- #1
dogma
- 35
- 0
Hello one and all. I could use a little guidance here on a probability problem.
Box #1 contains a black balls and b white balls while box #2 contains c black balls and d white 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 has a black balls and b white 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 has a black balls and b white balls?
Thanks in advance for your enlightenment (and do I need it).
dogma
Box #1 contains a black balls and b white balls while box #2 contains c black balls and d white 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 has a black balls and b white 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 has a black balls and b white balls?
Thanks in advance for your enlightenment (and do I need it).
dogma
Last edited: