Could someone explain this solution please?

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In summary, the conversation discusses the probability, P(N), in a given volume V for a certain number of molecules N. The probability is derived from the expression p=V/V0, where p represents the probability that one molecule is in V and q represents the probability that it is somewhere else. The conversation then goes on to explain how the probability of N particles being in V and the rest not being in V is calculated by multiplying p^N and q^(N0-N) and then summing over the possible ways to choose N particles among N0. Finally, the conversation touches on basic probability results and how the probability is ultimately calculated.
  • #1
mewmew
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http://img478.imageshack.us/img478/5606/part1bd8.jpg
http://img183.imageshack.us/img183/686/part2yu0.jpg

We got this in class from a TA and the professor is in China and not able to answer questions. I am confused by where the probability,P(N), comes from in part a. It looks like a multiplicity multiplied by some other stuff but I don't understand it at all. I haven't had any probability/statistics but I assume it's pretty basic. If anyone could help me understand how this probability comes about it would be much appreciated. Thanks
 
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  • #2
p=V/V0 is the probability that one molecule be in the volume V. q=1-p is the probablity that it be somwhere else. This expression of q comes from the fact that since the particle must be somewhere we must have p+q=1.

Choose N particles among N0. The probability that these N be in V and that the rest of them are NOT in V is [itex]p^Nq^{N_0-N}[/itex]. In general, the probability that exactly N particles be in V and the rest not in V is [itex]p^Nq^{N_0-N}[/itex] summed over as many ways there are to choose which N particles among N0 are going to be in V. And you probably at least know some basic probability results, among which that the number of ways to chose N amongst N0 is [tex]\binom{N_0}{N}=\frac{N_0!}{N!(N_0-N)!}[/tex]

So we have the result.
 
  • #3
Thanks, that makes it much clearer.
 

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