Let $P(E)=0.75, P(F)=0.26$, and $P(E U F)=0.77$. Find $(a) P(E \mid F)$ and $(b) P(F \mid E)$. $(\mathrm{a}) \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\mathrm{W}$ (Type an integer or decimal rounded to two decimal places as needed.) (b) $P(F \mid E)=\square$ (Type an integer or decimal rounded to wo decimal places as needed.)

Step 1 ：Given that \(P(E) = 0.75\), \(P(F) = 0.26\), and \(P(E \cup F) = 0.77\)

Step 2 ：We can find \(P(E \cap F)\) using the formula for the union of two events: \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\)

Step 3 ：Substituting the given values, we get \(0.77 = 0.75 + 0.26 - P(E \cap F)\)

Step 4 ：Solving for \(P(E \cap F)\), we get \(P(E \cap F) = 0.75 + 0.26 - 0.77 = 0.24\)

Step 5 ：Now we can find the conditional probabilities.

Step 6 ：For \(P(E|F)\), we use the formula for conditional probability: \(P(E|F) = \frac{P(E \cap F)}{P(F)}\)

Step 7 ：Substituting the values, we get \(P(E|F) = \frac{0.24}{0.26} = 0.92\)

Step 8 ：For \(P(F|E)\), we use the formula for conditional probability: \(P(F|E) = \frac{P(E \cap F)}{P(E)}\)

Step 9 ：Substituting the values, we get \(P(F|E) = \frac{0.24}{0.75} = 0.32\)

Step 10 ：So, the answers are \(\boxed{P(E|F) = 0.92}\) and \(\boxed{P(F|E) = 0.32}\)