Understanding Mutually Exclusive and Exhaustive Events

In probability, events are considered mutually exclusive if they cannot happen at the same time. For example, flipping a coin cannot result in both heads and tails simultaneously. These are mutually exclusive events.

Events are considered exhaustive if they cover all possible outcomes in the sample space. If you consider all possible outcomes of an experiment, they form an exhaustive set of events. For example, when rolling a standard six-sided die, the outcomes {1, 2, 3, 4, 5, 6} are exhaustive because one of them must occur.

When a set of events is both mutually exclusive and exhaustive, the sum of their individual probabilities must equal 1. This is a fundamental property used in solving this problem.

For events P, Q, R, and S being mutually exclusive and exhaustive, we have:

\(P(P) + P(Q) + P(R) + P(S) = 1\)

Analyzing the Given Probability Relations

The problem provides a relationship between the probabilities of the four events:

\(3P(P) = P(Q) = 2P(R) = 4P(S)\)

We want to find the value of \(P(Q)\). Let's express the probabilities of P, R, and S in terms of \(P(Q)\) using the given relations.

Let \(P(Q) = x\).

• From \(3P(P) = P(Q)\), we get \(3P(P) = x\). Dividing both sides by 3, we find \(P(P) = \frac{x}{3}\).
• We are given \(P(Q) = x\).
• From \(2P(R) = P(Q)\), we get \(2P(R) = x\). Dividing both sides by 2, we find \(P(R) = \frac{x}{2}\).
• From \(4P(S) = P(Q)\), we get \(4P(S) = x\). Dividing both sides by 4, we find \(P(S) = \frac{x}{4}\).

Calculating the Probability P(Q)

Since the events P, Q, R, and S are exhaustive, the sum of their probabilities is 1. We can substitute the expressions in terms of \(x\) into the equation \(P(P) + P(Q) + P(R) + P(S) = 1\):

\(\frac{x}{3} + x + \frac{x}{2} + \frac{x}{4} = 1\)

To solve for \(x\), we need to combine the terms on the left side by finding a common denominator. The denominators are 3, 1, 2, and 4. The least common multiple (LCM) of 3, 1, 2, and 4 is 12.

Convert each fraction to have a denominator of 12:

• \(\frac{x}{3} = \frac{x \times 4}{3 \times 4} = \frac{4x}{12}\)
• \(x = \frac{x \times 12}{1 \times 12} = \frac{12x}{12}\)
• \(\frac{x}{2} = \frac{x \times 6}{2 \times 6} = \frac{6x}{12}\)
• \(\frac{x}{4} = \frac{x \times 3}{4 \times 3} = \frac{3x}{12}\)

Now substitute these back into the equation:

\(\frac{4x}{12} + \frac{12x}{12} + \frac{6x}{12} + \frac{3x}{12} = 1\)

Combine the numerators over the common denominator:

\(\frac{4x + 12x + 6x + 3x}{12} = 1\)

\(\frac{25x}{12} = 1\)

To isolate \(x\), multiply both sides by 12:

\(25x = 1 \times 12\)

\(25x = 12\)

Finally, divide by 25 to find \(x\):

\(x = \frac{12}{25}\)

Since we defined \(x = P(Q)\), the probability of event Q is \(\frac{12}{25}\).

Verification

Let's check if the probabilities sum to 1 with \(P(Q) = \frac{12}{25}\):

• \(P(Q) = \frac{12}{25}\)
• \(P(P) = \frac{1}{3} P(Q) = \frac{1}{3} \times \frac{12}{25} = \frac{4}{25}\)
• \(P(R) = \frac{1}{2} P(Q) = \frac{1}{2} \times \frac{12}{25} = \frac{6}{25}\)
• \(P(S) = \frac{1}{4} P(Q) = \frac{1}{4} \times \frac{12}{25} = \frac{3}{25}\)

Sum of probabilities:

\(P(P) + P(Q) + P(R) + P(S) = \frac{4}{25} + \frac{12}{25} + \frac{6}{25} + \frac{3}{25} = \frac{4 + 12 + 6 + 3}{25} = \frac{25}{25} = 1\)

The probabilities sum to 1, which confirms our calculation for \(P(Q)\).

Revision Table: Key Concepts

Additional Information: Probability Principles

This problem utilizes fundamental principles of probability theory. The concept of a sample space (S) which contains all possible outcomes is crucial. Any event is a subset of the sample space. The probability of any event E, denoted P(E), is always between 0 and 1, inclusive (\(0 \le P(E) \le 1\)). The probability of the sample space S is always 1 (\(P(S) = 1\)).

For mutually exclusive events E₁, E₂, ..., E_n, the probability of their union (at least one of them occurring) is the sum of their individual probabilities:

\(P(E_1 \text{ or } E_2 \text{ or } \dots \text{ or } E_n) = P(E_1 \cup E_2 \cup \dots \cup E_n) = P(E_1) + P(E_2) + \dots + P(E_n)\)

When these events are also exhaustive, their union is the entire sample space S, so the sum of their probabilities must equal \(P(S)\), which is 1. This is exactly the property used in the solution: \(P(P) + P(Q) + P(R) + P(S) = 1\).