Two Heads Odds

Solution 1: Let $A$ be the probability of getting exactly $2$ heads. Let $B$ be the probability of getting at least $1$ heads. Applying Bayes' rule, $P(A | B) = \frac{P(A \cap B)}{P(B)}$ , which is the probability of getting $2$ heads given at least $1$ head.

We have $P(A) = \frac{{3\choose 2}}{2^3} = \frac{3}{8}$ and $P(B) = 1 - \left(\frac{1}{2}\right)^3 = \frac{7}{8}$ . $P(A \cap B) = P(A)$ as getting two heads means that we have at least one head.

Thus $P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{\frac{3}{8}}{\frac{7}{8}} = \frac{3}{7}$ .

Hence $a + b = 3 + 7 = 10$ .

Solution 2: This problem is easily solved by looking at the sample space. The sample space of flipping a coin 3 times is: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT.

Because we know at least one flip resulted in heads, we can rule out the possibility of TTT. Therefore there are $7$ possible outcomes. There are $3$ possible ways of getting exactly two heads: HHT, HTH and THH.

Therefore the probability is $\frac{3}{7}$ . Hence $a + b = 3 + 7 = 10$ .