Adding Probabilities of Mutually Exclusive Events

Why Add Probabilities?

Many probability questions ask whether one event or another will occur. A fair die might show an even number or a 5. A randomly selected survey respondent might choose bus travel or bike travel. If the events cannot happen at the same time in one trial, their probabilities can be added to find the probability that either event occurs.

The key condition is that the events have no outcomes in common. This is more precise than saying the events are “different”: two events can be described differently and still share possible outcomes. Before adding probabilities, identify the outcomes belonging to each event and check whether any outcome belongs to both.

Definition: Two events \(A\) and \(B\) are mutually exclusive (or disjoint) if they cannot both occur in the same trial. In terms of sets of outcomes, this means \(A\cap B=\varnothing\): they have no outcomes in common.

In probability, “\(A\) or \(B\)” ordinarily includes the possibility that both occur. That is called the inclusive meaning of “or.” When \(A\) and \(B\) are mutually exclusive, however, both cannot occur, so the inclusive and exclusive readings lead to the same event: an outcome in \(A\) or in \(B\).

The Addition Rule for Disjoint Events

For a discrete probability model, probabilities are assigned to outcomes in a finite or countable sample space. The probability of an event is the sum of the probabilities of the outcomes in that event. The individual outcome probabilities sum to 1 across the sample space. When the model is equally likely, each outcome has the same probability; in a finite sample space with \(N\) equally likely outcomes, each has probability \(1/N\).

The addition rule below applies when the events are disjoint. It does not require that all outcomes in the sample space be equally likely. It requires only that the probabilities are defined in a probability model and that the two events have no outcomes in common.

Theorem: If \(A\) and \(B\) are mutually exclusive events, then

P(A\text{ or }B)=P(A\cup B)=P(A)+P(B).

The symbol \(A\cup B\) means the event that an outcome belongs to \(A\), to \(B\), or to both. Because mutually exclusive events have no shared outcomes, adding their probabilities counts every outcome in their union exactly once.

Why the Rule Works

Here is a direct justification for a discrete model. Let the sample space be finite or countable, and let \(p(\omega)\) denote the probability assigned to outcome \(\omega\). The probability of an event is the sum of \(p(\omega)\) over the outcomes in that event. If \(A\) and \(B\) are disjoint, every outcome in \(A\cup B\) appears in exactly one of the two events. Thus the outcomes in the union can be split into the outcomes in \(A\) and the outcomes in \(B\), with no repetition:

\begin{aligned} P(A\cup B) &=\sum_{\omega\in A\cup B}p(\omega)\\ &=\sum_{\omega\in A}p(\omega)+\sum_{\omega\in B}p(\omega)\\ &=P(A)+P(B). \end{aligned}

The second equality depends on disjointness: there is no outcome that would be included in both sums. This argument also explains why the rule is about events, not merely about adding labels or descriptions. The outcomes must actually be separable into two non-overlapping groups.

Worked Example: Rolling an Even Number or a 5

Suppose a fair six-sided die is rolled once. Find the probability of rolling an even number or a 5.

Model and condition check. The outcomes are \(1,2,3,4,5,6\). The die is assumed fair, so each outcome has probability \(1/6\). Let \(A\) be rolling an even number, and let \(B\) be rolling a 5. Then \(A=\{2,4,6\}\) and \(B=\{5\}\). No outcome belongs to both sets, so \(A\) and \(B\) are mutually exclusive.

The probabilities are \(P(A)=3/6\) and \(P(B)=1/6\). Therefore,

P(A\text{ or }B)=P(A)+P(B) =\frac{3}{6}+\frac{1}{6} =\frac{4}{6} =\frac{2}{3} \approx 0.667.

The probability of rolling an even number or a 5 is \(2/3\), or about 0.667. The result also matches a direct count: four of the six equally likely outcomes, \(2,4,5,6\), satisfy at least one of the two conditions.

Worked Example: Choosing Between Two Survey Categories

Imagine a hypothetical survey with 200 completed responses. Each respondent gives one main mode of travel to campus. Among these responses, 54 list bus and 38 list bike. Suppose one of the 200 responses is selected at random. Find the probability that the selected response lists bus or bike.

Model and condition check. The possible outcomes are the 200 individual completed responses, treated as equally likely because one response is selected at random from the 200. Define \(A\) as selecting a response that lists bus and \(B\) as selecting a response that lists bike. Each response has exactly one main travel mode, so no response can be in both categories. The events are mutually exclusive.

The probability of selecting a bus response is \(P(A)=54/200=0.27\). The probability of selecting a bike response is \(P(B)=38/200=0.19\). Applying the addition rule,

P(A\text{ or }B)=0.27+0.19=0.46.

For a random selection from these 200 hypothetical responses, the probability of a bus or bike response is \(0.46\), or 46%. Equivalently, \(54+38=92\) of the 200 responses are in one of those two categories, and \(92/200=0.46\). This describes the hypothetical set of responses; it does not by itself establish the proportion for a broader population.

Worked Example: Spinner Outcomes in Two Separate Groups

A fair spinner has eight equal sectors numbered 1 through 8. Find the probability that it lands on a number less than 3 or on a number in the set \(\{6,8\}\).

Model and condition check. The sample space is \(\{1,2,3,4,5,6,7,8\}\), and the equal sectors make the eight outcomes equally likely. Let \(A=\{1,2\}\), the event of landing on a number less than 3, and let \(B=\{6,8\}\). These sets have no outcomes in common, so the events are mutually exclusive.

Each event contains two of the eight outcomes, so \(P(A)=2/8\) and \(P(B)=2/8\). Then

P(A\text{ or }B)=\frac{2}{8}+\frac{2}{8} =\frac{4}{8} =\frac{1}{2} =0.5.

The spinner has probability \(0.5\) of landing in either specified group. The four favorable outcomes are \(1,2,6,8\), so a direct count gives the same probability, \(4/8\).

Applying the Rule Carefully

A reliable solution has three parts: describe the events, verify they are disjoint, and then add their probabilities. The check is essential, not an optional detail. In each worked example, the events were mutually exclusive because one trial produced just one outcome, and the two sets of outcomes did not overlap.

For more than two events, probabilities can also be added when every pair of events is mutually exclusive. For example, if \(A\), \(B\), and \(C\) are pairwise disjoint, then no outcome is counted in more than one event, and

P(A\text{ or }B\text{ or }C)=P(A)+P(B)+P(C).

The same reasoning extends to any finite collection of pairwise disjoint events: their probabilities add because each outcome in the combined event belongs to exactly one of them. For instance, mutually exclusive response categories can be combined by summing their category probabilities, provided each individual response is assigned to only one category.

Common Mistake: Adding Before Checking for Shared Outcomes

A frequent error is to see two event descriptions and add their probabilities immediately. Suppose a single card is drawn from a standard deck. “A heart” and “a queen” are not mutually exclusive: the queen of hearts satisfies both descriptions. Simply adding the probability of a heart, \(13/52\), to the probability of a queen, \(4/52\), counts that one card twice.

The correction is to list or otherwise inspect the outcomes in both events before using the disjoint-events rule. If the sets share an outcome, the condition for this rule has not been met, so adding the two probabilities directly is not justified. The relevant addition rule for events that overlap is a separate result.

Another subtle mistake is treating “mutually exclusive” as meaning merely “not identical.” Events need not be identical to overlap. For example, on a die roll, “roll an even number” and “roll a number greater than 3” are different events, but outcomes 4 and 6 belong to both. What matters is whether at least one possible outcome lies in both event sets.

Key takeaway: Add \(P(A)\) and \(P(B)\) to find \(P(A\text{ or }B)\) only after verifying that \(A\) and \(B\) have no outcomes in common. For disjoint events, the sum counts each outcome in the combined event exactly once.

Check Your Understanding

For each question, identify the relevant events and check the condition for adding their probabilities.

A fair six-sided die is rolled once. What is the probability of rolling a 1 or a 4? Explain why the events meet the condition for the disjoint-events rule.
A fair spinner has ten equal sectors numbered 1 through 10. Find the probability of landing on 2 or on an odd number. Are these events mutually exclusive? Explain.
In a hypothetical set of 120 survey responses, 31 list walking and 29 list driving as the respondent’s one main commute mode. If one response is selected uniformly at random, find the probability it lists walking or driving.
In one card draw, explain why “a heart” and “a queen” are not mutually exclusive. Identify an outcome they share.
In your own words, explain why the probability of the union of two disjoint events is the sum of their probabilities.