## Probability: GR0568 #44

Forum for the GRE subject test in mathematics.
CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

### Probability: GR0568 #44

A fair coin is tossed 100 times, with each toss resulting in a head or tail. If H is the total number of heads and T is the total number of tails, which of the following events has the greatest probability?
A. H=50
B. T >=60
C. 51 <= H <= 55
D. H >= 48 and T >=48
E. H <=5 or H >=95

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
This is a classic question of bernoulli trials.

If the probability of success is "p" and the probabibility of failure is "q", then the probability of at least "k" successes on n trials is C(n, k) *
p^k * q^(n - k)

Using the above we find n = 100, p = q = 1/2 = 0.5 and,

A. H = 50,
P = C(100, 50) * (1/2)^50*(1/2)^(100-50) = C(100, 50)/2^100

B. T >= 60, P = C(100, 60) * (1/2)^40*(1/2)^60 + C(100, 61) * (1/2)^39 * (1/2)^ 61 + ...
P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100

C. 51 <= H <= 55,
P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. H >= 48 and T >=48
P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. H <=5 or H >=95
P = [ C(100, 0) + C(100, 1) + ... + C(100, 5) + C(100, 95) + ... + C(100, 100) ] / 2^100

Now using the properties of C(n, k) = n!/[k!(n - k)!], the above may be somewhat simplified to,

A. P = C(100, 50)/2^100

B. P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100
= 1 - [ C(100, 0) + ... C(100, 49) + C(100, 50) + C(100, 51) + ... + C(100, 59) ] / 2^100

C. P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. P = 2 [ C(100, 0) + C(100, 1) + ... C(100, 5) ] / 2^100

Which one is the greatest? Well to be precise use a calculator. Or...
Observe that C > A, D > A, E can easily be calculated to be small, B although large is far away from the middle value C(100, 50).

Elementarily, I know that the combinatorial function takes its greatest value at the midpoint.
E.g. C(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1
C(7, 0) = 1, C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35, C(7, 4) = 35, ....

So I would bet on D. as it takes the largest number of values from the middle of the range for C(100, k).

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
Oops, There were errors in my last post. The correct one is:

This is a classic question of bernoulli trials.

If the probability of success is "p" and the probabibility of failure is "q", then the probability of EXACTLY "k" successes on n trials is
C(n, k) * p^k * q^(n - k)

Using the above we find n = 100, p (heads) = q (tails) = 1/2 = 0.5 and,

A. H = 50,
P = C(100, 50) * (1/2)^50*(1/2)^(100-50) = C(100, 50)/2^100

B. T >= 60, P = C(100, 60) * (1/2)^40*(1/2)^60 + C(100, 61) * (1/2)^39 * (1/2)^ 61 + ...
P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100

C. 51 <= H <= 55,
P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. H >= 48 and T >=48
P = [ C(100, 48 ) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. H <=5 or H >=95
P = [ C(100, 0) + C(100, 1) + ... + C(100, 5) + C(100, 95) + ... + C(100, 100) ] / 2^100

Now using the properties of C(n, k) = n!/[k!(n - k)!], the above may be somewhat simplified to,

A. P = C(100, 50)/2^100

B. P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100
= 1 - [ C(100, 0) + ... C(100, 49) + C(100, 50) + C(100, 51) + ... + C(100, 59) ] / 2^100

C. P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. P = [ C(100, 48 ) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. P = 2 [ C(100, 0) + C(100, 1) + ... C(100, 5) ] / 2^100

Which one is the greatest? Well to be precise use a calculator. Or...
Observe that C > A, D > A, E can easily be calculated to be small, B although large is far away from the middle value C(100, 50).

Elementarily, I know that the combinatorial function takes its greatest value at the midpoint.
E.g. C(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1
C(7, 0) = 1, C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35, C(7, 4) = 35, ....

So I would bet on D. as it takes the largest number of values from the middle of the range for C(100, k).

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
I know very little about probablity, and with exactly 3 weeks until my exam, I have little chance of learning; therefore, I can be sure that there will be plenty of probability problems on my version of the exam. Good analysis. When you lign them up, it is easy to see D>A and D>C. One thing I saw online, which I gues I should remember, the variance of this, m trials with p=q=1/2, is \sigma^2=mpq=25, so the standard deviation is 5. So B is the upper tail above 2 standard deviations (approx 5%?), and E is both tails 9 standard deviations away (very close to zero?). I'll never remember it.
Last edited by CoCoA on Mon Sep 29, 2008 12:12 pm, edited 1 time in total.

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
I used a caluclator to check these values out:

A. 0.08 (0.07958 to be exact) = 8%

B. 0.03 = 3%

C. 0.32 = 32%

D. 0.38 = 38%

E. Almost negligible (0%).

My test is 6 weeks away (Nov 8th). It would be really helpful if you could post some feedback from your test.

Wonderacle
Posts: 5
Joined: Wed Oct 01, 2008 4:26 pm
you can first eliminate A, because C contains A

then obbiously it is a binomial experiment, we will use z-table to analyse,
n=100, p(H)=0.5, q=0.5
mean=50, sd=5,

B is equivalent to H</=40, 40 is 2 sd away from mean, so P=0.025,
E is too many sd away from the mean, just forget it,
D is equivalent to H ranged from 48 to 52,
C is ranged from 51 to 55,
so C and D both have the same amount interval difference,
however, D is in the middle, according to the z-table, it has the greatest probability, and it is obviously a lot more than 0.025

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
Thanks for the help.

But even if I remembered what a z-table was, I doubt if it is something that would help me on the exam that is now 2 weeks away. I guess I just have to accept that I will get Probability questions wrong.

JcraigMSU
Posts: 13
Joined: Wed Aug 06, 2008 9:25 am
I wouldn't get discouraged by thinking you will get some wrong. Acknowledging it will probably help you be more successful on the exam. I don't believe it is their intention people get 100% on the math GRE.