A box contains 17 yellow 25 green and 37 red jelly beans If

A box contains 17 yellow, 25 green and 37 red jelly beans. If 13 jelly beans are selected at random, what is the probability that:

5 are yellow----

5 are yellow and 7 are green...

At least 1 is yellow

Solution

There are 79 beans here.

Thus, there are 79C13 wasy to get 13 beans.

a)

There are 17C5 ways to get yellow, and 62C8 ways to get non yellow.

Thus,

P(5 are yellow) = (17C5)(62C8)/(79C13) = 0.079272929 [answer]

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b)

There are 17C5 ways to get yellow, 25C7 to get green, and 37C1 to get red.

Thus,

P(5Y,7G) = (17C5)(25C7)(37C1)/(79C13) = 0.000417007 [answer]

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c)

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

There are 62C13 ways to get non yellow.

THus,

P(no yellow) = 62C13 / 79C13 = 0.03147951

Thus,

P(at least 1 yellow) = 0.96852049 [ANSWER]