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July 29, 2016
Posted by **Hannah** on Wednesday, April 13, 2011 at 5:37pm.

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?

- probability and counting -
**MathMate**, Thursday, April 14, 2011 at 12:04pmIn total, there are 25+28+38=91 jelly beans.

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

=n!/(r!(n-r)!)

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)

=1-0.01

=0.99 (approximately)

See also following link for a detailed explanation:

http://mathforum.org/library/drmath/view/69151.html