A box contains N coins of which m are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 12, while it is 23 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. If the probability that the coin drawn is fair is f(N, m) ,then the value of 43 f(20,12) must be
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answer is 27.
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Detailed Solution
Let E1 E2 and A denote the following eventsE1: coin selected is fairE2: coin selected is biasedA:The first toss results in a head and the second toss results in a tail.PE1=mN,PE2=N−mN,PAE1=12×12=14 and PAE2=23×13=29By Baye's theorempE1A=PE1⋅PAE1PE1⋅PAE1+PE2⋅PAE2=mN×14mN×14+N−mN×29=9m8N+m=f(N,m)f20,12=9×128×20+12=108172=2743⇒43f(20,12)=27