Understanding Independent Events

The problem involves three independent events, A, B, and C. When events are independent, the occurrence of one event does not affect the probability of the others. This means the probability of their intersection is the product of their individual probabilities. For example, \(P(A \cap B) = P(A) \times P(B)\). Similarly, if we consider an event and the complement of another, like A and not B (\(B'\)), their intersection probability is \(P(A \cap B') = P(A) \times P(B')\).

We are given the probabilities of the three independent events:

• \(P(A) = \frac{1}{3}\)
• \(P(B) = \frac{1}{2}\)
• \(P(C) = \frac{1}{4}\)

Calculating Complement Probabilities

Since we need to consider cases where events do not occur, we first calculate the probabilities of the complements of events A, B, and C. The probability of a complement event \(E'\) is \(P(E') = 1 - P(E)\).

• \(P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{3-1}{3} = \frac{2}{3}\)
• \(P(B') = 1 - P(B) = 1 - \frac{1}{2} = \frac{2-1}{2} = \frac{1}{2}\)
• \(P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{4-1}{4} = \frac{3}{4}\)

Probability of Exactly 2 Events Occurring

We want to find the probability that exactly 2 out of the 3 events A, B, C occur. There are three possible scenarios for exactly two events to occur:

1. Event A and Event B occur, and Event C does not occur (\(A \cap B \cap C'\)).
2. Event A and Event C occur, and Event B does not occur (\(A \cap C \cap B'\)).
3. Event B and Event C occur, and Event A does not occur (\(B \cap C \cap A'\)).

Since the events A, B, C are independent, the events \(A\), \(B\), and \(C'\) are also independent. Similarly, \(A\), \(C\), and \(B'\) are independent, and \(B\), \(C\), and \(A'\) are independent. We can calculate the probability of each scenario by multiplying the probabilities of the individual events involved.

• Probability of \(A \cap B \cap C'\): \(P(A \cap B \cap C') = P(A) \times P(B) \times P(C') = \frac{1}{3} \times \frac{1}{2} \times \frac{3}{4} = \frac{1 \times 1 \times 3}{3 \times 2 \times 4} = \frac{3}{24} = \frac{1}{8}\)
• Probability of \(A \cap C \cap B'\): \(P(A \cap C \cap B') = P(A) \times P(C) \times P(B') = \frac{1}{3} \times \frac{1}{4} \times \frac{1}{2} = \frac{1 \times 1 \times 1}{3 \times 4 \times 2} = \frac{1}{24}\)
• Probability of \(B \cap C \cap A'\): \(P(B \cap C \cap A') = P(B) \times P(C) \times P(A') = \frac{1}{2} \times \frac{1}{4} \times \frac{2}{3} = \frac{1 \times 1 \times 2}{2 \times 4 \times 3} = \frac{2}{24} = \frac{1}{12}\)

These three scenarios are mutually exclusive (they cannot happen at the same time). Therefore, the probability of exactly 2 events occurring is the sum of the probabilities of these three scenarios.

\(P(\text{exactly 2 events}) = P(A \cap B \cap C') + P(A \cap C \cap B') + P(B \cap C \cap A')\)

\(P(\text{exactly 2 events}) = \frac{1}{8} + \frac{1}{24} + \frac{1}{12}\)

To add these fractions, we find a common denominator, which is 24.

• \(\frac{1}{8} = \frac{1 \times 3}{8 \times 3} = \frac{3}{24}\)
• \(\frac{1}{24}\) remains \(\frac{1}{24}\)
• \(\frac{1}{12} = \frac{1 \times 2}{12 \times 2} = \frac{2}{24}\)

Now, sum the fractions:

\(P(\text{exactly 2 events}) = \frac{3}{24} + \frac{1}{24} + \frac{2}{24} = \frac{3 + 1 + 2}{24} = \frac{6}{24}\)

Simplify the resulting fraction:

\(\frac{6}{24} = \frac{6 \div 6}{24 \div 6} = \frac{1}{4}\)

Thus, the probability of exactly 2 events occurring out of the three independent events is \(\frac{1}{4}\).