A box contains N coins, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is 1/2 while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair?
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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)`
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