Probability Coin Test

Question

Two biased coins are tossed 144 times. 50 trials are both heads and 39 trials are both tails out of these 144 trials. Among the rest, 25 trials have head in the first coin while tail in the second coin and 30 trials with tail in the first coin and head in the second coin. Calculate the marginal probabilities for both the coins.

Answer 1

Let H1 denote the event of the first coin landing heads side up, H2 the event of the second coin landing heads side up, T1 the event of the first coin landing tails side up, and H2 the event of the second coin landing tails side up. Then from the given information, we have

P(H1 ∩ H2) = 50/144

P(H1 ∩ T2) = 25/144

P(H2 ∩ T1) = 30/144

P(H2 ∩ T2) = 39/144

Thus, the marginal probabilities are

P(H1) = P((H1 ∩ (H2 ∪ T2)) = P((H1 ∩ H2)∪(H1 ∩ T2)) = P((H1 ∩ H2)∪(H1 ∩ T2)) = P(H1 ∩ H2)+P(H1 ∩ T2) = 50/144 + 25/144 = 75/144

P(H2) = P((H2 ∩ (H1 ∪ T1)) = P((H2 ∩ H1)∪(H2 ∩ T1)) = P((H2 ∩ H1)∪(H2 ∩ T1)) = P(H2 ∩ H1)+P(H2 ∩ T1) = 50/144 + 30/144 = 80/144

P(T1) = P((T1 ∩ (H2 ∪ T2)) = P((T1 ∩ H2)∪(T1 ∩ T2)) = P((T1 ∩ H2)∪(T1 ∩ T2)) = P(T1 ∩ H2)+P(T1 ∩ T2) = 30/144 + 39/144 = 69/144

P(T2) = P((T2 ∩ (H1 ∪ T1)) = P((T2 ∩ H1)∪(T2 ∩ T1)) = P((T2 ∩ H1)∪(T2 ∩ T1)) = P(T2 ∩ H1)+P(T2 ∩ T1) = 25/144 + 39/144 = 64/144