# Finding $P(D)$ given we know $P(C)$, $P(C|D')$ and $P(C|D)$

As the title states we need to find $P(D)$ given that $P(C)={9\over20}$, $P(C|D')={3\over7}$ and $P(C|D)={6\over13}$. I have been hitting my head against a brick wall for the past half an hour or so, unable to make much progress.

So far all that I have managed to do is some algebraic manipulation with ${6\over13}={P(C\cap D)\over P(D)}$ and ${3\over7}={P(C\cap D')\over P(D')}$. However I don't seem to be making any progress. Does anyone have any suggestions on a continuation?

-

I'll assume that $D'$ is the complement of the event $D$, since if not we don't have enough information to solve this problem. You have $$\mathbb P(C) = \mathbb P(C \, | \, D) \mathbb P(D) + \mathbb P(C \, | \, D') \mathbb P(D') = \mathbb P(C \, | \, D) \mathbb P(D) + \mathbb P(C \, | \, D') (1-\mathbb P(D)).$$ Now you have a linear equation in $\mathbb P(D)$, so it's pretty easy to find it. It gives you $$\frac 9{20} = \frac 6{13} \times \mathbb P(D) + \frac 37 (1 - \mathbb P(D)) = \frac 37 + \left( \frac 6{13} - \frac 37 \right) \mathbb P(D)$$ which leads to $\mathbb P(D) = \frac {13}{20}$.