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Problem

There are two bags, each contain black and white marbles.

Bag A contains 6 white marbles and 8 black marbles. Bag B contains 1 white and 8 black marbles. I randomly choose between bag A and B and pick a marble from that bag. The marble is white. Now, (without replacement) I randomly choose between two bags again, what's the probability that I will pick a black marble?

I'm trying to solve this using Bayes rule, but it seems that it can be solved intuitively fairly easily. Are either/both of the following approaches correct?

Using Bayes:

P(black|first marble was white) = P(black from bag A|first was white) + P(black from bag B|first was white)

= .5*[P(picked bag A for first marble|first marble was white)*P(pick white from bag A|picked first white marble from bag A) + P(picked bag B for first marble|first marble was white)*P(pick white from bag A|picked first white marble from bag B)]

+.5 * [P(picked bag A for first marble|first marble was white)*P(pick white from bag B|picked first white marble from bag A) + P(picked bag B for first marble|first marble was white)*P(pick white from bag B|picked first white marble from bag B)]

Using intuition:

You have a 1/7 chance of picking a white marble from bag B, and a 6/7 chance from picking a white marble from bag A. Therefore, I find the probability of picking the black marble from bag A and add the case where the white marble was picked from bag B + case where white marble was picked from bag A. Then, do the same for picking the black marble from bag B. This is as follows:

(1/2)[(1/7)(8/8) + (6/7)(8/9)] + (1/2)[(1/7)(8/14) + (6/7)(8/13)]

Thank you! Sorry for the extremely long post (I know it's confusing).

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1 Answer 1

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Pr(black on 2-nd|white on 1-st)=Pr(black on 2-nd&white on 1-st)/Pr(white on 1-st).$$*$$First we work on the numerator. There are 4 possible situations-they are disjoint and each has probability1/4 :(i) 1-st bag A, 2-nd bag B N.B. Pr(black on 2-nd&white on 1-st)=Pr(white on 1-st&black on 2-nd) $$=\frac{1}{4}\times \frac{6}{14} \times \frac{8}{13}$$ $$\text {(ii)first $A$ ,second $B:$ Pr(white on 1-st&black on 2-nd black) =}\frac{1}{4}\times \frac{6}{14} \times \frac{8}{9}$$ $$\text {(iii)first $B$ ,second $A:$ Pr(white on 1-st&black on 2-nd black) =}\frac{1}{4}\times \frac{1}{9} \times \frac{8}{14}$$ $$\text {(iv)first $B$ ,second $B:$ Pr(white on 1-st&black on 2-nd black) =}\frac{1}{4}\times \frac{1}{9} \times \frac{8}{8}$$ Pr(black on 2-nd & white on 1-st)=$$\frac{1}{4}\times \frac{6}{14} \times \frac{8}{13}+\frac{1}{4}\times \frac{6}{14} \times \frac{8}{9}+\frac{1}{4}\times \frac{1}{9} \times \frac{8}{14}+\frac{1}{4}\times \frac{1}{9} \times \frac{8}{8}$$ Now we work on the denominator of the original expression. Again there are two disjoint possible situations, each with equal probability, 1/2 in this case. Pr(white on first) = Pr(1 -st bag is A)Pr(white chosen|1-st bag is A)+ Pr(1 -st bag is B)Pr(white chosen|1-st bag is B) $$=\frac{1}{2} \times \frac{6}{14}+\frac{1}{2} \times \frac{1}{9}$$ Pr(black on 2-nd|white on 1-st) $$=\frac{\frac{1}{4}\times \frac{6}{14} \times \frac{8}{13}+\frac{1}{4}\times \frac{6}{14} \times \frac{8}{9}+\frac{1}{4}\times \frac{1}{9} \times \frac{8}{14}+\frac{1}{4}\times \frac{1}{9} \times \frac{8}{8}}{\frac{1}{2} \times \frac{6}{14}+\frac{1}{2} \times \frac{1}{9}}$$

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