E)=[P(A)×P(B)×P(C’)]+[P(A)×P(B’)×P(C)]+[P(A’)×P(B)×P(C)] [(0.4×0.3×0.8)+(0.4×0.7×0.2)+(0.6×0.3×0.2)] [0.96+0.056+0.036] 0.188 Hence, Probability of two hits is 0.188