w white balls, b black balls; 1 chosen randomly, returned w/

[spphat]

Hi everyone - thank you in advance for any possible help you can provide.

A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color is noted, and it is returned to the bowl along with n additional balls of the same color. Another single ball is randomly selected from the bowl (now containing w + b + n balls) and it is observed that the ball is black. Show that the (conditional) probability that the first ball selected was white is: w / (w + b + n).


For this, I was thinking I need to start with P(W1|B2), where W1 is white on the first selection, and B2 is black on the second selection. So, this equals [ P(W1 AND B2) ] / P(B2). I figured that P(B2) = B / (W + B + N), but then I don't know where to go from there. Please let me know if I am off base in my thinking. Thanks!

 

[spphat]

Re: Probability Question - Conditional Probability

Ok, well I figured it out. I will show a quick answer below, and if anyone wants to see it further I can do that.


But, I was right in that you need to solve it using P(W1|B2), which equals [ P(W1 AND B2) ] / [ P(B2) ]. The trick is to do the bottom, where P(B2) can be one of two choices, so you must include them both. So, P(B2) = P(1st was White AND 2nd was Black OR 1st was Black AND 2nd was Black) = P(W1 AND B2) + P(B1 AND B2). So on and so on.

I forgot that for P(B1 AND B2) that B2|B1 I needed to include B + n balls on the top, because since Black was picked, there are that many more black balls to choose from the second pick.