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Jim didn't study for his math test, and has to guess randomly on 10 multiple choice questions. If each question has 4 choices, what is the probability of gym getting 8 questions correct?

I'm assuming this is involving a binomial distribution, right? Any hints to approaching this will be greatly appreciated.

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The probability of getting exactly $k$ successes in $n$ trials is given by $$ f\left(k;n,p\right)=\left(\begin{array}{c} n\\ k \end{array}\right)p^{k}\left(1-p\right)^{n-k}. $$ Here, $k=8$, $n=10$ and $p=1/4$. Plug in and crunch. If however, you want to know the probability that Jim gets at least $8$ correct, you simply need to sum $$ f\left(8;10,1/4\right)+f\left(9;10,1/4\right)+f\left(10;10,1/4\right). $$

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