What is the probability of getting 3 heads in 4 coin tosses, given you get at least 2 heads?

Question

I came across this question while doing conditional probability questions and while I can get the answer by counting outcomes, I'm getting a different answers using Bayes' theorem. $\ P(2T)\ $ is the probability of getting at least 2 tails. $\ P(3T)\ $ is the probability of exactly 3 tails. We flip a fair coin 4 times and are looking for the probability of getting 3 heads given we get at least 2 heads. $$P(3T|2T) = \frac{P(2T|3T)P(3T)}{P(2T)}$$ $$P(2T|3T) = 1$$ $$P(2T) = \frac{1}{4}$$ $$P(3T) = \frac{3}{8}$$ $$P(3T|2T) = \frac{2}{3} $$

The answer is supposed to be $P(3T|2T) = \frac{4}{11}$ which is easy enough to see when the outcomes are counted but there's clearly a problem with the above method I'm using.

Answer 1

It is unclear from your question whether your asking for the probability of exactly three heads, but that's what I'll assume.

Given that there are at least two heads, the number of cases are as follows:

• Exactly $2$ heads: ${4 \choose 2} = 6$ cases
• Exactly $3$ heads: ${4 \choose 3} = 4$ cases
• Exactly $4$ heads: ${4 \choose 4} = 1$ case

Of these $11$ equally likely cases, four have exactly $3$ heads. Hence the probability is $P(3|2) = \frac{4}{11}$.

Answer 2

In my opinion, the tree diagram is the easiest way to tackle those problems (I wish I had learned it sooner)