There are three different types of circus prizes marked big (B), medium (M) and little (L). Each contains a certain number of red (R) and gold (G) balls, distributed as follows:
big prize (B) : 4R and 6G
medium prize (M) : 3R and 2G
little prize (L) : 2R and 1G
Your friend wins 3 big prizes, 1 medium prize and 2 little prizes. Without looking, you randomly reach into one of her prizes, and randomly take out one of its balls, which happens to be gold (G). Calculate the probability that you were choosing from a medium prize bag, i.e. calculate $$P(M|G)$$
My attempt:
$P(M|G)=\frac{P(M)P(G|M)}{P(G)}$
$P(M) = 1/6$ (1 medium/(3 big + 1 medium + 2 little))
$P(G|M)=2/5$ (2 gold balls out of 5 total in the medium prize bag)
$P(G) = 22/41$ (given 3 big prizes, 1 medium, and 2 little prizes, 22 balls are gold out of a total 41)
Therefore, $P(M|G)=41/330$, which is apparently incorrect. Help appreciated, thanks!
$P(G)=P(L)P(G|L)+P(M)P(G|M)+P(B)P(G|B)$, rather than P(G) = 22/41