To calculate \( P(A|B) \), which is the probability of event \( A \) given event \( B \), we can use the formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Step 1: Identify event \( B \)
Event \( B \) is defined as the set of even numbers from the set \( S \):
\[ B = {2, 4, 6, 8} \]
Step 2: Find \( P(B) \)
The total number of outcomes in the sample space \( S \) is 8. The number of favorable outcomes for \( B \) (the even numbers) is 4. Thus, we have:
\[ P(B) = \frac{\text{Number of favorable outcomes in } B}{\text{Total outcomes in } S} = \frac{4}{8} = \frac{1}{2} \]
Step 3: Find \( A \cap B \)
Now, we need to find the intersection of \( A \) and \( B \):
\[ A = {2, 5} \quad \text{and} \quad B = {2, 4, 6, 8} \]
The intersection \( A \cap B \) includes only the elements that are in both \( A \) and \( B \):
\[ A \cap B = {2} \]
Step 4: Calculate \( P(A \cap B) \)
The number of outcomes in \( A \cap B \) is 1 (only the number 2). Therefore, we have:
\[ P(A \cap B) = \frac{\text{Number of favorable outcomes in } A \cap B}{\text{Total outcomes in } S} = \frac{1}{8} \]
Step 5: Calculate \( P(A|B) \)
Now we can plug in the values into the conditional probability formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{8} \times \frac{2}{1} = \frac{2}{8} = \frac{1}{4} \]
Thus, the answer is:
\(\frac{1}{4}\)