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October 21st, 2015, 02:27 AM   #1
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Conditional probability, balls in jars

Setup: you have to jars. In the first jar are two white and three black balls (total 5). In the second jar is one white ball and one black ball (total 2). One ball is drawn at random from the first jar and put into the second jar.

Problem: If you draw one ball at random from the second jar and find that it is white, what is the conditional probability that you moved a black ball from the first jar?

My approach: Let us call the event that a black ball is moved A. Let us call the event that a white ball is picked from jar 2 B. We want to know P(B|A).

P(B|A)=P(A intersect B)/P(A)

We know P(A) is 1/3. But how do I calculate P(A intersect B)?
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October 21st, 2015, 06:34 AM   #2
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I would do this: Imagine doing this "experiment" 3000 times. (2/5)(3000)= 1200 times, you move a white ball from the first jar to the second, so there are now one white ball and three black balls in the first jar, two white balls and one black in the second jar. And (3/5)(3000)= 1800 times, you move a black ball from the first jar to the second so there are now two white and two black balls in the first jar, one white and two black balls in the second jar.

Of the 1200 times when there are two white and one black in the second jar, (2/3)(1200)= 800 times, you draw a white ball and (1/3)(1200)= 400 times, a black ball.

Of the 1800 times when there are one white and two black balls in the second jar, (1/3)(1800)= 600 times, you draw a white ball and (2/3)(1800)= 1200 times, a black ball.

So, of the 3000 experiments, you drew a white ball 800+ 600= 1400 times. Of those 1400 times, 800 were when you moved a white ball from one jar to another and 600 were when you moved a black ball. The probability that you moved a black ball, given that you drew a white ball, is 600/1400= 3/7.
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October 21st, 2015, 08:00 AM   #3
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Whoa, that was a new approach to me. I failed to consider the chance of moving a white ball to the second jar. I was stuck on the Bayes formula.
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