A box is seperated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature.

What is the probability that the molecules will be found in the same distribution as before the partition was punctured, that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half.

I got (1/2)^(100)*(1/2)^500 = 2.41*10^-181, but I think that's wrong. Seems to simplified.

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The answer at which you arrived at is correct. I think you didn't understand as to how you arrived at it.

Let the total volume of the box be V. So each side has a volume
(1/2)V.

We can calculate the propability that all the molecules at a certain arbitrary moment are situated in a smaller volume V1 within V .
The propability that one ...

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