A bowl has 7 red marbles 4 blue marbles and 8 green marbles

A bowl has 7 red marbles, 4 blue marbles, and 8 green marbles. You randomly pick five marbles from the bowl, with replacement:

a) How many ways could you get 3 red marbles and 2 blue marbles? Why?

b) What is the probability that you will get 2 blue marbles, 1 red marble, and 2 green marbles? Why?

Solution

a)

You replace them, so each time, there are 7 red marbles to choose from, and each time, 4 blue marbes to choose from.

Hence, there are 7*7*7*4*4 = 5488 ways, if you take them in that order.

However, you can still permute that in 5!/[3!2!] = 10 ways, by permutation of like objects.

Hence, there are 5488*10 = 54880 ways [ANSWER]

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b)

There are 19 marbles here.

If in that order, there are (4/19)(4/19)(7/19)(8/19)(8/19) = 0.002894876

However, you can permute the said balls in 5!/[2!1!2!] = 30 ways.

Thus,

P(2 blue, 1 red, 2 green) = 0.002894876*30 = 0.086846285 [answer]