Understanding Mutually Exclusive Events

In probability theory, events are considered mutually exclusive (also known as disjoint events) if they cannot happen at the same time. This means that the occurrence of one event makes the other event impossible. For instance, when rolling a single die, the event of rolling an even number and the event of rolling an odd number are mutually exclusive because you cannot roll a number that is both even and odd simultaneously.

For any two mutually exclusive events, say A and B, the probability that either A or B occurs (their union) is found by simply adding their individual probabilities. This is expressed by the formula:

\begin{equation*} P(A \cup B) = P(A) + P(B) \end{equation*}

Given Probability Relationship

The problem states that A and B are mutually exclusive events. It also provides a specific relationship between their probabilities:

\begin{equation*} P(A) = \frac{1}{2} P(B) \end{equation*}

To make the calculation easier, we can rearrange this equation to express \(P(B)\) in terms of \(P(A)\). By multiplying both sides of the equation by 2, we get:

\begin{equation*} P(B) = 2 P(A) \end{equation*}

Calculating Probability P(A)

To find the exact value of \(P(A)\), we need to assume a total probability for the union of these events. In the context of such problems, if A and B are the only events being discussed within a complete sample space, it's generally implied that their union covers the entire sample space, meaning the probability of their union is 1. Therefore, we will assume:

\begin{equation*} P(A \cup B) = 1 \end{equation*}

Now, we can use the formula for the union of mutually exclusive events and substitute the expression we found for \(P(B)\):

\begin{equation*} P(A \cup B) = P(A) + P(B) \end{equation*}

Substitute \(P(B) = 2 P(A)\) into the equation:

\begin{equation*} 1 = P(A) + 2 P(A) \end{equation*}

Combine the terms involving \(P(A)\) on the right side:

\begin{equation*} 1 = 3 P(A) \end{equation*}

Finally, to solve for \(P(A)\), divide both sides of the equation by 3:

\begin{equation*} P(A) = \frac{1}{3} \end{equation*}

Thus, the probability of event A is \(\frac{1}{3}\).