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An urn contains 4 red balls and 6 blue balls

three balls are drawn one at a time with replacement. what is the probability that 2 red balls and 1 blue ball are selected?

I did (0.4)^2(0.6) because they are independent events . but the solution is 3(0.4)^2(0.6).

Can someone please explain why is a three there?

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The probability of drawing RRB is $(0.4)^2(0.6)$, but so is the probability of drawing RBR, and so also is the probability of drawing BRR. Every one of those draws counts as $2$ reds and a blue, so the total probability of getting $2$ reds and a blue is the sum of those probabilities, which is $3(0.4)^2(0.6)$.

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HINT: Think of the binomial distribution.

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