Probability: GR0568 #44
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
A. H=50
B. T >=60
C. 51 <= H <= 55
D. H >= 48 and T >=48
E. H <=5 or H >=95
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)^(10050) = 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).
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)^(10050) = 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).
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)^(10050) = 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).
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)^(10050) = 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).
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.
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.

 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 ztable 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 ztable, it has the greatest probability, and it is obviously a lot more than 0.025
then obbiously it is a binomial experiment, we will use ztable 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 ztable, it has the greatest probability, and it is obviously a lot more than 0.025
For coin tosses, the probability of getting exactly k heads (or tails) in n tosses is: 1/(2^n)*C(n,k)...
For example the probability of getting between 51 and 55 heads is just the sum from k=51 to k=55 of the above... Using this you can get an expression for A, B, C and E. For D, you should note that if you get 48 head then you have 52 tails etc., so that the probability of D is:
2*[1/(2^100)*C(100,48)]+2*[1/(2^100)*C(100,49)]+2*[1/(2^100)*C(100,50)]
A is contained in the above, so the answer is not A... E is quite small compared to this, B contains a lot of terms, but is also comparatively small, and C will have 5 terms, two of which are contained in the above and 3 of which are smaller than any contained in the above:
Thus, the answer is D.
For example the probability of getting between 51 and 55 heads is just the sum from k=51 to k=55 of the above... Using this you can get an expression for A, B, C and E. For D, you should note that if you get 48 head then you have 52 tails etc., so that the probability of D is:
2*[1/(2^100)*C(100,48)]+2*[1/(2^100)*C(100,49)]+2*[1/(2^100)*C(100,50)]
A is contained in the above, so the answer is not A... E is quite small compared to this, B contains a lot of terms, but is also comparatively small, and C will have 5 terms, two of which are contained in the above and 3 of which are smaller than any contained in the above:
Thus, the answer is D.