A box contains 23 yellow 25 green and 29 red jelly beans If

A box contains 23 yellow, 25 green and 29 red jelly beans.
If 14 jelly beans are selected at random, what is the probability that:
a) 11 are yellow?  
b) 11 are yellow and 2 are green?  
c) At least one is yellow?

Solution

There are 77 beans.

Hence, there are 77C14 = 8.39984*10^14 ways to get 14 beans.

a)

There are (23C11)(54C3) = 33536942712 ways to choose 11 yellow and 3 non yellow.

Thus,

P(11 yellow) = 33536942712/8.39984*10^14 = 0.0000399257 [answer]

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b)

There are

(23C11)(25C2)(29C1) = 11763078600 ways to choose 11 yellow, 2 green, and 1 red.

Thus,

P(11Y, 2G, 1R) = 11763078600/8.39984*10^14 = 0.0000140039 [answer]

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c)

P(at least one yellow) = 1 - P(no yellow)

There are 54C14 = 3.24537*10^12 ways to choose 14 non-yellow beans.

Thus,

P(no yellow) = 3.24537*10^12 / 8.39984*10^14 = 0.003863615

Thus,

P(at least one yellow) = 0.996136385 [answer]