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Probability questions from Math Subject GRE

  1. Aug 22, 2009 #1


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    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) [tex]H = 50 [/tex]
    (B) [tex]T \ge 60 [/tex]
    (C) [tex]51 \le H \le 55 [/tex]
    (D) [tex]H \ge 48 [/tex] and [tex]T \ge 48[/tex]
    (E) [tex]H \le 5 [/tex] or [tex]H \ge 95 [/tex]

    2. Relevant equations

    3. The attempt at a solution

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

    [tex] \binom{100}{50}(1/2)^{100} [/tex]

    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.

    Last edited: Aug 22, 2009
  2. jcsd
  3. Aug 22, 2009 #2
    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.

  4. Aug 23, 2009 #3


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    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?
  5. Aug 23, 2009 #4
    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?
  6. Aug 23, 2009 #5


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    No table. You just get a pencil and scratch paper.
  7. Aug 23, 2009 #6
    Well that certainly puts a cramp in things.

    Relating all options to H we get:

    (a) H = 50
    (b) [itex]H \leq 39[/itex]
    (c) [itex]51 \leq H \leq 55[/itex]
    (d) [itex]48 \leq H \leq 52[/tex]
    (e) [itex]H \leq 5 \text{ or } H \geq 95[/itex]

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

    [tex]P(H=h) = \left( _h^{100} \right) \cdot \frac{1}{2^{100}}[/tex]

    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

    [tex]\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) ][/tex]

    [itex]P(H \leq 39) = P(H \leq 49) - P(40 \leq H \leq 49)[/itex]

    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 [itex]\sigma[/itex], but even that faces some wrinkles (like option d).

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