posterior predictive probability of heads on the third coin toss

Question

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)

Answer 1

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 $0.5\cdot P(A|HH) + 0.6 \cdot P(B|HH) + 0.9 \cdot P(C|HH)$.