Please answer the following question using set notation (i.e. defining events and all the assumption being used):

Given a class with equal number of male and female students, we have that

probability of male passing an exam is say, $p$ and probability of female passing an exam is say, $q$ what is the probability that a randomly selected student passes the third exam?

Note: student drops out if he/she fails an exam, thus cannot appear for the next exam.

  • $\begingroup$ Can you use $$\frac{N_a}{N}$$ where $N_a=(M*p)+(F*q)$ and N is the total number of students. $\endgroup$ – 123 Jan 13 '15 at 16:23
  • $\begingroup$ We are being tacitly invited to assume (unreasonably) that given that a randomly chosen female passes her first exam, the probability she passes the second is still $q$. It is also not clear whether we want the probability a randomly chosen student passes the third exam, or the conditional probability that the student passes the third given she/he passes the first two. $\endgroup$ – André Nicolas Jan 13 '15 at 17:04
  • $\begingroup$ wouldn't the probability of passing the third exam given atleast one of the first two is a fail, is automatically zero. so it must be that the conditional probability in this case is equal to the unconditional one. $\endgroup$ – user13892 Jan 13 '15 at 17:18

Consider that there are $x$ males and females in the class.

After the first exam: $px$ males are expected to pass while $qx$ females are expected to pass.

After the second exam: $p^2x$ males are expected to pass while $q^2x$ females are expected to pass.

Thus, when picking randomly for the third exam: Total number of students: $x(p^2 + q^2)$

The chance of picking a male student: $$\frac{p^2x}{x(p^2 + q^2)}$$

While chances of picking a female student: $$\frac{q^2x}{x(p^2 + q^2)}$$

From here, you can separately find the chances of picking a passing male student or a passing female student, and add them. The final answer should be: $$\frac{p^3 + q^3}{p^2 + q^2}$$

  • $\begingroup$ thank you very much gummy bear. now can i eat you :) $\endgroup$ – user13892 Jan 13 '15 at 17:22

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