Green Eyes

September 29, 2009

We haven’t done a math problem in a while. This one comes from my daughter’s high-school math class. She is never quite sure whether or not to ask me for help; sometimes she gets much more help than she really wants.

In a group of twenty-seven people, eleven have blue eyes, thirteen have brown eyes, and three have green eyes. If three people are randomly selected from the group, what is the probability that exactly one of them will have green eyes?

Your task is to find the probability. When you are finished, you are welcome to read or run a suggested solution, or to post your solution or discuss the exercise in the comments below.

Pages: 1 2

5 Responses to “Green Eyes”

  1. P. Riva said

    I’d say we have

    P = 3/27 * 24/27 * 24/27 * 3 =~ 26.34%

  2. P. Riva said

    OPS!!! I was a bit in a hurry:

    3/27 * 24/26 * 23/25 * 3 =~ 28.31%

  3. Peter said

    It may be easier to conceptualize as choosing one child w/ green eyes and two children with eyes of any other color. Your numerator is P(choosing one green of the 3 greens)*P(choosing 2 not greens of the 24 not greens). The denominator is the total number of combinations of choosing three children out of the 27.

    Number of ways of:
    -Choosing 1 green out of the 3 greens: 3C1 = 3
    -Choosing 2 non-greens out of the 24 non-greens: 24C2 = 276
    -Choosing 3 kids out of 27 kids: 27C3 = 2925

    So, the probability of choosing only one green eyed kid out of the three selected is:
    (3C1*24C2)/(27C3) = (3*276)/2925 = 0.283076923

  4. Graham said
    from operator import mul
    def falling_factorial(n, k):
        """(n)_k = n! / (n - k)! = n * (n - 1) * ... * (n - k + 1)"""
        return reduce(mul, xrange(n - k + 1, n + 1), 1)
    def binom(n, k):
        """n! / (k! * (n-k)!) = (n)_k / k!; k! = (k)_k"""
        return falling_factorial(n, k) / falling_factorial(k, k)
    if __name__ == "__main__":
        print 100 * binom(24, 2) * binom(3, 1) / float(binom(27, 3))

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