There are six female and four male mice in a group of 10. One of the female mice and two of the male mice have a particular disease. Suppose that two mice are selected at random from the group without replacement. The following events are defined: A: Both mice are female. B: Exactly one of the mice has the disease.
If exactly one of the mice has the disease, what is the probability that both are female?
Computing P(Both mice are female) $= {6 \over 10} \times {5 \over 9} = {1 \over 3}$
And P(Exactly one has disease) $= {3 \over 10} \times {7 \over 9} = {7 \over 30}$
So we use conditional probability to find P(A|B) which equals = $ P(A|B) = {{1 \over 3} + {7 \over 30} \over {7 \over 30} } = 2 {3 \over 7} $
Which is obviously incorrect as the probability cannot exceed 1, what am I doing wrong?
$P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$.
So you need $P(A \text{ and } B)$ as well. This is not the sum (which is an overestimation of the probability $P(A \text{ or } B)$).
Note that $P(B)$ is computed incorrectly in your case. It is $\frac{\binom{1}{3}\binom{1}{7}}{\binom{10}{2}}$. $P(A)$ is not needed.
To compute $P(A \cap B)$: we have to pick the only sick female mouse and one healthy female mouse, with probability $\frac{\binom{1}{1}\binom{5}{1}}{\binom{10}{2}}$.
Now use the formula.