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

Question

Philoid · Accepted Answer

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