
#1
Dec304, 05:44 AM

P: 35

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 



#2
Dec304, 06:18 AM

Sci Advisor
HW Helper
P: 9,398

You're on the right track by thinking conditionally.
Can I make some notational changes and abuses? P(correct allocation of balls) = P(correct number given a white transferred or correct n umber given a black transferred) mutuall exclusive =P(correct number given white transferred) +P(correct number given black transferred) =P(white put back given white taken) + P(balck put back given black taken) = somethings you've worked out. 



#3
Dec304, 09:36 AM

P: 35

First of all, thanks for your response. I greatly appreciate your help.
My mind is swimming…so hopefully I'm not going to make this worse. If I'm correctly utilizing the info you provided: P(correct allocation of balls) = P(a black balls and b white balls), since box #1 started off with a black balls and b white balls [and box #2 started off with c black balls and d white balls]. P(white ball chosen from box #1) =[tex]\frac{b}{a+b}[/tex] P(black ball chosen from box #1) =[tex]\frac{a}{a+b}[/tex] and P(white ball chosen from box #2 given a white ball chosen from box #1) =[tex]\frac{d+1}{c+d+1}[/tex] P(black ball chosen from box #2 given a black ball chosen from box #1) =[tex]\frac{c+1}{c+d+1}[/tex] and finally,the correct allocation of balls: P(a black balls and b white balls) = [tex]\frac{b}{a+b} \cdot \frac{d+1}{c+d+1}+\frac{a}{a+b} \cdot \frac{c+1}{c+d+1}[/tex] I'm still a little fuzzy about this…then again, using Playdoh is challenging for me. Am I warmer, colder, or way out in left field? Thanks again! dogma 



#4
Dec304, 09:58 AM

Sci Advisor
HW Helper
P: 9,398

Balls in Boxes, Probability Question
yp, that seems to be about right (and correcting any errors i may have made).




#5
Dec304, 10:39 AM

P: 35

thank you for your guidance and wisdom.
best of wishes, dogma 



#6
Dec304, 01:42 PM

P: 35

Follow on question:
Could I have used a hypergeometric distribution to figure this out? I guess I would just have to figure out how to set it up that way. dogma 



#7
Dec304, 11:40 PM

P: 4

One can use the hypergeometic distribution to get to the same result. Using the conditional probability approach, the probabilities (for choosing the black or white ball from box#1 )and the conditional probabilities (of choosing the black or white balls from box#2) are the same as the one you got earlier.
Lets see for one case: Box#1: choosing a black ball Probability = (a choose 1) * (b choose 0) / (a+b choose 1) = a/(a+b) (a choose 1 : ways of choosing 1 ball from a black balls) Similarly the probability of choosing a black ball from box# given black ball chosen from box#1 = (c+1)/ (c+d+1) This is exactly the first term in your equation. I would appreciate if someone could let me know how to handle mathematical notation in this text editor. 



#8
Dec404, 05:27 AM

P: 35

Check out this link on LaTex (it's in the General Physics forum): http://physicsforums.com/showthread.php?t=8997 It is a thread containing info about LaTex typesetting in a message. It's pretty easy to do. Thanks and good luck. dogma 


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