Probability of a certain ball drawn from one box given that other balls were drawn

Question

Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were drawn?

So in total there were exactly 2 green balls and 1 red ball drawn, from a different combinations of the 3 boxes. We could have selected the 2 greens from the first 2 boxes and a red from the last box, 2 greens from the last 2 boxes, or 2 greens from box 1 and 3. I get $\frac{2}{5} \frac{4}{6} \frac{3}{6} + \frac{2}{5} \frac{2}{6} \frac{3}{6} + \frac{3}{5} \frac{4}{6} \frac{3}{6} = \frac{2}{5}$. Now this is the probability of drawing 2 green balls. What do I do from here?

Answer 1

Hint: Let $G_1$ be the event a green was drawn from the first box, and let $T$ be the event two green were drawn. We want the conditional probability $\Pr(G_1|T)$, which is $\frac{\Pr(G_1\cap T)}{\Pr(T)}$.

Alternately, if the notation above is unfamiliar, you can use a "tree" argument.

Answer 2

E1 = Green from Box 1

E2 = 2 Greens Drawn

E1 and E2 = Green from Box 1 and 2 Greens Drawn

E1:

2 Green in Box 1

5 Total in Box 1

P(E1) = 2/5

E2:

Possible combinations are GGR and GRG. Add these two combinations together.

P(GGR) = (2/5) * (4/6) * (3/6) = 2/15

P(GRG) = (2/5) * (2/6) * (3/6) = 1/15

P(E1 and E2) = P(GGR) + P(GRG) = 3/15 = 1/5

P(E1 | E2)

P(E1 | E2) = P(E1 and E2) / P(E2)

P(E1 | E2) = (1/5) / (2/5) = 1/2