What Does “Given” Change?
In Exam Practice on Mutually Exclusive Events and Unions, you practiced deciding which outcomes belong to events and combining probabilities. Conditional probability asks a different question: how likely is event \(A\) when we know that event \(B\) has occurred? The condition narrows the outcomes we consider.
For example, suppose we select one student from a school. If we learn that the student owns a pet, we no longer consider all students equally as possible selections. We focus on the students who own a pet, then ask what fraction of that group owns a dog. That fraction is the probability of owning a dog given that the student owns a pet.
The vertical bar in \(P(A\mid B)\) means “given.” It is not a division sign by itself. The order matters: \(P(A\mid B)\) asks about \(A\) among cases in \(B\), while \(P(B\mid A)\) asks about \(B\) among cases in \(A\).
As in Using Two-Way Tables to Find Probabilities, an interior cell can represent a joint event, and a row or column total can represent a marginal event. Conditional probability uses both: the joint count or probability in the numerator, and the total for the condition in the denominator.
The Conditional Probability Formula
When \(P(B)>0\), the probability of \(A\) given \(B\) is the probability of both \(A\) and \(B\), divided by the probability of \(B\). The numerator keeps only the outcomes that satisfy both event descriptions. The denominator counts all outcomes that satisfy the condition.
The same idea can be expressed with counts when each person in a group is an equally likely selection. Divide the number of people in both \(A\) and \(B\) by the number of people in \(B\). This is often the quickest approach when working with a two-way table.
You can also picture the condition as temporarily setting aside every outcome outside \(B\). Within the remaining group, ask what fraction also belongs to \(A\). The conditional probability is between 0 and 1 because the shared outcomes \(A\cap B\) are part of the group \(B\).
Worked Example: Students and Pet Ownership
Worked Example: Students and Pet Ownership
A school survey asks 200 students whether they own any pet and whether they own a dog. The invented survey results are summarized below. Let \(D\) be the event that a randomly selected student owns a dog, and let \(P\) be the event that the student owns any pet. Find and interpret \(P(D\mid P)\).
| Owns a dog | Does not own a dog | Total | |
|---|---|---|---|
| Owns any pet | 45 | 75 | 120 |
| Does not own any pet | 0 | 80 | 80 |
| Total | 45 | 155 | 200 |
State: We want \(P(D\mid P)\): the probability that a student owns a dog, given that the student owns any pet.
Plan: The condition is \(P\), so use the 120 students who own any pet as the reference group. Of those students, 45 also own a dog. Divide the count in both events by the count in the condition group.
Do: The count in both events is 45, and the count in the condition group is 120. Therefore:
Using the probability formula gives the same result: \(P(D\cap P)=45/200=0.225\) and \(P(P)=120/200=0.60\), so \(P(D\mid P)=0.225/0.60=0.375\). The calculation checks because the factor of 200 cancels.
Conclude: Among the surveyed students who own any pet, the proportion who own a dog is 0.375, or 37.5%. Thus, the conditional probability that a randomly selected student owns a dog, given that the student owns any pet, is 0.375.
Notice how the denominator changed. If the question asked for the probability that a randomly selected student owns a dog without any condition, the relevant denominator would be all 200 students, giving \(45/200=0.225\). For \(P(D\mid P)\), students without pets are outside the reference group, so the denominator is 120 instead.
This is a new probability question, not a change to the survey results. The students who own dogs are still the same 45 students; the question changes which group we compare them with.
Worked Example: Rain and Delayed Deliveries
Worked Example: Rain and Delayed Deliveries
A delivery service reviews 100 randomly selected delivery days from a fictional year. On 30 days it rained; deliveries were delayed on 18 of those rainy days. On the 70 days without rain, deliveries were delayed on 14 days. Let \(R\) be the event that it rained and \(L\) the event that deliveries were delayed. Find \(P(L\mid R)\).
State: We seek the probability of a delay among rainy days, written \(P(L\mid R)\).
Plan: Use a four-step solution. The condition \(R\) defines the reference group, so the denominator is the 30 rainy days. The 18 rainy days with delays belong to both \(L\) and \(R\), so they form the numerator. The day counts are appropriate because the question considers a randomly selected day from this set of 100 days.
Do: Use the count ratio, or apply the conditional probability formula with probabilities based on all 100 days:
To verify with the formula, \(P(L\cap R)=18/100=0.18\) and \(P(R)=30/100=0.30\). Thus \(P(L\mid R)=0.18/0.30=0.60\), the same result.
Conclude: Among the rainy days reviewed, deliveries were delayed on 60% of days. The conditional probability of a delivery delay, given that it rained, is 0.60 for this set of days.
The condition determines the denominator even when the reverse question sounds similar. There were 32 delayed days altogether: 18 rainy and 14 dry. The probability of rain given that deliveries were delayed is \(P(R\mid L)=18/32=0.5625\), not 0.60. The two expressions use different reference groups: rainy days for \(P(L\mid R)\), and delayed days for \(P(R\mid L)\).
This difference is a useful reminder: changing the event after the bar changes the question. In general, there is no reason for \(P(A\mid B)\) and \(P(B\mid A)\) to be equal.
Worked Example: Using Given Probabilities
Worked Example: Using Given Probabilities
A community library models the experience of a randomly selected visitor from a particular month. Let \(A\) be the event that the visitor borrows a science book, and let \(B\) be the event that the visitor comes after school. Suppose \(P(A\cap B)=0.18\) and \(P(B)=0.40\). Find and interpret \(P(A\mid B)\).
State: We want the probability that the visitor borrows a science book, given that the visitor came after school.
Plan: The condition is \(B\), and its probability is positive. Use the conditional probability formula, placing the probability of both events over the probability of the condition.
Do: Substitute the given values:
The numerator represents visitors who both came after school and borrowed a science book. The denominator represents all visitors who came after school. As a check, 0.18 is less than 0.40, as it must be because the visitors in both events are included among the after-school visitors.
Conclude: In this probability model, the probability that a randomly selected visitor borrowed a science book, given that the visitor came after school, is 0.45. Equivalently, 45% of the visitors who came after school are modeled as borrowing a science book.
Interpreting Conditional Probability Carefully
A conditional probability describes a chance within a specified group. Keep the condition visible as you interpret the result: “Among the students who own any pet…” or “Of the rainy days…” Those phrases make clear which group is the denominator.
The formula itself does not say that the condition causes the event. For example, a conditional probability about delays on rainy days describes a relationship in the data or model. By itself, it does not establish that rain caused a delay. The interpretation should report what the probability means, not claim more than the calculation supports.
Before calculating, say the notation in words. For \(P(A\mid B)\), read “the probability of \(A\), given \(B\).” Then identify the group \(B\) and locate the outcomes that are both \(A\) and \(B\). This habit helps prevent choosing the wrong denominator.
Common Mistakes and AP Exam Tips
- Using the grand total as the denominator. For \(P(A\mid B)\), the denominator is the total in \(B\), not the entire table. A full-credit response identifies the condition group and divides the count in both events by that group’s total.
- Reversing the condition. \(P(A\mid B)\) and \(P(B\mid A)\) generally answer different questions. Read the expression from left to right: the event before the bar is what you are finding the probability of; the event after it specifies the reference group.
- Using the wrong table entry in the numerator. The numerator is the count or probability for \(A\cap B\), the outcomes in both events. It is not the total for \(A\) alone.
- Calling the bar a division sign. The bar means “given.” Division appears in the formula because the joint probability is divided by the probability of the condition.
- Leaving out the group in the interpretation. “The probability is 0.375” does not explain the result. State that 37.5% of the students who own any pet also own a dog.
- Ignoring a zero-probability condition. If \(P(B)=0\), \(P(A\mid B)\) is not defined by this formula. There are no outcomes in the condition group.
On an AP response, show the conditional expression, the numerator and denominator that match it, and a sentence in context. If a table is given, naming the relevant cell and margin makes the choice of counts transparent. A correct decimal without the reference group may leave the reader unsure whether the condition was understood.
Check Your Understanding
For each question, identify the condition group before calculating or interpreting.
- A school survey includes 160 students. Of the 64 students who ride the bus, 24 participate in a music club. Let \(B\) be riding the bus and \(M\) be participating in the club. Find and interpret \(P(M\mid B)\).
- In a two-way table, 36 people belong to both \(A\) and \(C\), and 90 people belong to \(C\). Find \(P(A\mid C)\) using counts.
- In a probability model, \(P(X\cap Y)=0.12\) and \(P(Y)=0.30\). Find \(P(X\mid Y)\).
- Explain why \(P(A\mid B)\) and \(P(B\mid A)\) may have different values.
- What does it mean if \(P(B)=0\) when trying to calculate \(P(A\mid B)\)?