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

AI Thread Summary
The discussion focuses on estimating the probability of finding all N air molecules in a volume of 3/4V while none are in the remaining 1/4V. The initial approach calculates the probability of finding N molecules in 3/4V as 4N/3V and the probability of finding none in 1/4V as (3V - 4N)/3V. A more intuitive method involves considering the room as two sections, where the probability of each molecule being in the larger section (3/4V) is consistently 3/4. This leads to the conclusion that the probability of all N molecules being in 3/4V is (3/4)^N. The discussion highlights the importance of understanding statistical mechanics concepts for accurate probability estimation.
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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.
 
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Neena Khan said:
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 = ?
 
Thx!
 
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