There are three coins. Coin 1 is fair (P(Heads)=1/2), coin 2 has…

Question

There are three coins. Coin 1 is fair (P(Heads)=1/2), coin 2 has P(Heads)=1/3, and coin 3 has P(Heads)=3/4. One coin is selected at random (each with probability 1/3) and flipped 4 times, yielding exactly 3 heads and 1 tail. Let p be the posterior probability that the chosen coin was coin 2. Write p = m/n in lowest terms and find m+n.

Accepted Answer

By Bayes' theorem with equal priors 1/3, the posterior is proportional to the likelihood of seeing exactly 3 heads in 4 flips, L_i = C(4,3) p_i^3 (1-p_i) = 4 p_i^3 (1-p_i).

Compute 4 p^3 (1-p) for each coin (the common factor 4 and the common prior 1/3 cancel in the ratio, but we keep them for clarity):
- Coin 1: 4*(1/2)^3*(1/2) = 4*(1/8)*(1/2) = 1/4.
- Coin 2: 4*(1/3)^3*(2/3) = 4*(1/27)*(2/3) = 8/81.
- Coin 3: 4*(3/4)^3*(1/4) = 4*(27/64)*(1/4) = 27/64.

Posterior(coin2) = (8/81) / (1/4 + 8/81 + 27/64).

Common denominator of 4, 81, 64 is 20736. Convert:
1/4 = 5184/20736, 8/81 = 2048/20736, 27/64 = 8748/20736.
Sum = (5184+2048+8748)/20736 = 15980/20736.

Posterior(coin2) = (2048/20736)/(15980/20736) = 2048/15980 = 512/3995 (dividing by 4).

Check lowest terms: 512 = 2^9; 3995 = 5*17*47, no factor of 2, so reduced.

m+n = 512 + 3995 = 4507.

Final answer: 4507

There are three coins. Coin 1 is fair (P(Heads)=1/2), coin 2 has…

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