Probability choosing more than one ball in a random box

Box $I$ contains two green and three red balls, box $II$ contains four green and two red balls, and box $III$ contains three green and three red balls. A box is chosen at random and three balls are drawn from that box. (I know this is without replacement).

(i) Find the probability that exactly two greens are chosen $[$$P(Exactly 2 greens)]. My attempt: [$$P($Exactly $2$ greens$)]$ $=$ $P[g_1r_2g_3]+P[g_1g_2r_3]+P[r_1g_2g_3]$

$=$ $(3/5)(2/6)(3/6)+(4/5)(4/6)(3/6)+(5/5)(4/6)(3/6)$

$=$ $7/10$

(ii) Find $P$ $[$Box $I$ is picked|Exactly $2$ greens$]$

My attempt:

I know this is conditional probability.To answer this, I know you need to answer part (i) of this question that I created.

$P$ $[$Box $I$ is picked|Exactly $2$ greens$]$ $=$ $P$ $($Exactly $2$ greens $\cap$ Box $I$ is picked$)$ $/$ $P$ $($ Exactly $2$ greens$)$

$=$ $(3/5)(2/6)(3/6)+(4/5)(4/6)(3/6)$

$=$ $11/30$

$(11/30)$ $/$ $(7/10)$ $=$ $11/21$

I know my calculations might be off because it has been a while. Can someone please help me to solve this problem. I know I have the correct idea and if I see it I will understand.

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You have do not have probability of $2$ green quite right. The same question was asked and answered quite recently. –  André Nicolas Mar 18 '14 at 23:02
Where did I go wrong? –  lone Mar 18 '14 at 23:23
I get the problem that Andre did there because I did that same problem before, but I was wondering what would happen if three balls were chosen and how the probability would look like. –  lone Mar 18 '14 at 23:50
By now the calculation has changed, so my earlier comment is no longer relevant. –  André Nicolas Mar 19 '14 at 2:57

(i) Write $G$ for the event that precisely two green balls are chosen, and $B_i$ for the event that box i is chosen ($i = 1,2,3$). Then $$P(G) = \sum_{i=1}^3P(G|B_i)P(B_i) = \frac{1}{60}{2 \choose 2}{3 \choose 1}\cdot \frac{1}{3} + \frac{1}{120} \cdot {4 \choose 2} \cdot {2 \choose 1} \cdot \frac{1}{3} + \frac{1}{120} \cdot {3 \choose 2}\cdot {3 \choose 1}\cdot\frac{1}{3}$$

$$= \frac{1}{60} +\frac{1}{30} + \frac{1}{40}$$

$$= \frac{9}{120} = \frac{3}{40}.$$

N.B. $1/60 = 1/5 \cdot 1/4 \cdot 1/3$, which is the probability of choosing three specified balls in a specified order out of a box containing 5 balls. Similarly $1/120 = 1/6 \cdot 1/5 \cdot 1/4$.

(ii) Continuing your formula we have

$$P(B_1|G) = \frac{P(B_1 \cap G)}{P(G)} = \frac{P(G|B1)P(B_1)}{P(G)} = \frac{\frac{1}{60} \cdot \frac{1}{3}}{\frac{3}{40}} = \frac{2}{27}.$$

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Thank you. I get it now. The only question I have is where did you get the $1/60$ from again. I suppose the $1/120$ came from adding $1/60$ twice. –  lone Mar 19 '14 at 2:14
It's a pleasure. I have now added an explanation of this point to my answer. If this answer has cleared up the issue for you, maybe you would consider accepting it (using the tick button) and/or upvoting it (using the uparrow button)? –  Frank Mar 19 '14 at 11:56