If P(A) = 0.4, P(B) = 0.2, and P(C) = 0.3, calculate the probability P(A ∪B ∪C), if the events A, B, 
and C are:
(i) Pairwise disjoint. (ii) Independen

To calculate the probability \( P(A \cup B \cup C) \), we'll handle the two cases separately: 

### (i) Pairwise Disjoint Events
When events are pairwise disjoint, it means that no two events can happen at the same time, i.e., the intersection of any two events is empty. Mathematically, this is expressed as:

\[
P(A \cap B) = P(B \cap C) = P(A \cap C) = 0
\]

For pairwise disjoint events, the probability of their union is simply the sum of their individual probabilities:

\[
P(A \cup B \cup C) = P(A) + P(B) + P(C)
\]

Substituting the given probabilities:

\[
P(A \cup B \cup C) = 0.4 + 0.2 + 0.3 = 0.9
\]

### (ii) Independent Events
When events are independent, the occurrence of one event does not affect the occurrence of the others. The probability of their union is calculated using the inclusion-exclusion principle:

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

For independent events:

\[
P(A \cap B) = P(A) \times P(B), \quad P(B \cap C) = P(B) \times P(C), \quad P(A \cap C) = P(A) \times P(C)
\]
\[
P(A \cap B \cap C) = P(A) \times P(B) \times P(C)
\]

Now, let's substitute the given probabilities:

\[
P(A \cap B) = 0.4 \times 0.2 = 0.08
\]
\[
P(B \cap C) = 0.2 \times 0.3 = 0.06
\]
\[
P(A \cap C) = 0.4 \times 0.3 = 0.12
\]
\[
P(A \cap B \cap C) = 0.4 \times 0.2 \times 0.3 = 0.024
\]

Finally, substitute these into the inclusion-exclusion formula:

\[
P(A \cup B \cup C) = 0.4 + 0.2 + 0.3 - 0.08 - 0.06 - 0.12 + 0.024 = 0.764
\]

### Summary of Results
- **(i) Pairwise disjoint events:** \( P(A \cup B \cup C) = 0.9 \)
- **(ii) Independent events:** \( P(A \cup B \cup C) = 0.764 \)

Would you like more details on any of these calculations?

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Here are some related questions for further practice:
1. What is the probability \( P(A \cup B) \) for disjoint events \( A \) and \( B \)?
2. How does the inclusion-exclusion principle apply when calculating probabilities for more than three events?
3. What changes in the probability calculation if events are mutually exclusive?
4. How would you calculate \( P(A \cup B \cup C) \) if two events are dependent?
5. Can the sum of individual probabilities of three events exceed 1? Why or why not?

**Tip:** Remember that disjoint events are always mutually exclusive, but independent events are not necessarily disjoint.

Learn how to calculate the probability of the union of events A, B, and C when they are pairwise disjoint or independent. Understand the application of the inclusion-exclusion principle in probability theory. Suitable for advanced high school students studying probability.

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