Probability questions from Math Subject GRE

[mrb]

This is from the practice Math Subject GRE 0568, problem 44.

Homework Statement

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

Homework Equations

The Attempt at a Solution

Well, I can set up expressions that give the probabilities. For instance, (A) is obviously

$$\binom{100}{50}(1/2)^{100}$$

And (B) just involves summing similar terms for values 60 through 100. But how to compare these? It's not too hard to figure out that A, C, or E can't be right, but how do I compare B vs. D? Another Math Subject GRE practice test I have has a similar problem.

Thanks.

 

[Elucidus]

Since the number of flips is sufficiently great, these probabilities can be approximated with a normal distribution. Don't forget to use continuity corrections.

I hope this is helpful.

--Elucidus

 

[mrb]

Thanks for your reply. Unfortunately I don't actually know any probability beyond the basics of counting. I am aware of the normal distribution and the fact that it can be used to approximate these kinds of probabilities, but I don't know how to go about doing that. Is this something that can be done in the 2-3 minutes I can afford to spend on a GRE question?

 

[mathie.girl]

I don't think the math to switch to the normal is hard (quite possibly easier than the counting), but you'd need a table to be able to get an answer, wouldn't you? Do you know if you get a table for the normal distribution during the test?

 

[mrb]

No table. You just get a pencil and scratch paper.

 

[Elucidus]

Well that certainly puts a cramp in things.

Relating all options to H we get:

(a) H = 50
(b) $H \leq 39$
(c) $51 \leq H \leq 55$
(d) [itex]48 \leq H \leq 52[/tex]<br /> (e) $H \leq 5 \text{ or } H \geq 95$<br /> <br /> Since this is a (fair) coin flipping problem, the probability for <i>H</i> = <i>h</i> flips for any <i>h</i> = 0, 1, 2, ..., 100 is:<br /> <br /> $$P(H=h) = \left( _h^{100} \right) \cdot \frac{1}{2^{100}}$$<br /> <br /> Since the second factor is always 1/2¹⁰⁰, then this really boils down to comparison of binomial coefficients.<br /> <br /> Part (a) is obviously the easiest. Part (b) seems the worst to evaluate while the rest are sums of 5 or so terms.<br /> <br /> Keep in mind<br /> <br /> $$\sum_{h=0}^{49} \left( _h^{100} \right) = \sum_{h=51}^{100} \left( _h^{100} \right) = \frac{1}{2} \cdot [1 - \left( _{50}^{100} \right) ]$$<br /> <br /> $P(H \leq 39) = P(H \leq 49) - P(40 \leq H \leq 49)$<br /> <br /> I may be having a rather opaque day, but this seems a really mucky exercise. The only other thing I can think of is that the standard deviation for <i>H</i> is 5, and one can exploit the approximate probability for falling in between multiples of $\sigma$, but even that faces some wrinkles (like option d).<br /> <br /> --Elucidus[/itex]