Hello one and all. I could use a little guidance here on a probability problem.(adsbygoogle = window.adsbygoogle || []).push({});

Box #1 containsablack balls andbwhite balls while box #2 containscblack balls anddwhite balls. A ball is chosen randomly from box #1 and placed in box #2. A ball is then randomly chosen from box #2 and placed in box #1. What is the probability that box #1 still hasablack balls andbwhite balls?

Okay, from that I come up with the following:

Let Random Variable X1 = a black ball is transferred to box #2 from box #1

Let R.V. X2 = a white ball is transferred to box #2 from box #1

Let R.V. Y1 = a black ball is transferred to box #1 from box #2

Let R.V. Y2 = a white ball is transferred to box #1 from box #2

[tex]P(X_1) = \frac{a}{a+b}[/tex] and [tex]P(X_2) = \frac{b}{a+b}[/tex]

[tex]P(Y_1 \mid X_1) = \frac{c+1}{c+d+1}[/tex] and [tex]P(Y_1 \mid X_2) = \frac{c}{c+d+1}[/tex]

[tex]P(Y_2 \mid X_1) = \frac{d}{c+d+1}[/tex] and [tex]P(Y_2 \mid X_2) = \frac{d+1}{c+d+1}[/tex]

This is where I get stuck.

I know (for example) that I can define another R.V. to represent, say, a black ball was selected from box #2 (I'll call it R.V. A), and...

[tex]P(A) = P(X_1) \cdot P(Y_1 \mid X_1) + P(X_2) \cdot P(Y_1 \mid X_2)[/tex]

Assuming I'm somewhat on the right track and haven't screwed things up, how would I go about determining the probability that box #1 still hasablack balls andbwhite balls?

Thanks in advance for your enlightenment (and do I need it).

dogma

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# Balls in Boxes, Probability Question

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