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(1 pt) A box contains 25 yellow, 28 green and 38 red jelly beans.

If 13 jelly beans are selected at random, what is the probability that:
a) 5 are yellow?
b) 5 are yellow and 7 are green?
c) At least one is yellow?

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1 answer

  1. In total, there are 25+28+38=91 jelly beans.

    Using (n,r) to stand for "n choose r"

    To choose y yellow, g green and r red jelly beans, the number of ways to choose is given by
    and the number of ways to choose the same number of jelly beans irrespective of colour is

    So the probability is:
    (25,y)*(28,g)*(38,r) / (91,y+g+r)

    For 5 yellow, we not worry about the other two colours, so they can be combined, call it x. We need 8 of x.
    =0.1592 (approx.)

    For (b), it is a similar expression. I get approx. 0.0012

    For (c), you would choose 13 out of green and red, and subtract from the probability of 1.
    =0.99 (approximately)

    See also following link for a detailed explanation:

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