There are $n$ balls in a drawer, of which $7$ are red. If $2$ balls are chosen randomly, then the probability that both are red is $\frac{3}{4}$. Find $n$.
$\textbf{I need help on how to solve this problem:}$
$P(E|F) = \frac{3}{4}$
$\textit{The number of possible ways there can be two red balls out of 7 balls in the drawer} = P(7,2) = 42$
$P(E \cap F) = \frac{42}{n}$
I'm unable to proceed beyond this. I've arrived at the answer as $56$ by diving $\frac{42}{\frac{3}{4}}$ and it is wrong!
IMO you didn’t need conditional probability for this, GIVEN that you’re selecting both balls at once and not one by one. Then you need to solve $$\frac{\binom{7}{2}}{\binom{n}{2}}=\frac 34$$ or $$\frac{21}{\frac{n(n-1)}{2}}=\frac 34$$ so that $\displaystyle n(n-1)=56$ and $$n=8.$$
Note:$\displaystyle\binom{n}{k}=C(n,k)=\ ^nC_k.$