0568 #44

Forum for the GRE subject test in mathematics.
joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

0568 #44

Postby joey » Tue Nov 03, 2009 9:23 pm

Quoth the GRE:
A fair coin is to be tossed 100 times, with each toss resulting in a head or a 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

Answer: D

Okay, so this is a binomial experiment, and we can easily translate all of the answer choices to statements about H alone by substituting T = 100 - H.
My inclination is to approximate the binomial by a normal distribution, except I don't have a table of values for the GRE. I can sort of make do by drawing a rough picture of a normal distribution with mean 50 and comparing areas, though. Doing this, I see that (E) is really small and (A) is less than (D) -- those are out. More, (C) is less than (D) because in both cases H ranges over five values, but in (D) the density function over that range is greater. This leaves (B) and (D). Depending on what I've drawn, I guess (D), but I'm not sure. Why must we disqualify (B)?

mtey
Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

Re: 0568 #44

Postby mtey » Wed Nov 04, 2009 5:02 am

Cause D is right in the center of the normal distribution, and B is slightly right. The area under the graph of the density of the normal distribution is greater in the center, which you can verify by simple drawing of the graph. And it is so, simply because e^{-x^2} in (-a, a) is greater than e^{-x^2} in (-a+h, a+h), h > 0.

EDIT: Oops sorry I thought you were refering to answer (C). Unfortunately we should memorize some quantiles of the normal distribution... at least \Phi(0), \Phi(.5), \Phi(1), \Phi(1.5), \Phi(2) In this case for (B) we want 1 - \Phi(2) = 1 - 0.97 = 0.03. Which is pretty small. The question is unfair I guess :)




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