Let `E_(1)` be the event that the coin drawn is fair and `E_(2)` e the event that the coin drawn is biased. Therefore,
`P(E_(1))=m/NandP(E_(2))=(N-m)/(N)`
A is the event that on tossing the coin, the head appears first and then appears the tail. therefore,
`P(A)=P(E_(1)nnA)+P(E_(2)nnA)`
`=P(E_(1))P(A//E_(1))+P(E_(2))P(A//E_(2))`
`=m/N((1)/(2))^(2)+((N-m)/(N))((2)/(3))((1)/(3))" "(1)`
We have to find the probability that A has happened because of `E_(1),` Therefore,
`P(E_(1)//A)=(P(E_(1)nnA))/(P(A))`
`=(m/N((1)/(2))^(2))/(m/N((1)/(2))^(2)+(N-m)/(N)((2)/(3))((1)/(3)))["using Eq."(1)]`
`=(m/4)/(m/4+(2(N-m))/(9))=(9m)/(m+8N)`