The probability of having the unfair coin is 1 3. The probability of having a fair coin and tossing heads twice in a row is 2 3 ⋅ 1 4 = 1 6. If you know that you’ve tossed heads twice in a row, then you know that you’re in one of these two cases; the probability that you’re in the first case (have the unfair coin) is
The rest are fair coins. So if three or unfair, the rest are eight coins. When this problem says that they are fair coins, it means that they have a 50/50 chance of coming up either heads or tails. You randomly choose one coin from the bag and flip it two times. What is the percent probability of getting two heads?
First, with your unfair coin, the probability of the coin landing on heads is P (H) = 2*P (T), (that is, 2 times the probability of landing on tails). Recall, the probabilities of exhausitve and mutually exclusive events must add to 1. So, P (H) + P (T) = 1, but P (H) = 2*P (T), so P (H) + P (T) = 2*P (T) + P (T) but this = 1.
Since in this problem the probability of picking your second coin depends on the first coin you picked, you use a 2 branch tree to pick your first coin, and then on each branch is another 2 branch tree. This leads to 4 possible outcomes: (Fair, Fair), (Fair, Unfair), (Unfair, Fair), (Unfair, Unfair).
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