Find the probability​ P(E or​ F) if E and F are mutually​ exclusive, P(E)=0.34, and P(F)=0.45

Since events \( E \) and \( F \) are mutually exclusive, the probability \( P(E \text{ or } F) \) is given by the sum of the probabilities of each event:

\[
P(E \text{ or } F) = P(E) + P(F)
\]

Given:
- \( P(E) = 0.34 \)
- \( P(F) = 0.45 \)

So,

\[
P(E \text{ or } F) = 0.34 + 0.45 = 0.79
\]

Thus, the probability \( P(E \text{ or } F) \) is \( 0.79 \).

Would you like more details or have any questions?

Here are some related questions:
1. What is the probability of either of two events happening if they are not mutually exclusive?
2. How do you calculate the probability of the intersection of two events?
3. What does it mean for two events to be mutually exclusive in probability theory?
4. Can the probability of mutually exclusive events ever exceed 1? Why or why not?
5. How would the probability change if \( P(E) = 0.6 \) and \( P(F) = 0.5 \)?

**Tip:** When dealing with mutually exclusive events, remember that their intersection is always zero, simplifying the calculation of the union's probability.

Understand how to calculate the probability of either event E or event F occurring when they are mutually exclusive. Given probabilities P(E) = 0.34 and P(F) = 0.45, learn the step-by-step method to find P(E or F). Ideal for students studying advanced probability theory.

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