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

Explanation

Let Xbe the number of times A shoots at the target to hit it for the first time and Y be the number of times B shoots at the target to hit for the first time. Then,
`P(X=m)=((2)/(5))^(m-1)((3)/(5))and P(Y=n)=((2)/(7))^(n-1)((5)/(7))`
We have `P(YgtX)=underset(m=1)overset(oo)sumunderset(n=m+1)overset(oo)sumP(X=m)P(Y=n)`
`" "[because"X and Y are independent"]`
`=underset(m=1)overset(oo)sum[{((2)/(5))^(m-1)((3)/(5))underset(n=m+1)overset(oo)sum{((2)/(7))^(n-1)((5)/(7))}]`
`=underset(m=1)overset(oo)sum((2)/(5))^(m-1)((3)/(5)){5/7(((2)/(7)))/(1-(2)/(3))}`
`=underset(m=1)overset(oo)sum((2)/(5))^(m-1)((3)/(5))((2)/(7))^(m)`
`=6/35underset(m=1)overset(oo)sum((4)/(35))^(m-1)=(6)/(35)(1)/(1-4/35)=6/31`