Let $S$ be a sample space and $E$ and $F$ be events associated with $S$. Suppose that $\operatorname{Pr}(E)=\frac{1}{3}, \operatorname{Pr}(F)=\frac{4}{9}$, and $\operatorname{Pr}(E U F)=\frac{3}{4}$. Calculate the following probabilities a. $\operatorname{Pr}(E \cap F)$ b. $\operatorname{Pr}(E \mid F)$ c. $\operatorname{Pr}(\mathrm{F} \mid \mathrm{E})$
Solution
Step 1 ：Let's denote the probability of event E as \(\operatorname{Pr}(E)\), the probability of event F as \(\operatorname{Pr}(F)\), and the probability of the union of E and F as \(\operatorname{Pr}(E \cup F)\). Given that \(\operatorname{Pr}(E) = \frac{1}{3}\), \(\operatorname{Pr}(F) = \frac{4}{9}\), and \(\operatorname{Pr}(E \cup F) = \frac{3}{4}\).
Step 2 ：We can find the probability of the intersection of E and F, denoted as \(\operatorname{Pr}(E \cap F)\), by rearranging the formula for the probability of the union of two events: \(\operatorname{Pr}(E \cup F) = \operatorname{Pr}(E) + \operatorname{Pr}(F) - \operatorname{Pr}(E \cap F)\). Solving for \(\operatorname{Pr}(E \cap F)\) gives us \(\operatorname{Pr}(E \cap F) = \operatorname{Pr}(E) + \operatorname{Pr}(F) - \operatorname{Pr}(E \cup F)\), which simplifies to \(\operatorname{Pr}(E \cap F) = \frac{1}{3} + \frac{4}{9} - \frac{3}{4} = \boxed{0.02777777777777768}\).
Step 3 ：Next, we can find the conditional probability of E given F, denoted as \(\operatorname{Pr}(E | F)\), using the formula for conditional probability: \(\operatorname{Pr}(E | F) = \frac{\operatorname{Pr}(E \cap F)}{\operatorname{Pr}(F)}\). Substituting the values we have, we get \(\operatorname{Pr}(E | F) = \frac{0.02777777777777768}{\frac{4}{9}} = \boxed{0.06249999999999978}\).
Step 4 ：Finally, we can find the conditional probability of F given E, denoted as \(\operatorname{Pr}(F | E)\), using the formula for conditional probability: \(\operatorname{Pr}(F | E) = \frac{\operatorname{Pr}(E \cap F)}{\operatorname{Pr}(E)}\). Substituting the values we have, we get \(\operatorname{Pr}(F | E) = \frac{0.02777777777777768}{\frac{1}{3}} = \boxed{0.08333333333333304}\).
