Suppose A and B shoot independently until each hits his target. They have probabilities 3/5 and 5/7 of hitting the targets at each shot. Find the probability than B will require more shots than A.

Options

A. `5//21`
B. `6//31`
C. `7//41`
D. none of these

Answer

B

Explanation

`(b)` If `A` takes `r` shot then `B` will take more than `r` shots so required probablities
`=sumP(A')^(r-1)*P(A)[P(B')^(r )*P(B)+P(B')^(r+1)*P(B)+....]`
`=sum(P(A'))^(r-1)*P(A)*((P(B'))^(r )P(B))/(1-P(B'))`
`=sum_(r=1)^(oo)(P(A'))^(r-1)*P(A)*(P(B'))^(r )`
`=(P(A)*P(B))/(1-P(A')*P(B'))=(3//5*2//7)/(1-2//5*2//7)=(6)/(31)`