Question 726821: A box contains 21 yellow, 28 green and 38 red jelly beans. If 14 jelly beans are selected at random, find each of the following probabilities.
a) The probability that 2 are yellow is: .
b) The probability that 2 are yellow and 11 are green is:
c) The probability that at least one is yellow is:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Final answer to each part is in red
For each part, I'm using the formula
n C r = (n!)/(r!*(n-r)!)
which is the combination formula
For example, if n = 10 and r = 2, then
10 C 2 = (10!)/(2!*(10-2)!)
10 C 2 = (10!)/(2!*8!)
10 C 2 = (10*9*8!)/(2!*8!)
10 C 2 = (10*9)/(2!)
10 C 2 = (10*9)/(2*1)
10 C 2 = 45
This problem uses this combination formula and idea extensively. Because of this and because you'll be using this formula a lot in statistics, it will help to know how to compute n C r for any n,r combo (either by hand, table/tree, or a calculator). I won't go into the steps used to compute each combination since there are a lot of them below, but let me know if you need any extra help finding them.
Anyways, onto the actual problem
a)
There are 21 C 2 ways to pick 2 yellow and 66 C 12 ways to pick the remaining colors. This is out of 87 C 14 ways total (to pick 14 jelly beans of any color and any count)
Note: 21+28+38 = 87
So the probability is
((21 C 2)*(66 C 12))/(87 C 14) = 0.19113709882654
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b)
There are 21 C 2 ways to pick 2 yellow, 28 C 11 ways to pick 11 green, and 38 C 1 ways to pick red. This is out of 87 C 14 ways total
So the probability is
((21 C 2)*(28 C 11)*(38 C 1))/(87 C 14) = 0.00003168297624
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c)
There are 28+38 = 66 jelly beans that are NOT yellow. So there are 66 C 14 ways to pick 14 jelly beans that are NOT yellow. There are 87 C 14 ways to pick 14 jelly beans (that include yellow).
So the probability of picking 14 jelly beans...and not a single one of them is yellow...is
(66 C 14)/(87 C 14) = 0.014312778044
Now subtract this from one to get
1 - 0.014312778044 = 0.985687221956
So the probability that at least one is yellow is 0.985687221956
Why did I subtract that second to last result from 1? Because the probabilities P(at least one yellow) and P(no yellows) add to 1 (those two events are complementary events...one or the other must happen...but not both at the same time)
So
P(at least one yellow) + P(no yellows) = 1
----> P(at least one yellow) = 1 - P(no yellows)
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