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There are $10$ questions in exam. Student will pass the exam if he/she knows the answer to $2$ questions or if he/she gives the correct answer to one of them, and then gives the correct answer to one additional question (that is not in the group of $10$ questions). What is the number of questions that student needs to answer correctly such that he/she will pass the exam with probability greater than $0,8$.

We consider two cases:

$A:$ "Student knows the answer to $2$ questions from the group of $10$."

$$P(A)>0.8$$ $$\frac{x}{10}\cdot\frac{x-1}{9}>0.8\Rightarrow x\in(-\infty,-8)\cup(9,+\infty)\Rightarrow x_1=10$$

We get that for case $A$, student needs to answer all $10$ questions from the group of $10$.

Second case:

$B$: "Student knows the answer to exactly one of the $10$ questions from the group of $10$ questions, plus one additional question."

$P(B)>0.8$

How can we solve this case?

  • 0
    So if I understood well : He first receives a test with $10$ questions. He will immediately fails if he answer correctly to $0$ question. If he answers correctly two or more, he will immediately succeed. However if he answers correctly to exactly $1$ question, he will be taken apart after the exam in a small dark and stinky room with the professor who will ask him $1$ new question (never seen on the exam) on which his entire success now relies. Right ? – 2017-03-31
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    Something in missing in the question. "Student will pass the exam if he/she knows the answer to 2 questions ... What is the number of questions that student needs to answer correctly such that he/she will pass the exam" It's just 2, isn't it. The student needs to answer 2 questions correctly to pass the exam. – 2017-03-31

1 Answers 1

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I don't quite understand what the set-up is, but here is my interpretation based on what you say about case A. There are ten questions. The student will be asked two of them. If he gets them both right, he passes. If he gets them both wrong, he fails. If he gets one right and one wrong, he is asked another question, and passes if he gets that right.

If this is the correct interpretation, the probability of passing given that he knows $k$ answers out of $10$ is $$\frac{k}{10}\times\frac{k-1}9+2\times\frac{k}{10}\times\frac{10-k}{9}\times\frac{k-1}{8}$$ (the first term is the probability of getting the first two correct; the second term is the probability of right-wrong-right or wrong-right-right (these are the same as each other so it is just twice P(right-wrong-right)). This simplifies to $$\frac{15k^2-k^3-14k}{360},$$ which is bigger than $0.8$ if $15k^2-k^3-14k>288$. Putting $k=6$ gives $240$ and $k=7$ gives $294$, so he needs to know $7$ answers.