A Complete Answer Needs a Reason for Disjointness
In Reading a Probability Table for “Or” Questions, you practiced translating “or” into a union and identifying which outcomes belong to it. This tutorial focuses on a related free-response skill: when a question says two events are mutually exclusive, or when you need to decide whether they are, explain why. Then use the correct probability rule and notation to find the union.
A calculation alone may not answer the whole question. If the addition rule for mutually exclusive events is used, the response should establish that the events have no outcomes in common. Naming the events and describing why one outcome cannot meet both definitions makes the reasoning clear to an AP reader.
The central question is about the possible outcomes, not just the event labels. An answer such as “the events are disjoint because they are different” is not enough: two differently named events can still share outcomes. Instead, use the definitions to show that an outcome in one event cannot also be in the other.
For example, if one outcome records the result of a single roll of a die, “the result is even” and “the result is odd” cannot both be true. But “the result is even” and “the result is greater than 3” can both be true, since a roll of 4 is in both events. Checking the possible outcomes is the reliable way to justify a claim of disjointness.
Notation and the Addition Rule
The union \(A\cup B\) contains outcomes in \(A\), in \(B\), or in both. As covered in The Addition Rule for Mutually Exclusive Events, when \(A\) and \(B\) are disjoint, there are no outcomes in both events to count twice. Therefore, the probability of the union is the sum of the individual probabilities.
Be precise about the symbols. \(A\cap B\) is the event that both \(A\) and \(B\) occur; \(P(A\cap B)\) is the probability of that event. \(A\cup B\) is the event that \(A\) or \(B\) or both occur; \(P(A\cup B)\) is its probability. If the events are mutually exclusive, “or both” contributes no possible outcomes, but “or” still means the union.
A useful response has a connected chain: define the events, establish whether they can overlap, select the applicable rule, calculate, and state what the probability means in context. The worked examples use that order.
Worked Example: A Randomly Generated Digit
Worked Example: A Randomly Generated Digit
A computer generates one digit at random from 0 through 9, with all ten digits equally likely. Let \(A\) be the event that the digit is less than 5, and let \(B\) be the event that the digit is greater than 6. Are \(A\) and \(B\) mutually exclusive? Find the probability that the digit is less than 5 or greater than 6.
State: The sample space is \(S=\{0,1,2,3,4,5,6,7,8,9\}\). The events are \(A=\{0,1,2,3,4\}\) and \(B=\{7,8,9\}\).
Plan: Compare the outcomes in the two events. If they share no digit, they are mutually exclusive and the mutually exclusive addition rule applies. Because the ten digits are equally likely, count the outcomes in each event and divide by 10 to find their probabilities.
Do: No digit is both less than 5 and greater than 6. Thus \(A\cap B=\varnothing\), so \(A\) and \(B\) are mutually exclusive. Event \(A\) has five outcomes, giving \(P(A)=5/10\); event \(B\) has three outcomes, giving \(P(B)=3/10\). Add those probabilities:
As a check, the union contains the eight distinct digits \(0,1,2,3,4,7,8,9\). The only digits outside the union are 5 and 6, so the count check gives \(8/10=0.8\) as well.
Conclude: The events are mutually exclusive because no digit can be both less than 5 and greater than 6. The probability that the generated digit is less than 5 or greater than 6 is 0.8.
Make the Justification Specific to the Chance Process
A good disjointness argument identifies what one outcome represents and then uses the event definitions. In the digit example, one outcome is a single generated digit. The same digit cannot satisfy both “less than 5” and “greater than 6.” That is stronger reasoning than simply saying the events are “separate.”
Sometimes event definitions already restrict an outcome to exactly one category. In that case, name the restriction. If a process assigns each account exactly one subscription tier, for instance, an account cannot be in two different tiers at the same time. The phrase “exactly one” is part of the chance process and supports the disjointness claim.
Do not confuse a statement about one outcome with a statement about different repetitions. Two events that cannot both occur on one trial are mutually exclusive. Whether events on separate trials are independent is a different question, covered in Mutually Exclusive Versus Independent Events. For a union question about a single selected outcome, focus on whether that outcome can belong to both events.
Worked Example: One Subscription Tier per Account
Worked Example: One Subscription Tier per Account
A probability model describes the active subscription tier of a randomly selected account. Every account has exactly one tier: basic, standard, or premium. The probabilities are \(0.46\) for basic, \(0.35\) for standard, and \(0.19\) for premium. Let \(B\) be the event that the account has a basic tier and \(S\) the event that it has a standard tier. Find \(P(B\cup S)\), and justify whether the events are mutually exclusive.
State: \(B\) means the selected account has a basic tier; \(S\) means it has a standard tier.
Plan: The model says each account has exactly one tier. Therefore, a single account cannot have both the basic and standard tiers. The events are disjoint, so add their probabilities to find the union.
Do: Since no account belongs to both tiers, \(B\cap S=\varnothing\), and \(P(B\cap S)=0\). Apply the mutually exclusive addition rule:
A check uses the remaining tier: \(0.46+0.35+0.19=1.00\), so the probability of basic or standard is also \(1-0.19=0.81\). This confirms the addition result using the probability assigned to the only tier outside the union.
Conclude: The events are mutually exclusive because each selected account has exactly one tier. The probability that a randomly selected account has a basic or standard subscription is 0.81.
When You Must Check Rather Than Assume
A prompt may provide probabilities for two events without saying whether the events are mutually exclusive. Do not assume they are disjoint merely because they have different names or describe different characteristics. Use the situation or any provided information about their intersection.
If the problem gives event descriptions and a sample space, compare the outcomes in the events. If it gives a value for \(P(A\cap B)\), a value of zero means the intersection has probability zero, but it does not necessarily mean that the events have no shared outcomes. It establishes that they have no shared outcomes only when every outcome has positive probability; otherwise, check directly whether \(A\cap B=\varnothing\). If the intersection probability is positive, the events are not mutually exclusive, and the general addition rule—not the disjoint-events shortcut—is needed. As discussed in The General Addition Rule, that rule subtracts the intersection once.
For a free-response question, it is often helpful to write the intersection statement before the calculation. For instance: “Because one selected account has exactly one tier, it cannot be both basic and standard; therefore \(B\cap S=\varnothing\), and the events are mutually exclusive.” That sentence connects the contextual reason to the notation and the rule.
Worked Example: Multiples of Four or Odd Results
Worked Example: Multiples of Four or Odd Results
A fair 12-sided die has faces numbered 1 through 12. Let \(M\) be the event that the result is a multiple of 4, and let \(O\) be the event that the result is odd. Determine whether \(M\) and \(O\) are mutually exclusive, then find the probability of a multiple of 4 or an odd result.
State: The equally likely sample space is \(S=\{1,2,\ldots,12\}\). The events are \(M=\{4,8,12\}\) and \(O=\{1,3,5,7,9,11\}\).
Plan: Check whether any multiple of 4 is also odd. If the event sets do not overlap, use the mutually exclusive addition rule. Since the die is fair, each event probability is its number of outcomes divided by 12.
Do: The multiples of 4 listed in \(M\) are all even, while every outcome in \(O\) is odd. Thus \(M\cap O=\varnothing\), so the events are mutually exclusive. There are three outcomes in \(M\) and six in \(O\): \(P(M)=3/12\) and \(P(O)=6/12\). Therefore:
The count check agrees: the union has the three multiples of 4 and the six odd results, with no shared result, for nine outcomes out of 12. The other three faces are 2, 6, and 10, so \(1-3/12=9/12=0.75\) gives the same probability.
Conclude: The events are mutually exclusive because no result can be both odd and a multiple of 4. The probability of rolling a multiple of 4 or an odd result is 0.75.
A Free-Response Routine
When a question asks for a union and disjointness matters, use a short sequence that keeps the logic visible. The justification belongs before the addition rule, not as an unsupported claim after the answer.
Give each event a symbol and state what it means in the situation.
Use the sample space or the rules of the chance process to decide whether one outcome can satisfy both event definitions.
If there is no shared outcome, state \(A\cap B=\varnothing\) or \(P(A\cap B)=0\), and explain why in context.
For disjoint events, find \(P(A\cup B)=P(A)+P(B)\), then state what that probability means for the specified selection or trial.
Common Mistakes and AP Exam Tips
- Claiming disjointness without evidence. “They are different events” does not prove that they cannot occur together. A full-credit justification identifies the restriction or incompatible conditions that prevent one outcome from belonging to both.
- Confusing intersection and union notation. \(A\cap B\) means both events occur; \(A\cup B\) means at least one occurs. State the union when answering an “or” question.
- Adding probabilities before checking overlap. The shortcut \(P(A\cup B)=P(A)+P(B)\) requires mutually exclusive events. If they overlap, use the general addition rule and subtract \(P(A\cap B)\).
- Using “exactly one” to explain an inclusive “or.” A union includes outcomes in both events too. When the events are disjoint, there happen to be no such shared outcomes, but the meaning of union has not changed.
- Giving only a decimal. A calculation without event definitions or reasoning may not demonstrate that the correct rule applies. Show the disjointness justification and name the rule through its equation.
- Leaving out the context in the conclusion. “The answer is 0.81” is less informative than “The probability that a randomly selected account has a basic or standard subscription is 0.81.”
A final check is to make sure a union probability is between 0 and 1 and is at least as large as either individual event probability. For disjoint events, the sum \(P(A)+P(B)\) cannot exceed 1. These checks can catch arithmetic or model errors, but they do not replace a clear justification of disjointness.
Check Your Understanding
For each question, define the events and explain what happens to one outcome before deciding whether the events are mutually exclusive.
- A fair coin is flipped once. Let \(H\) be heads and \(T\) be tails. Are \(H\) and \(T\) mutually exclusive? State the intersection using notation.
- A fair six-sided die is rolled once. Let \(E\) be an even result and \(G\) be a result greater than 3. Are \(E\) and \(G\) mutually exclusive? Identify an outcome that supports your answer.
- A chance model gives \(P(A)=0.28\) and \(P(B)=0.17\), and the events are known to be mutually exclusive. Find \(P(A\cup B)\) and give a one-sentence interpretation if \(A\) and \(B\) describe two outcomes for one randomly selected item.
- A randomly selected account has exactly one of three plan types. Explain why the events “basic plan” and “premium plan” are mutually exclusive, using both words and notation.
- Why is the statement “\(A\) and \(B\) have different names, so they are disjoint” not a sufficient justification?