Part a A box contains 3 coins two of which are fair and the

A box contains 3 coins, two of which are fair and the third has probability 3/4 of coming up heads. A coin is chosen randomly from the box and tossed 3 times. (a) What is the probability that all 3 tosses are heads? (b) Given that all three tosses are heads, what is the probability that the biased coin was chosen?

Solution

(a)

Let F shows the event that a fair coin is selected and U shows the event that an unfair coin is selected. When we choose a coin from the box either a fair coin will come with probability P(F)= 2/3 or a unfair coin will come with probability P(U)= 1/3. Let H shows the event that three heads will come. If a fair coin comes, then the probability that in 3 tosses 3 heads will come is P(H|F) =(1/2)*(1/2)*(1/2)=1/8. And if unfair coin comes then probability that in 3 tosses 3 heads will come is P(H|U)=(3/4)*(3/4)*(3/4)=27/64. So the probability that all 3 tosses are heads is

P(H)=P(H|F)P(F) + P(H|U)P(U) = (1/8)*(2/3) + (27/64)*(1/3) = 0.224

Hence, required probability is 0.224.