If E and F are independent events with P(E) = 0.31 and P(F) = 0.4, then what is P(Ec ∪ F)? round to the 4th decimal place

To find \( P(E^c \cup F) \), we can use the following rule for the probability of the union of two events:

\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

Here, \( A = E^c \) (the complement of \( E \)), and \( B = F \).

First, we need to calculate each probability step-by-step:

1. **Find \( P(E^c) \)**:
   \[
   P(E^c) = 1 - P(E) = 1 - 0.31 = 0.69
   \]

2. **Since \( E \) and \( F \) are independent, the intersection of \( E^c \) and \( F \) is given by**:
   \[
   P(E^c \cap F) = P(E^c) \cdot P(F) = 0.69 \cdot 0.4 = 0.276
   \]

3. **Now apply the formula for the union**:
   \[
   P(E^c \cup F) = P(E^c) + P(F) - P(E^c \cap F)
   \]
   Substituting the values:
   \[
   P(E^c \cup F) = 0.69 + 0.4 - 0.276 = 0.814
   \]

Thus, \( P(E^c \cup F) = 0.8140 \) when rounded to the fourth decimal place.

Would you like further clarification on these steps?

Here are 5 questions to explore:

1. What is the complement rule in probability?
2. How do we handle probabilities of independent events?
3. Can the union rule be applied to dependent events, and how does it differ?
4. What is the difference between union and intersection in probability?
5. How does the complement of an event affect its probability?

**Tip**: The complement of an event \( E \), denoted as \( E^c \), represents all outcomes in the sample space that are not in \( E \). The probability of the complement is always \( 1 - P(E) \).

If E and F are independent events with P(E) = 0.31 and P(F) = 0.4, then what is P(Ec ∪ F)? Round to the 4th decimal place.

This problem involves calculating the probability of the union of independent events E and F, where P(E) = 0.31 and P(F) = 0.4. Using complement and union rules, we solve for P(Ec ∪ F) step by step. Suitable for students in Grades 10-12.

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