Posted by **Christine** on Sunday, May 11, 2008 at 8:42pm.

you have 8 blue marbles, 7 red marbles, and 5 green marbles. What is the probability of obtaining at least 2 red marbles in three draws with replacements?

I have the answer, just can't figure out how to get it.

Thanks!

- statistics -
**drwls**, Sunday, May 11, 2008 at 9:04pm
Seven out of the twenty are red. The probability of NOT getting ANY red in three tries is (13/20)^3 = 0.2746

The probability of getting only one red is 3*(13/20)^2*(7/20) = 0.4436

The probability of getting two or more is 1 - 0.2736 - 0.4436 = 0.2828

- statistics -
**Damon**, Monday, May 12, 2008 at 5:53pm
3*(13/20)^2*(7/20)

means

three times (13/20) squared times (7/20)

in general the probability of getting r successes out of n tries in binomial distribution if probability of success is p is

P(r) = C(n,r) p(r)^r (1-p)^(n-r)

- statistics -
**Christine**, Monday, May 12, 2008 at 8:04pm
That is even more confusing. I am totally lost on this problem.

I know that I have to multiply the three possibilities, so (7/20)(7/20)(13/20) three times (same numbers, different order) and I get 0.07963. I thought I would add those three numbers or essentially multiply that number by three, however, I did not get the right answer doing it that way.

Thanks for any help. I am totally confused.

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