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a bag contains 3 red balls and 2 blue balls. A ball is taken out at random and put back. A second ball is chosen and put back. I've already drawn the tree diagram for this. What is the probability of at least one red ball?

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Both events are independent of each other. That is an important first step. Now notice that the probability of getting at least one red ball is the same as the probability of not having both blue balls, so our final probability will be 1 - P(drawing blue balls both times). P(drawing blue balls both times) is 2/5*2/5, which is 4/25, so your final probability will be 1 - 4/25 = 21/25

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Note that ${\text{Prob(at least 1 in 2)}} = {\text{Prob(exactly 1)}} + {\text{Prob(exactly 2)}}$ and the number of ways to draw $r$ red balls in 2 draws is the same as the number of ways of choosing $r$ objects from 2 i.e. $$\frac{{2!}}{{r!(r - 2)!}}.$$ So using binomial distribution we know that the probability of exactly $r$ red balls in 2 draws is $$\frac{{2!}}{{r!(r - 2)!}}{p^r}{(1 - p)^{2 - r}}$$ where $p = 3/5$. Now add these up with $r = 1$ and $r = 2$.

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The expression above seems to reduce to $p^2$, which, for $p=3/5$ evaluates to 9/25, which is at variance with mathguy's result. – alancalvitti Nov 3 '12 at 3:16
Which expression reduces to $p^2$? As far as I know this approach should give the correct answer. – glebovg Nov 4 '12 at 1:08
Three things: this answer is the correct answer to the question if we are working without replacement, but the OP specified that it was with replacement. Second, you forgot the case of drawing exactly three red balls. And third, the above comment is hinting at the fact that the term (1-p)^(2-r) should be (1-p)^(5-r). – Paul Apr 16 at 18:22

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