Question

If $P(E)=\frac{1}{3}$, $P(F)=\frac{1}{5}$ and $P(E\cup F)=\frac{1}{2}$ then $P(E|\overline{F})+P(F|\overline{E})$ is equal to

• A. $\frac{5}{4}$
• B. $\frac{5}{8}$
• C. $\frac{3}{4}$
• D. $\frac{7}{8}$

Hint:

$P(A \cap \overline{B})$ is the probability of "A only." It's always calculated by subtracting the shared part ($A \cap B$) from the total probability of $A$. Visualizing this with a Venn diagram can prevent common subtraction errors.

Answer 1

The Correct Option is B

We need to evaluate the sum of two conditional probabilities involving complements.

Step 1: \color{redFind the Probability of the Intersection
Using the Addition Theorem: $P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
$\frac{1}{2} = \frac{1}{3} + \frac{1}{5} - P(E \cap F)$
$P(E \cap F) = \frac{1}{3} + \frac{1}{5} - \frac{1}{2}$
Common denominator is 30:
$P(E \cap F) = \frac{10 + 6 - 15}{30} = \frac{1}{30}$.

Step 2: \color{redCalculate Component Probabilities
$P(\overline{E}) = 1 - 1/3 = 2/3$.
$P(\overline{F}) = 1 - 1/5 = 4/5$.
$P(E \cap \overline{F}) = P(E) - P(E \cap F) = 1/3 - 1/30 = (10-1)/30 = 9/30 = 3/10$.
$P(F \cap \overline{E}) = P(F) - P(E \cap F) = 1/5 - 1/30 = (6-1)/30 = 5/30 = 1/6$.

Step 3: \color{redEvaluate the Individual Conditional Probabilities
$P(E|\overline{F}) = \frac{P(E \cap \overline{F})}{P(\overline{F})} = \frac{3/10}{4/5} = \frac{3}{10} \times \frac{5}{4} = \frac{15}{40} = \frac{3}{8}$.
$P(F|\overline{E}) = \frac{P(F \cap \overline{E})}{P(\overline{E})} = \frac{1/6}{2/3} = \frac{1}{6} \times \frac{3}{2} = \frac{3}{12} = \frac{1}{4}$.

Step 4: \color{redCompute the Final Sum
Sum $= \frac{3}{8} + \frac{1}{4}$
Sum $= \frac{3}{8} + \frac{2}{8} = \frac{5}{8}$.
This matches Option (2).