Step 1: Understanding the Question:
The question asks to find the probability of the union of two events, \(A\) and \(C\), given relationships between the probabilities of three mutually exclusive and exhaustive events \(A\), \(B\), and \(C\).
Step 2: Key Formula or Approach:
Because \(A\), \(B\), and \(C\) are mutually exclusive and exhaustive events, their individual probabilities sum to one:
\[ P(A) + P(B) + P(C) = 1 \]
Additionally, because they are mutually exclusive, the probability of the union of any two is simply the sum of their individual probabilities:
\[ P(A \cup C) = P(A) + P(C) \]
Step 3: Detailed Explanation:
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Step 3.1: Express all probabilities in terms of \(P(A)\):
We are given:
\[ P(A) = \frac{1}{3} P(B) \implies P(B) = 3 P(A) \]
We are also given:
\[ P(B) = 2 P(C) \implies P(C) = \frac{1}{2} P(B) \]
Substitute the expression for \(P(B)\) to find \(P(C)\) in terms of \(P(A)\):
\[ P(C) = \frac{3}{2} P(A) \]
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Step 3.2: Solve for \(P(A)\) using the sum of probabilities:
\[ P(A) + 3 P(A) + \frac{3}{2} P(A) = 1 \]
Combine like terms:
\[ \left( 1 + 3 + 1.5 \right) P(A) = 1 \implies 5.5 P(A) = 1 \implies \frac{11}{2} P(A) = 1 \]
\[ P(A) = \frac{2}{11} \]
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Step 3.3: Calculate \(P(C)\) and \(P(A \cup C)\):
\[ P(C) = \frac{3}{2} P(A) = \frac{3}{2} \left(\frac{2}{11}\right) = \frac{3}{11} \]
Calculate the union probability:
\[ P(A \cup C) = P(A) + P(C) = \frac{2}{11} + \frac{3}{11} = \frac{5}{11} \]
Step 4: Final Answer:
The probability \(P(A \cup C)\) is equal to \(\frac{5}{11}\).