probability and counting
posted by Hannah .
(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?

In total, there are 25+28+38=91 jelly beans.
Using (n,r) to stand for "n choose r"
=n!/(r!(nr)!)
To choose y yellow, g green and r red jelly beans, the number of ways to choose is given by
(25,y)*(28,g)*(38,r)
and the number of ways to choose the same number of jelly beans irrespective of colour is
(25+28+38,y+g+r)=(91,y+g+r)
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.
Probability
=(25,5)*(66,8)/(91,13)
=53130*5743572120/1917283000904460
=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.
1(66,13)(25,0)/(91,13)
=10.01
=0.99 (approximately)
See also following link for a detailed explanation:
http://mathforum.org/library/drmath/view/69151.html