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{\bf Question}

Assume that $10\%$ of the balls in a certain box are red and that
20 balls are selected from the box at random, with replacement.
Find the probability that more than 3 red balls will be obtained
by using the binomial distribution.


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{\bf Answer}

Let $X$ denote the number of red balls in the 20 balls selected,
then $X \sim Binomial(20,0.1)$. So

$$P\{X=k\}=\left(\begin{array}{c}20\\k\end{array}\right)0.1^k0.9^{20-k},\
\ \ k=0,1,...,20$$ and
$$P\{X>3\}=1-P\{X=0\}-P\{X=1\}-P\{X=2\}-P\{X=3\}=?$$


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