Calculating the Probability of Exactly One Event Occurring

We are given two independent events, A and B, with their respective probabilities:

• Probability of event A, \({\rm{P}}\left( {\rm{A}} \right) = \frac{1}{3}\)
• Probability of event B, \({\rm{P}}\left( {\rm{B}} \right) = \frac{3}{4}\)

We need to find the probability that exactly one of these two events occurs. This means either event A occurs and event B does not occur, OR event B occurs and event A does not occur.

Let's denote the event that B does not occur as \({\rm{B}}^{\rm{c}}\), and the event that A does not occur as \({\rm{A}}^{\rm{c}}\). The probability of the complement of an event is \({\rm{P}}({\rm{E}}^{\rm{c}}) = 1 - {\rm{P}}({\rm{E}})\).

So, we can calculate:

• Probability of not A, \({\rm{P}}\left( {\rm{A}}^{\rm{c}} \right) = 1 - {\rm{P}}\left( {\rm{A}} \right) = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}\)
• Probability of not B, \({\rm{P}}\left( {\rm{B}}^{\rm{c}} \right) = 1 - {\rm{P}}\left( {\rm{B}} \right) = 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4}\)

Since events A and B are independent, the occurrence of one event does not affect the probability of the other. This means:

• Event A occurring and Event B not occurring (\({\rm{A}} \cap {\rm{B}}^{\rm{c}}\)) has probability \({\rm{P}}({\rm{A}} \cap {\rm{B}}^{\rm{c}}) = {\rm{P}}({\rm{A}}) \times {\rm{P}}({\rm{B}}^{\rm{c}})\).
• Event B occurring and Event A not occurring (\({\rm{B}} \cap {\rm{A}}^{\rm{c}}\)) has probability \({\rm{P}}({\rm{B}} \cap {\rm{A}}^{\rm{c}}) = {\rm{P}}({\rm{B}}) \times {\rm{P}}({\rm{A}}^{\rm{c}})\).

The event "exactly one of the two events A or B occurs" is the union of two mutually exclusive events: (\({\rm{A}} \cap {\rm{B}}^{\rm{c}}\)) and (\({\rm{B}} \cap {\rm{A}}^{\rm{c}}\)). Therefore, the probability of this event is the sum of their individual probabilities:

\({\rm{P}}({\text{exactly one}}) = {\rm{P}}({\rm{A}} \cap {\rm{B}}^{\rm{c}}) + {\rm{P}}({\rm{B}} \cap {\rm{A}}^{\rm{c}})\)

Substitute the calculated probabilities:

\({\rm{P}}({\text{exactly one}}) = \left({\rm{P}}({\rm{A}}) \times {\rm{P}}({\rm{B}}^{\rm{c}})\right) + \left({\rm{P}}({\rm{B}}) \times {\rm{P}}({\rm{A}}^{\rm{c}})\right)\)

\({\rm{P}}({\text{exactly one}}) = \left(\frac{1}{3} \times \frac{1}{4}\right) + \left(\frac{3}{4} \times \frac{2}{3}\right)\)

\({\rm{P}}({\text{exactly one}}) = \frac{1 \times 1}{3 \times 4} + \frac{3 \times 2}{4 \times 3}\)

\({\rm{P}}({\text{exactly one}}) = \frac{1}{12} + \frac{6}{12}\)

\({\rm{P}}({\text{exactly one}}) = \frac{1 + 6}{12}\)

\({\rm{P}}({\text{exactly one}}) = \frac{7}{12}\)

Alternatively, we could calculate the probability of exactly one event occurring as the probability of the union of A and B minus the probability of the intersection of A and B: \({\rm{P}}({\rm{A}} \cup {\rm{B}}) - {\rm{P}}({\rm{A}} \cap {\rm{B}})\).

For independent events:

• \({\rm{P}}({\rm{A}} \cap {\rm{B}}) = {\rm{P}}({\rm{A}}) \times {\rm{P}}({\rm{B}}) = \frac{1}{3} \times \frac{3}{4} = \frac{3}{12} = \frac{1}{4}\)
• \({\rm{P}}({\rm{A}} \cup {\rm{B}}) = {\rm{P}}({\rm{A}}) + {\rm{P}}({\rm{B}}) - {\rm{P}}({\rm{A}} \cap {\rm{B}}) = \frac{1}{3} + \frac{3}{4} - \frac{1}{4}\)
• \({\rm{P}}({\rm{A}} \cup {\rm{B}}) = \frac{1}{3} + \left(\frac{3}{4} - \frac{1}{4}\right) = \frac{1}{3} + \frac{2}{4} = \frac{1}{3} + \frac{1}{2}\)
• To add these fractions, find a common denominator (which is 6): \(\frac{1 \times 2}{3 \times 2} + \frac{1 \times 3}{2 \times 3} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}\)

Now, subtract \({\rm{P}}({\rm{A}} \cap {\rm{B}})\) from \({\rm{P}}({\rm{A}} \cup {\rm{B}})\):

\({\rm{P}}({\text{exactly one}}) = {\rm{P}}({\rm{A}} \cup {\rm{B}}) - {\rm{P}}({\rm{A}} \cap {\rm{B}}) = \frac{5}{6} - \frac{1}{4}\)

Find a common denominator (which is 12):

\({\rm{P}}({\text{exactly one}}) = \frac{5 \times 2}{6 \times 2} - \frac{1 \times 3}{4 \times 3} = \frac{10}{12} - \frac{3}{12}\)

\({\rm{P}}({\text{exactly one}}) = \frac{10 - 3}{12} = \frac{7}{12}\)

Both methods yield the same result. The probability that exactly one of the two independent events A or B occurs is \(\frac{7}{{12}}\).

Revision Table: Probability Concepts

Additional Information on Independent Events Probability

Independent events are fundamental in probability theory. Two events, A and B, are considered independent if the probability of one event occurring does not influence the probability of the other event occurring. The key property used in this problem is that if A and B are independent, then:

• The probability of both A and B occurring is the product of their individual probabilities: \({\rm{P}}({\rm{A}} \cap {\rm{B}}) = {\rm{P}}({\rm{A}}) \times {\rm{P}}({\rm{B}})\).
• Similarly, if A and B are independent, then A and the complement of B (\({\rm{B}}^{\rm{c}}\)) are also independent. The same applies to \({\rm{A}}^{\rm{c}}\) and B, and \({\rm{A}}^{\rm{c}}\) and \({\rm{B}}^{\rm{c}}\). This allows us to use the product rule for probabilities involving complements as we did in the solution: \({\rm{P}}({\rm{A}} \cap {\rm{B}}^{\rm{c}}) = {\rm{P}}({\rm{A}}) \times {\rm{P}}({\rm{B}}^{\rm{c}})\) and \({\rm{P}}({\rm{B}} \cap {\rm{A}}^{\rm{c}}) = {\rm{P}}({\rm{B}}) \times {\rm{P}}({\rm{A}}^{\rm{c}})\).

Understanding the difference between independent and mutually exclusive (or disjoint) events is crucial. Mutually exclusive events cannot happen at the same time, so \({\rm{P}}({\rm{A}} \cap {\rm{B}}) = 0\). Independent events CAN happen at the same time, and their joint probability is the product of individual probabilities. For non-zero probability events, independent events are not mutually exclusive, and mutually exclusive events are not independent.

The problem required the probability of "exactly one" event. This scenario covers two distinct possibilities:

1. Event A happens AND Event B does NOT happen.
2. Event B happens AND Event A does NOT happen.

Since these two scenarios cannot occur simultaneously (you can't have A happen and B not, AND B happen and A not, at the exact same time), they are mutually exclusive outcomes. Therefore, we can simply add their probabilities.

The final result, \(\frac{7}{12}\), represents the chance that when the experiment is performed, either A occurs but B doesn't, or B occurs but A doesn't.