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Suppose there are $3$ red balls and $1$ blue ball. What's the probability that if I draw a ball $10$ times, I don't draw a blue ball?

Is it $1-\Pr(\text{no red ball}) = 1-\big(\frac{3}{4}\big)^3$?

Also, what is the chance that I draw at least $1$ blue ball within the $10$ draws?

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Do you mean 1 - (3/4)^$\color{red}{10}$? – user2468 Aug 13 '12 at 0:15

Assuming that you are drawing indpendently and with replacement. The probability would be $$(\hbox{probability of not drawing a blue ball})^{10}=(3/4)^{10}$$ the probability of drawing a red ball all 10 times.

To draw at least one blue ball in 10 draws all that is required is that you do not draw a red ball all 10 times.

So it is $${1-(3/4)^{10}}$$.

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