Question

A black bag contains two coins: one fair, and the other biased (with probability 3/4 of landing heads). Suppose you pick a coin from the bag — you are twice as likely to pick the fair coin as the biased one — and flip it 8 times. Given that three of the first four flips land heads, what is the expected number of heads in the 8 flips?

Answer 1

Here there are two coins

one fair and other biased with probability 3/4 of landing heads.

Pr(choosing fair coin) = 2 Pr(Choosing Biased coin)

so as these two events are complementary

Pr(Choosing fair coin) = 2/3

Pr(Choosing Biased Coin) = 1/3

Now we fipped the coin 8 times with fair coin

and we get the three of first three coins as heads but next four flips doesn't depend on previous 4 flips.

Expected number of heads in next 4 coins (Choosen coin is fair) = 4 * 1/2 = 2

Total expected number of heads in the 8 flips = 3 + 2 = 5

Now, we choose biased coin,

Expected number of heads in next 4 coins (Choosen coin is fair) = 4 * 3/4 = 3

Total expected number of heads in the 8 flips = 3 + 3 = 6

Now,

Expected number of heads in the 8 flips = 5 * 2/3 + 6 * 1/3 = 16/3 = 5.33