a) Students have two quizzes that they must take in their undergraduate studies unit. The probability of getting a perfect score on the first quiz is 68%, the probability of getting a perfect score on the second quiz is 74%, and the probability of falling short of a perfect score on both quizzes is 19%. What is the probability of scoring a perfect score on both quizzes?
My attempt: Pr(Scoring a perfect score on both quizes) $= 0.68 \cdot 0.74 = 0.5032$
b) Refer to the probabilities described in question a. If a student has achieved a perfect score on the first quiz, what is the conditional probability of getting a perfect score on the second quiz?
My attempt:
$$\frac{\text{Pr(Scoring a perfect score on both quizzes)}}{\text{Pr(Perfect score on second Quiz)}} = 0.68$$
but I feel like there is more to both questions
Best Answer
Let $A$ be the event that a perfect score is obtained on the first quiz; let $B$ be the event that a perfect score is obtained on the second quiz. We are given \begin{align*} P(A) & = 0.68\\ P(B) & = 0.74\\ P(A^C \cap B^C) & = 0.19 \end{align*} Hence, \begin{align*} P(A^C) & = 1 - P(A) = 0.32\\ P(B^C) & = 1 - P(B) = 0.74 \end{align*} The probability of not obtaining a perfect score on at least one of the quizzes is $$P(A^C \cup B^C) = P(A^C) + P(B^C) - P(A^C \cap B^C) = 0.32 + 0.26 - 0.19 = 0.39$$ Thus, the probability of obtaining a perfect score on both quizzes is $$P(A \cap B) = 1 - P(A^C \cup B^C) = 1 - 0.39 = 0.61$$ so your assumption that the probabilities of obtaining a perfect score on each quiz are independent was false.
The probability that a student who received a perfect mark on the first quiz obtains a perfect mark on the second quiz is $$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.61}{0.68} = \frac{61}{68}$$