1
$\begingroup$

In the class of Brad, a student will be dropped into his/her course if he/she has $3$ failed exams. The exam of Brad has $10$ multiple choice questions. Each question has 2 possible answer. To receive a passing grade, one must answer $90$% and above each of the questions correctly. Assuming that $5$ exams are taken by a student. What is the probability of a student dropping the course if he/she answered all the questions randomly? How about passing it? I just want to clarify my assumption that I would use the binomial probability distribution here but I don't know how to execute it. I've come up with an answer of $0.50$ both if the said formula is used.

$\endgroup$
  • $\begingroup$ This cannot be answered without knowing the probability of the student answering any given question correctly! Are you assuming that the student is just "guessing" so has probability 0.5 of answering any question correctly? If so, that would be odd so you really should say that. Also you say that a student is dropped if they "fail" 3 of the exams but you don't say what "failing" means. Is the alternative to "receive a high grade" to fail the test? $\endgroup$ – user247327 Feb 11 '17 at 10:40
  • $\begingroup$ Noted. I Edited it so the students will answer the questions randomly or will guess them and its a passing grade $\endgroup$ – Codex Feb 11 '17 at 10:49
1
$\begingroup$

X~binom$(10,\frac12)$, P(pass one exam) $=\binom{10}{9}\cdot(\frac12)^{10} + \binom{10}{10}\cdot(\frac12)^{10} = \frac{11}{1024}$

Similarly, for # of exams passed, Y~binom$(5,\frac{11}{1024})$

P(get dropped) $$= \sum_{k=0}^2\binom5{k}\cdot\left(\frac{11}{1024}\right)^k\cdot\left(\frac{1013}{1024}\right)^{5-k}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.