Probability of N molecules in a volume of 3/4V and none in 1/4V

[curiousKevo]

Estimate the probability that in a room of volume V, all the N molecules of air are found in volume $\frac{3V}{4}$ and none at all in the reaining volume V/4
My take to the solution is

Pr of finding N molecules in $\frac{3V}{4}$ : $\frac{N}{3V/4}$= $\frac{4N}{3V}$
Pr finding none in V/4: 1-$\frac{4N}{3V}$ = $\frac{3V-4N}{3V}$

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I used coin toss logic. No idea i did it correct. I'm new at staistical mechanics. I'm kinda lost just give me hints.

 

[stevendaryl]

Estimate the probability that in a room of volume V, all the N molecules of air are found in volume${3V} \div {4}$ and none at all in the reaining volume ${V}\div{4}$.

My take to the solution is

Pr of finding N molecules in ${3V} \div {4}$ : ${N} \div {3V/4}$ = ${4N} \div {3V}$

Pr finding none in V/4: $1- {4N} \div {3V}$ = ${3V-4N} \div {3V}$

That's not a very good estimate. Think about it this way: You divide the room into two parts, call them section A and section B. Section A is 3 times as large in volume as section B.

Suppose you start off with an empty room (no air at all). Then you introduce a single molecule into the room at a random location. Then the probability will be 3/4 that the molecule will be placed into section A. Now introduce a second molecule. The probability that it will be in section A is again 3/4. So the probability that the first two molecules will be released into section A is:

$P_2 = (3/4)*(3/4) = (3/4)^2$

So if you keep on releasing molecules into the room at a random location, then the probability that the first $N$ molecules will all be in section A is:

$P_N = ?$

 

[curiousKevo]

Thx!