Among $33$ students in a class, $17$ of them earned A's on the midterm, $14$ earned A's on the final exam, $11$ of them did not earn A on either exam.
What is the probability that a randomly selected student from this class earned A's on both exams?
My attempt:
$P(\text{earned A on midterm}) = \frac{17}{33}$
$P(\text{earned A on final}) = \frac{14}{33}$
$P(\text{didn't earn A on either exam}) = \frac{11}{33}$
NEED: $P(A_m \cap A_f)$
$$P(A_m \cap A_f) = P(A_m) + P(P_f) – P(A_m \cup A_f)$$
Stuck. Please help
Best Answer
$A_m \cup A_f$ is the set of students who got an $A$ on either the midterm or the final; its complement consists of the students who got an $A$ on neither exam, which you know refers to $11$ students. Since there are $33$ students in total, it must be that $|A_m \cup A_f| = 33 - 11 = 22$.
This means that $P(A_m \cup A_f) = 22/33 = 2/3$, and you now have numerical values for all of the terms on the right hand side of the equation that you provided. Plugging them in allows you to solve for the value of the left hand side of the equation, which is the probability about which you are asking.