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Hi I am having a difficult time to understand how to solve this problem although I get the concept of conditional probability but this is a tricky one:

suppose we have 4 balls that can be placed into 4 boxes one after another with equally likely possibilities. the question is what is the probability that a box contains exactly 3 balls given that the first 2 balls have been placed into different boxes. The solution I tried is as below but I am not convince that is done correctly: I take these events: PA(2 balls in the box) = 2/4 = 1/2 PB(1 ball in the box) = 1/4 then, calculate the prob that these two events happened PA and PB = 1/2* 1/4=1/8

is this correct? and appreciate for some explanation as I don't quite understand how this works.

Thanks.

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You're correct.

Another way of looking at it: Two of the boxes have one ball, so both remaining balls will need to go into just one of those boxes to win.

There are $2$ ways to do this, and $4^2 = 16$ ways to distribute the remaining two balls without restriction. So your probability is $2/16 = 1/8$.

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  • $\begingroup$ thank you, by using the second method i can better understand how this works. I am trying to associate what i did above with the conditional probability formula as i wanted to calculate it using these prob events instead P(box contains 3 balls| 2 balls placed into 2 different boxes). How can we calculate it in this way? $\endgroup$ – MugB Jan 13 '17 at 19:30

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