B2. Alice places a coin, heads up, on a table then turns off the light and leaves the room. Bill enters the room with 2 coins and flips them onto the table and leaves. Carl enters the room, in the dark, and removes a coin at random. Alice reenters the room, turns on the light and notices that both coins are heads. What is the probability that the coin Carl removed was also heads?
Solution :
Since we started with one coin heads up, there are 4 possible configurations before Carl enters the room: HHH, HHT, HTH, HTT. Since there are two heads when Carl left, the configuration HTT was impossible. For the remaining configurations:
Case 1: HHH There are 3 ways Carl could leave 2 heads, they all involve him removing a head.
Case 2: HHT or HTH There is only one way for each of these that would leave 2 heads, both involve removing a tail.
Therefore, the probability that Carl removes a head is $\frac{3}{5}$.
Answer: $\frac{3}{5}$