Suppose you have three boxes The first contains 1 green marb

Suppose you have three boxes. The first contains 1 green marble and 3 yellow marble, the second contains 2 green marbles and 2 yellow marble, and the third contains 3 green marbles and 1 yellow marble. A box is selected at random, and then a marble is drawn at random. What is the probability that the marble is green? What is the probability that the third box was the one selected given that the marble is green?

Solution

Let 1,2,3 = represent the boxes
Y = yllow
G = green

a)

P(G) = P(1) P(G|1) + P(2) P(G|2) + P(3) P(G|3)

= (1/3)*(1/4) + (1/3)*(2/4) + (1/3)*(3/4)

= 1/2 [ANSWER]

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b)

P(2|G) = P(2) P(G|2) / P(G)

= (1/3)*(2/4)/(1/2)

= 1/3 [ANSWER]