posterior predictive probability of heads on the third coin toss

There are three types of coins which are indistinguishable apart from their probability of landing heads when tossed.

• Type A coins are fair, with probability .5 of heads
• Type B coins have probability .6 of heads
• Type C coins have probability .9 of heads

You have a drawer containing 4 coins: 2 of type A, 1 of type B, and 1 of type C. You reach into the drawer and pick a coin at random. Assuming the first two tosses were heads, what is the posterior predictive probability of heads on the third toss?

(the answer is 0.71522 but I can't figure out how)

• Please edit the question and show us what you have tried. Mar 19, 2021 at 10:23
• Re Math Lover's comment, in particular, you have tagged your query as bayesian, and in fact, the solution does involve using Bayes Theorem. So, why not include your understanding of Bayes Theorem in your edited query, along with your attempt to use that theorem? Mar 19, 2021 at 10:43

Since you have provided the answer, I suppose it is an exercise for self-learning, so I think it's fine to help a bit. In order to solve this, we need to compute the following conditional probability: given two heads, the probability of the coin being A, B, or C given the first two tosses are heads (we denote this as $$P(A|HH), P(B|HH), P(C|HH)$$ respectively). This can be computed using the definition of conditional probability, which yields (for $$B$$ and $$C$$ analogically) $$P(A|HH) = \frac{P(A \cap HH)}{P(HH)}$$ where P(HH) can be computed using Law of total probability, which gives $$P(HH) = P(H|A) \cdot P(A) + P(H|B) \cdot P(B) + P(H|C) \cdot P(C)$$. Probability of head on the third toss depends on what coin do we have in hand, which is $$0.5\cdot P(A|HH) + 0.6 \cdot P(B|HH) + 0.9 \cdot P(C|HH)$$.