Given the first draw was RED, find the probability that the second draw will be

Question

Given the first draw was RED, find the probability that the second draw will be GREEN. (*keep in mind we draw once from each box*)

or

P(G2 | R1) =?

Consider two boxes containing marbles of differing colors BOXA 2 red 3 green BOXB 3 red lgreen_ Total: 5 marbles Total: 4 marbles The exterior of the boxes are identical and we are never allowed to inspect the interior of the boxes. The boxes are randomized so that we don't know which box is which. We will make one draw at random from EACH box. Answer the following Define events: Ai: the first draw will be made from box A Bi: the first draw will be made from box B Ri: the first draw will be red R2: the second draw will be red Gi: the first draw will be green G2: the second draw will be green

Explanation / Answer

Given that first draw was red, probability that the second draw will be a green is computed as:

Using bayes theorem we get:

P(G2 | R1) = P(R1 and G2) / P(R1)

Now as both boxes are equally likely to be selected initially, therefore,

P(R1) = P(R | Box A)P(Box A) + P(R | Box B) P(Box B)

Here P(Box A) = P(Box B)= 0.5, therefore, we get:

P(R1) = (2/5)*0.5 + (3/4)*0.5 = 0.575

Now computing P(R1 and G2), we can compute this as:

P(R1 and G2) = P( R from box A )P(G from box B)*0.5 + P( R from box B )P(G from box A)*0.5

P(R1 and G2) = (2/5)*(1/4)*0.5 + (3/4)*(3/5)*0.5 = 0.05 + 0.225 = 0.275

Therefore, now the required probability can be computed as:

P(G2 | R1) = P(R1 and G2) / P(R1) = 0.275 / 0.575 = 0.4783

Therefore 0.4783 is the required probability here.

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