If E and F are two disjoint events with P(E) = 0.16 and P(F) = 0.32, what is P(E ∪ FC)? (Enter answer as a decimal with at least 2 correct decimal places)

Given that \( E \) and \( F \) are two disjoint events, we know:

- \( P(E) = 0.16 \)
- \( P(F) = 0.32 \)
- \( P(E \cap F) = 0 \) (since disjoint events have no overlap)

We are tasked with finding \( P(E \cup F^c) \), where \( F^c \) is the complement of event \( F \). Recall that:

\[
P(E \cup F^c) = P(E) + P(F^c) - P(E \cap F^c)
\]

### Step 1: Calculate \( P(F^c) \)
The probability of the complement of \( F \) is:

\[
P(F^c) = 1 - P(F) = 1 - 0.32 = 0.68
\]

### Step 2: Find \( P(E \cap F^c) \)
Since \( E \) and \( F \) are disjoint, they cannot both occur. Therefore, the intersection \( E \cap F^c \) is simply \( P(E) \). This is because if \( F \) does not occur, \( E \) can occur:

\[
P(E \cap F^c) = P(E) = 0.16
\]

### Step 3: Calculate \( P(E \cup F^c) \)
Now we can apply the formula for the union:

\[
P(E \cup F^c) = P(E) + P(F^c) - P(E \cap F^c) = 0.16 + 0.68 - 0.16 = 0.68
\]

Thus, the probability \( P(E \cup F^c) \) is:

\[
\boxed{0.68}
\]

Let me know if you'd like further clarification or details!

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Here are 5 related questions you might consider:
1. What is the probability of \( F \cup F^c \)?
2. How would the result change if \( E \) and \( F \) were not disjoint?
3. What is \( P(F^c) \cap E^c \)?
4. How would you compute \( P(E \cup F) \)?
5. What is the interpretation of \( P(E \cap F) \) when \( E \) and \( F \) are independent?

**Tip:** Always remember the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) for union probabilities!

Learn how to calculate P(E ∪ FC) when E and F are disjoint events with P(E) = 0.16 and P(F) = 0.32. This problem involves key probability concepts, including the union of events and complementary events. Suitable for high school students.

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