An urn has 20 red balls 30 white balls and 50 black balls A

An urn has 20 red balls, 30 white balls, and 50 black balls. A sample of ten balls is selected at random without replacement without ordering from the urn. Find the probability (without simplification) that:

a.) the sample has 2 red balls, 3 white balls and 5 black balls.

b.) the sample has at least one black ball.

c.) the sample has at most one red ball.

d.) the sample has either 1 black ball or 2 black balls.

e.) the sample has either 1 black ball or 1 white ball.

Solution

a)

There are 100C10 ways to get any 10 balls.

Here, there are (20C2)(30C3)(50C5) ways to do the task.

Thus,

P = (20C2)(30C3)(50C5)/100C10 = 0.094418385 [ANSWER]

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b)

P(at least one black) = 1 - P(0 black)

There are 50C10 ways to draw no black balls.

Thus,

P = 1 - (50C10)/(100C10) = 0.99940658 [ANSWER]

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c)

There are 80C10 ways for 0 red, and (80C9)(20C1) ways for 1 red.

Thus,

P = (80C10 + (80C9)(20C1))/(100C10) = 0.363049434 [ANSWER]

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d)

There are (50C9)(50C1) for 1 black, and (50C8)(50C2) for 2 black.

Thus,

P = ((50C9)(50C1) + (50C8)(50C2))/(100C10) = 0.045230158 [ANSWER]

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e)

There are (50C9)(50C1) ways for 1 balck, and (70C9)(30C1) ways for 1 white,

Thus,

P = ((50C9)(50C1) + (70C9)(30C1))/(100C10) = 0.119944565 [ANSWER]