Use the Addition Rule in Reverse
In Why We Subtract the Overlap, you saw that the general addition rule corrects for outcomes counted in both events. That rule is usually used to find the probability of “A or B” when the individual probabilities and overlap are known. But sometimes the union probability is known instead, and the overlap is the missing value.
The notation \(A\cap B\) means that both event A and event B occur. The phrase “A and B” refers to this intersection. If we know \(P(A)\), \(P(B)\), and \(P(A\cup B)\), we can rearrange the general addition rule to solve for \(P(A\cap B)\).
To see the rearrangement, add \(P(A\cap B)\) to both sides of the general addition rule. Then subtract \(P(A\cup B)\) from both sides. The result says that the probability of both events is the sum of the individual event probabilities minus the probability of their union.
This makes sense from the region-counting explanation. Adding \(P(A)\) and \(P(B)\) counts the overlap twice. The union contains the overlap once, so the difference between the sum and the union is exactly one copy of the overlap. Here, instead of using the overlap to find the union, we use the union to recover the overlap.
A Reliable Calculation and Check
The calculation has three parts: identify the requested intersection, substitute the three known probabilities, and check that the result is a possible overlap. Keep the event labels and the union notation clear, especially when a problem uses “or” and “and” in words.
For “A and B,” write \(P(A\cap B)\). For “A or B,” write \(P(A\cup B)\).
Add \(P(A)\) and \(P(B)\), then subtract \(P(A\cup B)\).
The overlap cannot be negative or larger than either individual event probability. State what the result means in the given situation.
A calculated intersection should be no greater than \(P(A)\), because every outcome in both events is also in A. It should also be no greater than \(P(B)\). The union should be at least as large as either individual event, and no probability can exceed 1. These are useful checks on the inputs and the arithmetic.
The events do not have to be mutually exclusive. In fact, if they were mutually exclusive, their intersection probability would be 0. The rearranged rule works for any two events, including overlapping events, as long as the probabilities refer to events in the same chance process and sample space.
Worked Examples
Worked Example: Two Features in a Garden Survey
In an invented survey of community gardeners, let \(A\) be the event that a randomly selected gardener grows tomatoes, and let \(B\) be the event that the gardener grows peppers. Suppose \(P(A)=0.62\), \(P(B)=0.47\), and \(P(A\cup B)=0.81\). Find and interpret the probability that a selected gardener grows both.
State: The question asks for the probability of both events, so the target is \(P(A\cap B)\). The given union probability is \(P(A\cup B)=0.81\).
Plan: Use the rearranged general addition rule. All three probabilities describe gardeners selected from the same population, so they can be used together in this rule.
Do: Substitute the known values:
Check the result: \(0.28\) is nonnegative and is less than both \(0.62\) and \(0.47\), so it is a possible overlap. Substituting back into the original rule gives \(0.62+0.47-0.28=0.81\), the stated union probability.
Conclude: The probability that a randomly selected gardener grows both tomatoes and peppers is \(0.28\), or 28%.
Notice that the individual probabilities add to \(1.09\), which is more than 1. That does not mean the information is impossible. The two events can overlap, and subtracting their shared probability gives a union probability of \(0.81\), which is within the allowed range from 0 to 1.
Worked Example: Morning and Evening Fitness Classes
An invented recreation-center summary describes a randomly selected registered participant. Let \(M\) be the event that the participant attends a morning fitness class, and let \(E\) be the event that the participant attends an evening fitness class. Suppose \(P(M)=0.38\), \(P(E)=0.52\), and \(P(M\cup E)=0.70\). Find the probability that the participant attends both types of class.
Identify the events: “Both types” means morning and evening, so we want \(P(M\cap E)\), not \(P(M\cup E)\). The union is already given.
Calculate: Rearrange the addition rule and substitute:
The arithmetic can be checked another way: \(0.38+0.52=0.90\), and \(0.90-0.70=0.20\). The result is at most \(0.38\) and \(0.52\), as an intersection must be. Putting it back into the addition rule gives \(0.38+0.52-0.20=0.70\).
Interpret: The probability that a randomly selected registered participant attends both a morning and an evening fitness class is \(0.20\), or 20%.
The word “both” is a useful signal to look for an intersection, but always connect that word to the event definitions. Someone might attend both classes on different days; the intersection does not mean the classes happen simultaneously. It means the selected participant meets both event descriptions.
Worked Example: E-Book and Audiobook Borrowing
In an invented library-use summary, let \(D\) be the event that a randomly selected library member borrowed an e-book during a month, and let \(L\) be the event that the member borrowed an audiobook. Suppose \(P(D)=0.44\), \(P(L)=0.35\), and \(P(D\cup L)=0.61\). Find \(P(D\cap L)\).
The target is the intersection because we want members who borrowed an e-book and an audiobook. Use the union and two marginal probabilities in the rearranged rule:
The result is possible: \(0.18\) is greater than or equal to 0 and no larger than either \(0.44\) or \(0.35\). Also, \(0.44+0.35-0.18=0.61\), which recovers the given union probability.
Conclusion: The probability that a randomly selected library member borrowed both an e-book and an audiobook during the month is \(0.18\), or 18%.
Check Whether the Given Probabilities Fit
The rearrangement is also a way to detect inconsistent information. Suppose a problem states \(P(A)=0.30\), \(P(B)=0.25\), and \(P(A\cup B)=0.60\). Applying the formula gives:
A negative probability is impossible. Therefore, those three values cannot all describe the same two events in a valid probability model. The issue is not that the overlap is “less than zero”; rather, at least one stated value or assumption must be wrong. In particular, a union cannot be larger than the sum of the two individual probabilities.
There are other quick checks. The intersection cannot exceed either event, and the union cannot be smaller than either event. If a result fails one of these checks, revisit the labels and arithmetic before interpreting it. Do not change a negative result to zero or silently adjust a provided probability.
- \(0\leq P(A\cap B)\).
- \(P(A\cap B)\leq P(A)\) and \(P(A\cap B)\leq P(B)\).
- \(P(A\cup B)\geq P(A)\) and \(P(A\cup B)\geq P(B)\).
- Every probability, including the union, is at most 1.
Common Mistakes and AP Exam Tips
- Using the wrong target. “A and B” asks for \(P(A\cap B)\); “A or B” asks for \(P(A\cup B)\). Write the requested notation before substituting numbers.
- Subtracting the wrong probability. To find the intersection, subtract the union from the sum of the individual event probabilities: \(P(A)+P(B)-P(A\cup B)\). Do not subtract \(P(A)\) from \(P(B)\).
- Treating the events as disjoint without evidence. The mutually exclusive addition rule applies only when the events cannot occur together. The general addition rule and its rearrangement allow overlap.
- Assuming a large sum of the individual probabilities is impossible. \(P(A)+P(B)\) can exceed 1 when some outcomes belong to both events. The union, after correcting for overlap, must still be a valid probability.
- Reporting a number without context. A full-credit response identifies the events, shows the substitution, and describes what the intersection probability means for the selected person or outcome.
- Ignoring an impossible result. If the calculation gives a negative intersection or an intersection larger than an individual event probability, explain that the values are inconsistent instead of presenting the result as a valid probability.
A clear AP-style response can be concise while still showing the reasoning: “Let \(A\) be the event that a selected member borrowed an e-book and \(B\) the event that the member borrowed an audiobook. By the general addition rule, \(P(A\cap B)=0.44+0.35-0.61=0.18\). Thus, the probability that a randomly selected member borrowed both is 0.18.” This identifies the target, names the rule, substitutes the supplied values, and interprets the answer in context.
Check Your Understanding
For each question, identify the requested probability, show the substitution, and check whether the result is possible.
- Suppose \(P(A)=0.55\), \(P(B)=0.40\), and \(P(A\cup B)=0.75\). Find \(P(A\cap B)\).
- In an invented survey, 0.31 of residents use a neighborhood pool, 0.46 use a recreation path, and 0.63 use the pool or path. What is the probability a resident uses both?
- Explain why \(P(A)+P(B)\) can be greater than 1 even though each probability is between 0 and 1.
- Suppose \(P(A)=0.22\), \(P(B)=0.50\), and \(P(A\cup B)=0.60\). Use the rearranged rule and explain why the answer signals inconsistent information.
- A problem asks for “A or B” and gives \(P(A)\), \(P(B)\), and \(P(A\cup B)\). Which probability is missing, and which version of the addition rule should you use?