I just asked this question, but worded it wrong so while the given answers are useful, they still leave me confused for where I am in the progression through my stats book.
My problem is I've got 3 balls chosen (with replacement) out of 6, of which 2 of the 6 are blue. What is the expectation and variance of the number of blue balls picked?
In my original question, I included the idea of "chance" (see here: Probability of particular subset of balls occurring in a larger set chosen from a total?), but that lead to a discussion of the binomial theorem, which I have not yet come to in my book, so it can't be expected in the answer (and is a bit ahead of me).
So far, I've covered expectation and variance of random variables, and probability theory. I have a very hard time with probability theory, so that's what is killing me here.
My approach here is that this uses linear combinations of element expectations and variances, but I'm at an almost total loss as to how to find the probabilities of these blue balls in the selection of 3, to establish a probability mass function that can be used for expectation and variance calculations.
In organizing the problem, I've got four conditions, where the number of blue balls is 0, 1, 2, and 3, but how to use combinations for finding the relative probabilities of each leaves me at a blank.
I know I'm choosing 3 out of the 6, so the possible combinations are: 6!/(3!(6-3)!) = 20. However, this being "with replacement" means that we'd be using permutations for establishing the number of possibilities of choosing 3, and not combinations, so this would be 6!/3! = 120, right?
I'm guessing this means that we'd need to find the possible permutations of 3 that include 0 balls, and divide this by 120 to get the probability, and then repeat for 1, 2, and 3? I'm not seeing how to do this part...