# How do i calculate the probability of at least one red ball?

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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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 $r = 2$.
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