A box contains 15 yellow, 29 green and 36 red jelly beans.
If 10 jelly beans are selected at random, what is the probability that:
a) 6 are yellow?
That's (15 yellows, choose 6) AND (29+36=65 non-yellows choose 4)
out of
(15+29+36=80 jelly beans choose 10)
That's or .0020580634
b) 6 are yellow and 3 are green?
That's (15 yellows, choose 6) AND (29 greens choose 3) AND (36 reds choose 1)
out of
(15+29+36=80 jelly beans choose 10)
That's or .0003998668146
c) At least one is yellow?
This is much easier to do by first finding the probability of the
complement event and then subtracting from 1
The complement event is to choose all non-yellows, that is,
only greens and reds.
That's 29+36=65 non-yellows, choose 10
out of
(15+29+36=80 jelly beans choose 10)
That's or .10872439
But that's the probability of the complement event.
The probability that you want is 1 - .10872439 or .891275641
Edwin