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