SOLUTION: A box contains 23 yellow, 30 green and 31 red jelly beans. If 14 jelly beans are selected at random, what is the probability that: a) 12 are yellow? b) 12 are yellow and 1 a

Algebra ->  Probability-and-statistics -> SOLUTION: A box contains 23 yellow, 30 green and 31 red jelly beans. If 14 jelly beans are selected at random, what is the probability that: a) 12 are yellow? b) 12 are yellow and 1 a      Log On


   

Question 669084: A box contains 23 yellow, 30 green and 31 red jelly beans.
If 14 jelly beans are selected at random, what is the probability that:
a) 12 are yellow?
b) 12 are yellow and 1 are green?
c) At least one is yellow?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
A box contains 23 yellow, 30 green and 31 red jelly beans.
If 14 jelly beans are selected at random, what is the probability that:
a) 12 are yellow?
I assume you mean EXACTLY 12 yellow and EXACTLY 2 non-yellow ones

%28%22C%2823%2C12%29%22%2A%22C%2861%2C2%29%22%29%2F%22C%2884%2C14%29%22 = .0000007796187072

b) 12 are yellow and 1 are green?
I assume you mean EXACTLY 12 yellow, EXACTLY 1 green and 
EXACTLY 1 red

 = .0000003352458754

c) At least one is yellow?
We calculate the probability of the complement event, that none are
yellow, and subtract from 1:

1 - %22C%2861%2C14%29%22%2F%22C%2884%2C14%29%22 =.9929065388

Edwin