Box I contains 3 fair coins and 2 double-headed coins. Box II contains n fair coins and 1 double-headed coin. A box is chosen at random and a single coin is drawn from it at random. The coin is tossed twice and shows Heads on both tosses. If the probability that the coin came from Box I is 22 37 , then the value of n is

Step-by-step solution

Let E₁ be the event that Box I is chosen and E₂ be the event that Box II is chosen. We have P(E₁) = 1 2 and P(E₂) = 1 2 . Let A be the event that the drawn coin shows Heads on both tosses. For Box I, the probability of drawing a fair coin is 3 5 and a double-headed coin is 2 5 . P(A|E₁) = 3 5 ( 1 2 )^2 + 2 5 (1)^2 = 3 20 + 8 20 = 11 20 For Box II, the probability of drawing a fair coin is n n+1 and a double-headed coin is 1 n+1 . P(A|E₂) = n n+1 ( 1 2 )^2 + 1 n+1 (1)^2 = n 4(n+1) + 4 4(n+1) = n+4 4(n+1) By Bayes' T