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Let's solve the given problem step by step. We are given that \( A \), \( B \), and \( C \) are mutually exclusive events, meaning they cannot occur simultaneously. The probabilities are:

- \( P(A) = 0.23 \)
- \( P(B) = 0.34 \)
- \( P(C) = 0.41 \)

We need to find the following probabilities:

### (a) \( P(A \cup B \cup C) \)

Since \( A \), \( B \), and \( C \) are mutually exclusive, the probability of their union is the sum of their individual probabilities:

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

Substituting the given values:

\[
P(A \cup B \cup C) = 0.23 + 0.34 + 0.41 = 0.98
\]

### (b) \( P(A \cap B \cap C) \)

Because \( A \), \( B \), and \( C \) are mutually exclusive, they cannot occur together. Hence, the intersection of these events is 0:

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

### (c) \( P(A \cap B) \)

For mutually exclusive events, the intersection between any two events is also 0:

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

### (d) \( P((A \cup B) \cap C) \)

This is asking for the probability of the intersection of \( C \) with the union of \( A \) and \( B \). Since \( A \), \( B \), and \( C \) are mutually exclusive, the intersection of \( C \) with \( A \cup B \) is 0:

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

### (e) \( P(A' \cap B' \cap C') \)

This is the probability that none of the events \( A \), \( B \), or \( C \) occur. To calculate this, we first find the complement of each event:

\[
P(A') = 1 - P(A) = 1 - 0.23 = 0.77
\]
\[
P(B') = 1 - P(B) = 1 - 0.34 = 0.66
\]
\[
P(C') = 1 - P(C) = 1 - 0.41 = 0.59
\]

Since \( A \), \( B \), and \( C \) are mutually exclusive, their complements are independent. Thus, we can multiply the probabilities:

\[
P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') = 0.77 \times 0.66 \times 0.59 = 0.300
\]

Rounding to two decimal places:

\[
P(A' \cap B' \cap C') = 0.30
\]

### Summary of Results:
- (a) \( P(A \cup B \cup C) = 0.98 \)
- (b) \( P(A \cap B \cap C) = 0 \)
- (c) \( P(A \cap B) = 0 \)
- (d) \( P((A \cup B) \cap C) = 0 \)
- (e) \( P(A' \cap B' \cap C') = 0.30 \)

Would you like further clarification or details?

Here are five related questions:
1. How would the probabilities change if \( A \), \( B \), and \( C \) were not mutually exclusive?
2. What is the general formula for \( P(A \cup B) \) when events are not mutually exclusive?
3. How do you calculate the complement of the union of three events?
4. If \( P(A) \), \( P(B) \), and \( P(C) \) were dependent, how would you approach calculating \( P(A \cap B) \)?
5. Can mutually exclusive events ever be independent?

**Tip:** Mutually exclusive events cannot occur at the same time, while independent events do not influence each other's occurrence.

If A, B, and C are mutually exclusive events with P(A) = 0.23, P(B) = 0.34, and P(C) = 0.41, determine the following probabilities: (a) P(A ∪ B ∪ C), (b) P(A ∩ B ∩ C), (c) P(A ∩ B), (d) P((A ∪ B) ∩ C), (e) P(A' ∩ B' ∩ C').

Learn how to calculate the probability of mutually exclusive events with given probabilities. This problem covers key probability concepts including union, intersection, and complement of events. Suitable for Grades 9-11.

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