Probability of Finding n Particles in a Volume
Probability of Finding n Particles in a Volume
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.
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.
Re: Probability of Finding n Particles in a Volume
how much accuracy do you need?
e ~ 2,7 = 30.3 =3 (10.1)
3^10~10^5
(10.1)^10~(1+0.1)^10~1+10*0.1=2
you get ~ 2 * 10^5
e ~ 2,7 = 30.3 =3 (10.1)
3^10~10^5
(10.1)^10~(1+0.1)^10~1+10*0.1=2
you get ~ 2 * 10^5
Re: Probability of Finding n Particles in a Volume
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 thatnegru wrote:how much accuracy do you need?
e ~ 2,7 = 30.3 =3 (10.1)
3^10~10^5
(10.1)^10~(1+0.1)^10~1+10*0.1=2
you get ~ 2 * 10^5
$$3^{10} \approx 10^{5}?$$
Re: Probability of Finding n Particles in a Volume
I guess you just noted that $$3^{10} = (3^2)^{5}$$
Re: Probability of Finding n Particles in a Volume
ok then you can do
3^(10)=(101)^5=10^(5) (10.1)^(5)
so you need to compute (10.1)^(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
3^(10)=(101)^5=10^(5) (10.1)^(5)
so you need to compute (10.1)^(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
Re: Probability of Finding n Particles in a Volume
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!
Re: Probability of Finding n Particles in a Volume
it's way over GRE levels anyway
GRE answers would've probably been separated by at least one or two orders of magnitude
GRE answers would've probably been separated by at least one or two orders of magnitude
Re: Probability of Finding n Particles in a Volume
That's the problem. There were three different answers on the order of $$10^{5}$$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
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.

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Re: Probability of Finding n Particles in a Volume
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.
$$e^{10} = 10^x$$
$$10 = x \ln(10)$$
$$10 = 2.3 x$$
$$10/2.3 = 4  0.8/2.3 = 4.3 = 5 + 0.7$$
This is enough to distinguish the answers.

 Posts: 1203
 Joined: Sat Nov 07, 2009 11:44 am
Re: Probability of Finding n Particles in a Volume
None of the published study guides actually do a good job simulating PGREtype 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.
It's not a problem, since there are now 5 released actual PGRE's.
Re: Probability of Finding n Particles in a Volume
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 $$10^{0.7} \approx 4.5$$.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.