GR8677 #15



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yosun2015tester 20150722 12:26:56  what   yosun2015tester 20150722 12:25:20  testing123 july 22 2015   mpdude8 20120415 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.   livieratos 20111107 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) = 1E6/1 = 1E6?
Ajith 20151014 09:30:40 
I guess \r\nHere, No. of atoms in = N\r\nNo. of atoms in = No. of atoms in =\r\nThus, =

  ubaraj 20071101 01:15:13  the volume is 1. There can be no He atoms in the volume 10E6. So all N atoms have to be within the volume (110E6), and so the probability becomes (110E6)^N. The choice is C!!   Furious 20070817 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 1e6 area.
For B, the probability is 1e6, 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.   wishIwasaphysicist 20060124 11:27:10  How did you connect P(Ngasatoms) = (11E6)^N to answer C?
yosun 20060201 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 .

yosun2015tester 20150722 12:31:28 
testing123

yosun2015tester 20150722 12:37:21 
testering12345

yosun2015tester 20150722 12:49:08 

 

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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 1e6 area.
For B, the probability is 1e6, 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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