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There are $40$ questions in a test, $30$ questions are easy such that everyone can solve, $10$ questions are difficult such that no one can solve. $25$ students are going to do the test and each one of them choose randomly $5$ different questions such that different students are choosing the questions independently. To pass the exam the student must answer at least four questions.

  • $\color{gray}{\text{A. Find the distribution of the number of the correct answers of one specific student}}$

  • B. Find the distribution of the number of students (from the $25 $) that will fail in the exam

The answer for A. is here


My attempt for B:

Because that the distribution of the number of the correct answers of one specific student is $P(X) = \dfrac{\binom{30}{X}\binom{10}{5-X}}{\binom{40}{5}}$

and there are $25$ students so omega will be all the students $|\Omega|=25$ so the answer for B. will be

$P(Y)=\frac{1-P(Y) = \dfrac{\binom{30}{Y}\binom{10}{5-Y}}{\binom{40}{5}}}{25}$

$Y\in \{0,1,...,25\}$

Is it correct?

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  • $\begingroup$ Hint : The number of students failing in the exam is binomial distributed : $Y$~$B(25;p)$ , where $p$ denotes the probability that a specific student fails, which is $p=P(X\le 3)$, where $X$ is the number of questions the student can answer. $\endgroup$ – Peter Jan 22 '16 at 13:58
  • $\begingroup$ What does your last formula mean? It looks like you are dividing an entire equation by 25. I would expect something in which the variable is the number of students who fail, so if $Y$ is the number of failing students I can set $Y=6$ and plug that into the formula to find out how likely it is that $6$ students will fail. $\endgroup$ – David K Jan 22 '16 at 14:01
  • $\begingroup$ @DavidK I etited $\endgroup$ – user306663 Jan 22 '16 at 14:03
  • $\begingroup$ I still don't know what you mean by dividing an entire equation by 25. How can $P(Y)$ equal 1/25 of an equation? Also, what about the probability that $6$ students fail: then $Y=6$, and what is $\binom{10}{5-6}$? Do you get a value of $P(6)$ that seems reasonable? $\endgroup$ – David K Jan 22 '16 at 14:11
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Probability of picking $4$ or $5$ easy questions is $$ p={\binom{10}{1}\binom{30}{4}\over\binom{40}{5}}+{\binom{30}{5}\over\binom{40}{5}} $$ So propbability of failing the test is $f=1-p$. If $F$ is the number of students that fail the test, $$ P(F=n)=\binom{25}{n}f^np^{25-n} $$ which is a Binomial distribution.

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