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Suppose a student chooses the answer to each question in an exam randomly, with equal probability for each option and with choices independent of each other. What is the probability that they will get at least $6$ out of $12$, and thus pass the exam? Note: for each question there are $5$ answers, (a $1/5$ chance).

I found this question on my physics tutorial sheet and tried to do the following:

$$P(X=6) = nCk \cdot p^k \cdot (1-p)^{n-k}$$

However, when I substituted $n=12, k=6, p=1/5$ my answer was $0.0155$, but apparently it was incorrect. Any help would be much appreciated.

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    $\begingroup$ The question asks for the probability of getting at least six correct. You calculated the probability of getting exactly six correct. $\endgroup$ – N. F. Taussig Sep 12 '19 at 0:18
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    $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 12 '19 at 0:21
  • $\begingroup$ You need to compute the probability of getting 6,7,8.....12 questiosn right; they are all passing scores. if you are using a graphing calculator it might have a binomcdf function; you can use that for these problems $\endgroup$ – Saketh Malyala Sep 12 '19 at 0:26
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The question asks for the probability that the student gets at least six questions correct by randomly guessing. You used the binomial distribution to compute the probability that the student gets exactly six answers correct. Since we want to find the probability that the student gets at least six answers correct, you should have computed $$\Pr(X \geq 6) = \sum_{k = 6}^{12} \binom{12}{k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{12 - k}$$ I will leave the details to you.

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