statistics
posted by Christine .
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!

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 
I thought I was understanding this but I am really not sure about the symbols.
Does the 3*(13/20)^2*(7/20) translate to 3 times 13/20 times 2 times 7/20?
Sorry I am not up on the symbols...
I will keep trying. 
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 (1p)^(nr) 
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.