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GR8677 #15
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Comments |
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!! |  | 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. | | 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  .
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