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
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.
First of all, thanks for your response. I greatly appreciate your help. My mind is swimmingâ€¦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 thisâ€¦then again, using Playdoh is challenging for me. Am I warmer, colder, or way out in left field? Thanks again! dogma
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
Hypergeometric 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.
Lele, Check out this link on LaTex (it's in the General Physics forum): https://www.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