Three Mutually Exclusive Events in a Row

Adding Probabilities for Three Disjoint Events

In Testing Mutual Exclusivity with Probabilities, you used the intersection of two events to decide whether they were mutually exclusive. Now extend that idea to three events. If the events describe separate categories—such as three blood types for one randomly selected person—then the probability of getting one of those categories is the sum of their probabilities, provided no outcome belongs to more than one category.

For three events, checking only one pair is not enough. Events \(A\), \(B\), and \(C\) are pairwise mutually exclusive when \(A\) and \(B\) have no outcomes in common, \(A\) and \(C\) have no outcomes in common, and \(B\) and \(C\) have no outcomes in common. In symbols, all three pairwise intersections are empty. The word “pairwise” emphasizes that every pair must be checked.

The union \(A\cup B\cup C\) means that at least one of the three events occurs. For pairwise disjoint events, no outcome is counted in more than one event, so adding the three probabilities counts every outcome in the union exactly once. This is the three-event extension of the addition rule for two mutually exclusive events.

This rule is useful when a question asks for the probability of being in any of several distinct categories. Define the events for the same chance process or randomly selected individual, verify that they cannot occur together, and add their probabilities. Keep the context attached to the result: the sum is the chance of belonging to at least one of the named categories.

Disjoint Does Not Necessarily Mean Exhaustive

A set of events is exhaustive if every outcome in the sample space belongs to at least one of the events. Disjointness and exhaustiveness describe different properties. Disjoint events do not overlap, while exhaustive events leave no outcome out. Three events can be disjoint without being exhaustive.

Because the three events are disjoint, their probabilities add to the probability of their union. That sum must be between 0 and 1. If the events are exhaustive, their union is the entire sample space, so their probabilities sum to 1. If the events are not exhaustive, the sum is less than 1 when the complement of their union has positive probability. However, the sum can still equal 1 when the complement of their union has probability zero.

For example, suppose a model lists four outcomes: red, blue, green, and black. Let \(A\), \(B\), and \(C\) be the events of red, blue, and green, respectively. If the model assigns probability zero to black, then the three events are not exhaustive because black is not included, but their probabilities can still sum to 1. In contrast, if the omitted outcome has a positive probability, the sum of the three event probabilities is less than 1.

Do not assume that categories are disjoint just because their names sound different. Ask whether a single outcome or individual could meet two event definitions at once. For categories that assign each person to exactly one group, the definitions often make disjointness clear. For other descriptions, such as “uses a phone app,” “rides a bike,” and “walks to school,” one person might fit more than one event. In that case, the simple three-event addition rule does not apply.

A Reliable Process

A clear solution starts with the meaning of each event, not just the arithmetic. Then verify that the categories do not overlap, add the probabilities, and interpret what the total represents. Finally, decide whether the events cover all possible outcomes. That final check tells you whether the sum should be 1; it is not required for adding probabilities of disjoint events.

Worked Examples

This example illustrates two separate checks. The events are disjoint, which justifies adding their probabilities. They are not exhaustive, so their union probability is less than 1. The sum describes the chance of being in one of these three categories—not the chance of having any ABO blood type.

Common Mistakes and AP Exam Tips

• Checking only two of the three pairs. To establish pairwise mutual exclusivity, check \(A\) with \(B\), \(A\) with \(C\), and \(B\) with \(C\). One overlapping pair is enough to make the simple addition rule invalid.
• Adding probabilities for overlapping events. If an outcome can satisfy more than one event, adding the three probabilities may count it multiple times. Establish disjointness before using the formula.
• Confusing “or” with “and.” \(P(A\cup B\cup C)\) is the probability that at least one event occurs. It is not the probability that all three occur at once.
• Assuming disjoint categories must add to 1. Disjointness only says that categories do not overlap. Categories whose omitted set has positive probability have a sum below 1. A sum of 1 is guaranteed when the events are exhaustive, but it can also occur when the complement of their union has probability zero.
• Calling the categories exhaustive because the sum is 1. A sum of 1 alone does not prove that every outcome is included if zero-probability outcomes are possible. Use the event definitions and the sample space to judge exhaustiveness.
• Giving only a decimal. A full-credit explanation identifies the union in context, such as “the chance the selected student is in the A, B, or C band,” and shows that the events are disjoint before adding.

A strong response might say: “The categories are mutually exclusive because one selected student can be in only one of these grade bands. Thus, \(P(A\cup B\cup C)=0.30+0.40+0.20=0.90\). The probability that the student is in one of the three bands is 0.90.” If asked whether the events cover every possibility, answer that separately from whether they overlap.