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Conditional Probability- balls and bags
To be archived: Conditional Probability
By goodone on Thursday, November 08, 2001 - 07:29 am:

A bag contains one ball which is either black or white with equal probability. An engineer drops an extra (identical) black ball into the bag and then reaches in and extracts one ball, which turns out to be black. What is the probability that the ball remaining in the bag is black?

By Dan Goodman on Thursday, November 22, 2001 - 01:58 pm:

First, draw a probability tree, the first two branches are "First ball is black" with probability 1/2 and "First ball is white" with probability 1/2. If we are in the first branch and we add a black ball then we will always pick a black ball from the bag, so at the end of the first branch we can write "Picked a black ball, probability 1/2". We add two branches to the end of the second branch; these are "Picked a black ball, probability 1/2 (given that the first ball is white)" and "Picked a white ball, probability 1/2 (given that the first ball is white)" because if the first ball was white and the second white, picking a black or white ball is equally likely. Now, at the end of these branches we write "Picked a black ball, probability 1/4" and "Picked a white ball, probability 1/4". So, we have 3 events in total:

(1) With probability 1/2 the first ball was black and we picked a black ball
(2) With probability 1/4 the first ball was white and we picked a black ball
(3) With probability 1/4 the first ball was white and we picked a white ball

Finally, we know that we picked a black ball, so we must be in case (1) or (2). So, the probability that the first ball was black is the probability of (1) divided by the probability of (1) or (2). The probability of (1) is 1/2 and the probability of (1) or (2) is 1/2+1/4=3/4. So, the probability that the first ball was black given that we picked a black ball is (1/2)/(3/4)=2/3.

Hope that is right and that it makes sense...