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One box contains 125 pearls, 65 red, 35 green and 25 black. There are 7 pearls chosen randomly.   What is the probability of exactly 3 red, 1 green and 3 black pearls being chosen?

Solution: $\frac{430612}{9765625}$

The black pearls are removed. How big is the probability to get exactly 2 red and 1 green pearl instead 3 black?

Solution: $\frac{3549}{8000}$

I wanted to use binomial distribution, but I am not sure what is excatly here n and k, and how to combine these conditions?

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    For the first question, you might want to look at the multinomial distribution, it might give you some hints on how to solve it. For the second, I believe it's just a little bit different from the binomial – 2017-02-24
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    Hint: the stated results assume that the pearls are put back in the urn after each draw. If they are not, the results are $7032480\over159481267$ and $104\over231$ – 2017-02-24
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    In first case binomial distribution not works. Because binomial distribution works on whether event happens or not. In second case you can use. – 2017-02-24

1 Answers 1

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In first case using multinomial distribution,

$\frac{7!}{3! \times 1! \times 3!} \left(\frac{65}{125}\right)^3 \left(\frac{35}{125}\right)^1 \left(\frac{25}{125}\right)^3$

$= \frac{430612}{9765625}$

In second case n=3 and k=2.

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    If any doubt I am here for you. – 2017-02-24