If A, B and C are mutually exclusive and exhaustive events of a random experiment such that \(P(A) = \frac{1{3} P(B)\) and \(P(B) = 2 P(C)\), then \(P(A \cup C)\) is equal to _______}
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Using variables can make this calculation easier.
Let $P(A) = 2x$. Then $P(B) = 6x$ and $P(C) = 3x$.
Summing them gives $2x + 6x + 3x = 11x = 1 \implies x = 1/11$.
Then $P(A \cup C) = 2x + 3x = 5x = 5/11$. This approach avoids working with fractions directly.