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.
A
`5//21`
B
`6//31`
C
`7//41`
D
none of these
Text Solution
Verified by Experts
The correct Answer is:
B
`(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)`
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