Probability Urn problem - application of Bayes thm

Question

I am learning statistics and I am trying to solve this problem:

Players A and B draw balls in turn, without replacement from and urn containing three red and four green balls. A draws first. The winner is the person who draws the first red ball. Given that A won, what is the probability that A drew a red ball on the first draw? (Bartoszyński, 4.4.7)

So given from this information, I suppose I am asking what is the probability $\mathbb{P}(A_1)$ given $\mathbb{P}(A_i)$

$A_i:=\{\text{red drawn on i-th turn}\}$

$\mathbb{P}(A_1|A_i)=\mathbb{P}(A_i|A_1)\mathbb{P}(A_1) / \mathbb{P}(A_i|A_1)\mathbb{P}(A_1)+...$

I have found the probabilities of individual draws

$\mathbb{P}(A_1)={3\over7}$

...

$\mathbb{P}(A_3)={3\over 3}$

I have tried more, but I don't know what all to put in the denominator.

Could you please give me at least a hint?

Answer 1

Let $E = $ the event that A wins and $F = $ the event that A draws a red ball on the first turn. Then $$P(F|E) = \frac{P(E|F)P(F)}{P(E)}$$ $$P(E|F) = 1$$ $$P(F) = \frac37$$ $$P(E) = \frac37 + \frac47*\frac36*\frac35 + \frac47*\frac36*\frac25*\frac14*\frac33$$