# Probability questions from Math Subject GRE

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1. Aug 22, 2009

### mrb

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

1. The problem statement, all variables and given/known data

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 $$H \ge 95$$

2. Relevant equations

3. 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.

Last edited: Aug 22, 2009
2. Aug 22, 2009

### 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

3. Aug 23, 2009

### 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?

4. Aug 23, 2009

### 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?

5. Aug 23, 2009

### mrb

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

6. Aug 23, 2009

### 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) $48 \leq H \leq 52[/tex] (e) [itex]H \leq 5 \text{ or } H \geq 95$

Since this is a (fair) coin flipping problem, the probability for H = h flips for any h = 0, 1, 2, ..., 100 is:

$$P(H=h) = \left( _h^{100} \right) \cdot \frac{1}{2^{100}}$$

Since the second factor is always 1/2100, then this really boils down to comparison of binomial coefficients.

Part (a) is obviously the easiest. Part (b) seems the worst to evaluate while the rest are sums of 5 or so terms.

Keep in mind

$$\sum_{h=0}^{49} \left( _h^{100} \right) = \sum_{h=51}^{100} \left( _h^{100} \right) = \frac{1}{2} \cdot [1 - \left( _{50}^{100} \right) ]$$

$P(H \leq 39) = P(H \leq 49) - P(40 \leq H \leq 49)$

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 H 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).

--Elucidus