Probability of Finding n Particles in a Volume

Moonjob
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Joined: Fri Dec 02, 2011 4:28 pm

Probability of Finding n Particles in a Volume

Postby Moonjob » Fri Dec 02, 2011 4:36 pm

In Molitoris' GRE Practice book, Test 2 Problem 22, the reader is asked to find the probability of finding n particles of N in a subvolume V_o of V.

The an approximation to the answer is e^{-10}, which the solution tells you is 4.5 \times 10^{-5}, but I don't know how I would calculate that value in 1.5 minutes. Brute force, Taylor expansion, and limits seem to be too slow and/or unwieldy to use on the test. Any help would be appreciated.

negru
Posts: 308
Joined: Wed Oct 13, 2010 3:49 pm

Re: Probability of Finding n Particles in a Volume

Postby negru » Fri Dec 02, 2011 4:49 pm

how much accuracy do you need?

e ~ 2,7 = 3-0.3 =3 (1-0.1)

3^-10~10^-5

(1-0.1)^-10~(1+0.1)^10~1+10*0.1=2

you get ~ 2 * 10^-5

Moonjob
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Joined: Fri Dec 02, 2011 4:28 pm

Re: Probability of Finding n Particles in a Volume

Postby Moonjob » Fri Dec 02, 2011 4:54 pm

negru wrote:how much accuracy do you need?

e ~ 2,7 = 3-0.3 =3 (1-0.1)

3^-10~10^-5

(1-0.1)^-10~(1+0.1)^10~1+10*0.1=2

you get ~ 2 * 10^-5

The other options on that problem are 2.2 \times 10^{-5} and 9.0 \times 10^{-5}, so unfortunately, I need more accuracy than that. So, how did you conclude that
3^{-10} \approx 10^{-5}?

Moonjob
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Joined: Fri Dec 02, 2011 4:28 pm

Re: Probability of Finding n Particles in a Volume

Postby Moonjob » Fri Dec 02, 2011 4:57 pm

I guess you just noted that 3^{-10} = (3^2)^{-5}

negru
Posts: 308
Joined: Wed Oct 13, 2010 3:49 pm

Re: Probability of Finding n Particles in a Volume

Postby negru » Fri Dec 02, 2011 5:08 pm

ok then you can do

3^(-10)=(10-1)^-5=10^(-5) (1-0.1)^(-5)

so you need to compute (1-0.1)^(-15). to a pretty good approximation that is (1+0.1)^(15), but you need to expand this to second order via taylor. can't think of any easier way

Moonjob
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Joined: Fri Dec 02, 2011 4:28 pm

Re: Probability of Finding n Particles in a Volume

Postby Moonjob » Fri Dec 02, 2011 5:11 pm

Okay thank you. I don't know if I'll be able to pull that all off in 1.5 minutes, especially since e^-10 was already two approximations deep, but maybe I can guess it right!

negru
Posts: 308
Joined: Wed Oct 13, 2010 3:49 pm

Re: Probability of Finding n Particles in a Volume

Postby negru » Fri Dec 02, 2011 5:15 pm

it's way over GRE levels anyway

GRE answers would've probably been separated by at least one or two orders of magnitude

Moonjob
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Joined: Fri Dec 02, 2011 4:28 pm

Re: Probability of Finding n Particles in a Volume

Postby Moonjob » Fri Dec 02, 2011 5:28 pm

negru wrote:it's way over GRE levels anyway

GRE answers would've probably been separated by at least one or two orders of magnitude


That's the problem. There were three different answers on the order of 10^{-5}
The choices were
9.0 \times 10^{-5}
4.5 \times 10^{-5}
and
2.2 \times 10^{-5}.

I don't know if this book gets its problems from an actual GRE, so I'll just bank on them not giving such similar answers in the real exam.

CarlBrannen
Posts: 381
Joined: Mon May 24, 2010 11:34 pm

Re: Probability of Finding n Particles in a Volume

Postby CarlBrannen » Fri Dec 02, 2011 9:01 pm

You should know that \ln(10) = 2.3. Then:

e^{-10} = 10^x
-10 = x \ln(10)
-10 = 2.3 x
-10/2.3 = -4 - 0.8/2.3 = -4.3 = -5 + 0.7

This is enough to distinguish the answers.

bfollinprm
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Joined: Sat Nov 07, 2009 11:44 am

Re: Probability of Finding n Particles in a Volume

Postby bfollinprm » Sat Dec 03, 2011 11:55 am

None of the published study guides actually do a good job simulating PGRE-type questions. They're good for review of the material, but the people who write these study books must not look at the PGRE very closely when writing practice tests.

It's not a problem, since there are now 5 released actual PGRE's.

Moonjob
Posts: 6
Joined: Fri Dec 02, 2011 4:28 pm

Re: Probability of Finding n Particles in a Volume

Postby Moonjob » Sat Dec 03, 2011 5:36 pm

CarlBrannen wrote:You should know that \ln(10) = 2.3. Then:

e^{-10} = 10^x
-10 = x \ln(10)
-10 = 2.3 x
-10/2.3 = -4 - 0.8/2.3 = -4.3 = -5 + 0.7

This is enough to distinguish the answers.


Thank you Carl. This is a great approach! It's not too difficult to account for the translation of powers of 10 into literal values, i.e. to intuit that 10^{0.7} \approx 4.5.




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