Use the Overlap Probability to Decide
In Union Probabilities from a Two-Way Table, you found the probability of “A or B” by counting outcomes in either event, including the overlap once. Now we can use the same addition rule in reverse to answer a different question: do events \(A\) and \(B\) have any outcomes in common?
As in What Mutually Exclusive Events Mean, two events are mutually exclusive, or disjoint, when they cannot both occur on the same trial. Their intersection, written \(A\cap B\) or “A and B,” contains the outcomes shared by both events. In a probability model where each possible outcome has positive probability, the events are disjoint exactly when the probability of that intersection is zero.
Often, a problem gives \(P(A)\), \(P(B)\), and \(P(A\cup B)\), rather than giving the overlap directly. The general addition rule from The General Addition Rule lets us calculate the missing intersection probability. Subtracting the union probability removes the part that was counted in both individual event probabilities.
After calculating, compare \(P(A\cap B)\) with zero. A result of exactly zero means the events are mutually exclusive in the stated model; a positive result means they are not.
The word exactly matters. A probability such as \(0.0000\) may be zero, or it may be a small positive value that was rounded. If a problem supplies rounded probabilities, a calculation that appears to give zero may not establish that the true overlap is exactly zero. Use the precision and assumptions stated in the question.
A Reliable Decision Process
Start by identifying what \(A\) and \(B\) mean and what one outcome or trial represents. Then use the supplied probabilities to find the overlap, if it is not given directly. Finally, connect the numerical result back to the event descriptions. The calculation answers whether the overlap has positive probability; the context explains what that means.
State what \(A\) and \(B\) mean for the same chance process or randomly selected individual.
Use a supplied value for \(P(A\cap B)\), or calculate it with \(P(A)+P(B)-P(A\cup B)\).
If the intersection probability is exactly zero, the events are mutually exclusive in the model. If it is positive, they are not.
Explain whether both event descriptions can occur together and refer to the probability that supports your decision.
Before interpreting the result, check that the supplied probabilities make sense. Every probability must be between 0 and 1, and the intersection cannot be larger than either individual event probability. Also, the calculated intersection must not be negative. A negative result does not mean “negative overlap”; it indicates that the given numbers are inconsistent, perhaps because of an error or rounding.
Worked Examples
Worked Example: Two Prize Types in a Drawing
In an invented drawing, one ticket is selected and receives one prize. Let \(A\) be the event that the ticket wins a meal voucher, and \(B\) the event that it wins a theater pass. Suppose \(P(A)=0.38\), \(P(B)=0.27\), and \(P(A\cup B)=0.65\). Are \(A\) and \(B\) mutually exclusive?
State: We need to determine whether a ticket can win both types of prize. In probability terms, the question is whether \(P(A\cap B)=0\).
Plan: The individual event probabilities and the union probability are given. Use the rearranged general addition rule to find the probability of both events, then compare it with zero. These probabilities refer to the same ticket selection and the same drawing.
Do: Substitute the supplied probabilities:
Check the arithmetic: \(0.38+0.27=0.65\), and \(0.65-0.65=0\). The result is between 0 and the smaller event probability, \(0.27\), so it is a possible intersection probability.
Conclude: The probability that a ticket wins both a meal voucher and a theater pass is zero. Therefore, the events are mutually exclusive in this drawing model. This conclusion agrees with the description that each selected ticket receives one prize.
Here, the sum of the two individual probabilities is exactly the union probability. That equality is another way to recognize a zero overlap: there was no overlap to subtract from the sum. However, when you are asked to decide whether events are mutually exclusive, show the intersection calculation or state clearly why it equals zero.
Worked Example: Attending a Workshop and Volunteering
In an invented community survey, one resident is selected at random. Let \(A\) mean the resident attended a weekend workshop, and \(B\) mean the resident volunteered at a neighborhood event during the same month. Suppose \(P(A)=0.52\), \(P(B)=0.29\), and \(P(A\cup B)=0.68\). Determine whether these events are mutually exclusive.
State: We are checking whether a resident can meet both descriptions, which is determined by \(P(A\cap B)\).
Plan: Use the general addition rule rearranged to calculate the intersection. A positive result will mean some residents can be in both events.
Do: Substitute and calculate:
Check: \(0.52+0.29=0.81\), and \(0.81-0.68=0.13\). The result is positive and is no larger than the smaller of \(0.52\) and \(0.29\), so it is consistent with the individual probabilities.
Conclude: The probability that a randomly selected resident both attended the workshop and volunteered is \(0.13\), or 13%. Because the intersection probability is positive, the events are not mutually exclusive. The event descriptions can occur together for some residents.
Notice that the conclusion depends on the overlap, not on whether the individual event probabilities are large or small. Two events can each be fairly common and still be disjoint, or be uncommon and still have a positive overlap. The decisive value is \(P(A\cap B)\).
Worked Example: Morning Bus and After-School Tutoring
An invented school summary describes 240 students. Of these, 96 ride the morning bus, 72 attend after-school tutoring, and 72 do neither. Let \(A\) mean a randomly selected student rides the morning bus, and \(B\) mean the student attends after-school tutoring. Decide whether \(A\) and \(B\) are mutually exclusive.
State: We want to know whether any students are in both events. The summary gives the number in neither event, so first find the number in the union.
Plan: “Neither” is the complement of \(A\cup B\). Find the union count by subtracting the neither count from the total, convert the relevant counts to probabilities, and then use the general addition rule to find the intersection.
Do: There are \(240-72=168\) students in at least one of the two events. Therefore:
Now find the intersection probability:
The count-based check gives the same result. The bus and tutoring counts add to \(96+72=168\), exactly the union count. Thus no students are counted in both groups.
Conclude: The events are mutually exclusive in this summary: no student is in both the morning-bus group and the after-school-tutoring group. The probability that a randomly selected student belongs to both groups is zero.
Common Mistakes and AP Exam Tips
- Confusing “or” with “and.” \(P(A\cup B)\) is the probability of at least one event occurring; \(P(A\cap B)\) is the probability that both occur. To test mutual exclusivity, focus on the intersection.
- Stopping after writing the formula. A full-credit response substitutes the given values, calculates the intersection, and states what the result means for the events in context.
- Calling events disjoint because their descriptions sound different. Different descriptions do not guarantee different outcomes. Check whether one outcome or individual could satisfy both event definitions.
- Accepting a negative intersection probability. For example, a negative result from \(P(A)+P(B)-P(A\cup B)\) signals inconsistent inputs or a calculation problem. An intersection probability cannot be below zero.
- Treating a rounded zero as certain proof. If probabilities are reported to limited precision, a computed value of \(0.00\) might conceal a small positive overlap. State the conclusion at the precision justified by the problem.
- Giving a conclusion without context. Say what “both” means for the individuals or outcomes in the problem, not only that the events are “disjoint.”
A strong AP-style response is direct: “Using the addition rule, \(P(A\cap B)=\ldots=0\). Therefore, \(A\) and \(B\) are mutually exclusive in this model.” For a positive result, state that the probability of both events is positive and explain that the events can occur together. Keep the numerical evidence and the contextual conclusion connected.
Check Your Understanding
For each situation, calculate or identify the intersection probability and explain whether the events are mutually exclusive.
- \(P(A)=0.41\), \(P(B)=0.24\), and \(P(A\cup B)=0.65\). Find \(P(A\cap B)\) and state your conclusion.
- \(P(C)=0.60\), \(P(D)=0.35\), and \(P(C\cup D)=0.80\). Find the probability that both events occur. Are they mutually exclusive?
- A group has 180 people. Of these, 54 are in event \(A\), 36 are in event \(B\), and 90 are in neither. Find \(P(A\cap B)\) and decide whether \(A\) and \(B\) are disjoint.
- Explain why a negative value for \(P(A)+P(B)-P(A\cup B)\) cannot be a valid intersection probability.
- If a calculation using rounded probabilities gives \(P(A\cap B)=0.00\), what should you consider before claiming the events are definitely mutually exclusive?