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Verbatim question for GR8677 #15
Statistical Mechanics}Probability

Probability is mostly common sense and adhering to definitions.

The probability that a gas molecule or atom is in the small cube is P(in)=1E-6. The probability that it's not in that small cube is P(not)=1-1E-6. Assuming independent gas molecules or atoms, i.e., the usual assumption of randomness in Stat Mech, one gets, P(N gas atoms)=\left(1-1E-6\right)^N.

The answer is thus C.
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Comments
ernest21
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danielsw98667
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NEC
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yosun2015tester
2015-07-22 12:26:56		whatNEC
yosun2015tester
2015-07-22 12:25:20		testing123 july 22 2015NEC
mpdude8
2012-04-15 20:09:22		For this, I just looked at it logically. Think about N = a small number, like 2. A and E are out -- there's always at least a chance that neither one will be in that subset, or one of them will. Also, when N = 2, the probability that both particles are outside of such a small subset of the total volume should be almost 1. B is out, as even for N = 2, the probability is extremely small, and races towards zero as you increase N slightly.

D is also out, as it is the exact opposite effect of what you want, logically. Probability, as you add more particles, should tend to 0. C is the only choice that has the correct tendency as N grows, and at a reasonable rate of increase.		NEC
livieratos
2011-11-07 03:54:30		so the probability of one atom to be in the given small volume is the ratio of the small volume to the entire volume? P(yes) = 1E-6/1 = 1E-6?
Ajith
2015-10-14 09:30:40		 I guess P(an atom in 10^{-6})=\\frac{No. of atoms in that volume}{tolal number of atoms}\r\nHere, No. of atoms in 1 m^3= N\r\nNo. of atoms in 10^{-6} m^3= No. of atoms in 1 m^3 * 10^{-6}= N *10^{-6}\r\nThus, P(an atom in 10^{-6})=\\frac{N*10^{-6}}{N}=10^{-6}
NEC
ubaraj
2007-11-01 01:15:13		the volume is 1. There can be no He atoms in the volume 10E-6. So all N atoms have to be within the volume (1-10E-6), and so the probability becomes (1-10E-6)^N. The choice is C!!NEC
Furious
2007-08-17 15:40:58		You really just need to look at the limiting factors in this problem.

When N is 0 the probability should be 1. That eliminates A and D.

When N is extremely high, the probability should go to 0. That eliminates E, (not that any of us thought it was E)

So that leaves us with B and C. Which both match the limiting cases, but just think about it when N is 1. When N is one there should be a very high chance that it is not in the specific 1e-6 area.

For B, the probability is 1e-6, not very high at all.
For C, the probability is 0.99999, pretty close to 1.

This helped me since, I get super confused whenever I think about probability.		NEC
wishIwasaphysicist
2006-01-24 11:27:10		How did you connect P(Ngasatoms) = (1-1E-6)^N to answer C?
yosun
2006-02-01 22:02:05		 hi wishIwasaphysicist, because each of the N molecules are independent, the probability would be multiplicative. Thus, if the particle for each particle is P, the probability for N particles would be P^N.
yosun2015tester
2015-07-22 12:31:28		 testing123
yosun2015tester
2015-07-22 12:37:21		 testering12345

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yosun2015tester
2015-07-22 12:49:08		 \\langle my \\rangle
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You really just need to look at the limiting factors in this problem. When N is 0 the probability should be 1. That eliminates A and D. When N is extremely high, the probability should go to 0. That eliminates E, (not that any of us thought it was E) So that leaves us with B and C. Which both match the limiting cases, but just think about it when N is 1. When N is one there should be a very high chance that it is not in the specific 1e-6 area. For B, the probability is 1e-6, not very high at all. For C, the probability is 0.99999, pretty close to 1. This helped me since, I get super confused whenever I think about probability.

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