We are given three independent events A, B, and C with probabilities $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, and $P(C) = \frac{1}{4}$. We want to find the probability of the event $(A \cap B) \cup C$. We need to provide the answer as an irreducible fraction.

1. Problem Description

We are given three independent events A, B, and C with probabilities $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, and $P(C) = \frac{1}{4}$. We want to find the probability of the event $(A \cap B) \cup C$. We need to provide the answer as an irreducible fraction.

2. Solution Steps

Since A, B, and C are independent events, $A \cap B$ is also independent of C.

Therefore, we can use the formula for the probability of the union of two events:

$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$.

In our case, $X = A \cap B$ and $Y = C$.

Thus, we have:

$P((A \cap B) \cup C) = P(A \cap B) + P(C) - P((A \cap B) \cap C)$.

Since A, B, and C are independent,

$P(A \cap B) = P(A)P(B) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$.

Also, since A, B, and C are independent,

$P((A \cap B) \cap C) = P(A \cap B \cap C) = P(A)P(B)P(C) = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{24}$.

Now we can substitute these values into the formula:

$P((A \cap B) \cup C) = \frac{1}{6} + \frac{1}{4} - \frac{1}{24}$.

To add these fractions, we need a common denominator, which is

2

4. $P((A \cap B) \cup C) = \frac{4}{24} + \frac{6}{24} - \frac{1}{24} = \frac{4 + 6 - 1}{24} = \frac{9}{24}$.

Now we simplify the fraction:

$\frac{9}{24} = \frac{3 \cdot 3}{3 \cdot 8} = \frac{3}{8}$.

3. Final Answer

The probability $P((A \cap B) \cup C)$ is $\frac{3}{8}$.