When New Information Changes a Model
A model may describe the chance of an event before we know anything else about a particular case. Once additional information is available, the relevant group of outcomes can change, and so can the probability we use. For example, a delivery delay might be more likely on rainy days than on dry days. The overall delay probability is useful before the weather is known; a conditional probability is more relevant once the forecast is known.
In Interpreting a Model Prediction in Context, you practiced explaining a probability as a chance under a stated model. Here, we refine that interpretation: we specify the information already known and calculate the chance within the outcomes consistent with that information. This is called a conditional probability.
The denominator is the probability of the condition we are told has occurred. The numerator is the probability that both the condition and the event of interest occur. Think of the condition as narrowing the reference group: instead of considering every possible outcome, consider only outcomes in \(B\), and find what fraction of those are also in \(A\).
Conditional probability does not necessarily mean that one event causes another. It describes a chance after information is given. Whether the information makes the chance larger, smaller, or unchanged depends on the model. As discussed in Independence Assumptions in Real Settings, if \(A\) and \(B\) are independent, knowing that \(B\) occurred does not change the probability of \(A\), so \(P(A\mid B)=P(A)\). If they are not independent, the conditional probability may differ from the original probability.
Refining a Probability with a Condition
To calculate a conditional probability, name the event of interest and the condition, then use the joint probability of both events as the numerator and the probability of the condition as the denominator. The condition must have positive probability; otherwise, there are no outcomes in that group from which to calculate a fraction.
The order matters. \(P(A\mid B)\) asks for the chance of \(A\) when \(B\) is known. The reversed expression \(P(B\mid A)\) asks for the chance of \(B\) when \(A\) is known. They share the same joint event in the numerator, but generally have different denominators and different meanings.
Worked Example: Revising Delivery Predictions for Rain
A delivery service’s model assigns probability \(0.30\) to rain on a delivery day. It assigns probability \(0.09\) to both rain and a delayed delivery. The model’s overall probability of a delay is \(0.12\). Find and compare the model’s probability of a delay on a rainy day and on a day without rain.
State. Let \(R\) be the event that it rains on a delivery day, and let \(D\) be the event that a delivery is delayed. We need \(P(D\mid R)\) and \(P(D\mid R^c)\), where \(R^c\) means that it does not rain.
Plan. Use the conditional-probability formula. The information supplied includes \(P(R)=0.30\), \(P(D\text{ and }R)=0.09\), and \(P(D)=0.12\). Both conditioning events have positive probability: \(P(R)=0.30\) and \(P(R^c)=1-0.30=0.70\).
Do. For rainy days:
For dry days, first find the probability of a delay and no rain. A delayed day is either rainy or dry, so subtract the rainy-delay probability from the overall delay probability:
Then divide by the probability of a dry day:
As a check, the overall delay probability is the sum of the rainy-delay and dry-delay probabilities: \(0.09+0.03=0.12\), matching the model’s stated overall probability.
Conclude. According to this model, the probability of a delivery delay is \(0.30\), or 30%, on a rainy day, compared with about \(0.0429\), or 4.29%, on a dry day. These are model-based chances for delivery days in the two weather groups. The difference does not, by itself, prove that rain causes delays; it reports how the model’s probabilities vary with the weather condition.
Using a Two-Way Table
A two-way table organizes counts according to two categorical variables. To find a conditional probability from the table, focus on the row or column specified by the condition. The number of cases in that condition group is the denominator, and the number in that group that also satisfy the event is the numerator.
This is the count version of the same formula. When all cases are equally weighted, dividing the joint count by the condition-group count gives the fraction in that group that meets the event. Be careful not to divide by the table’s overall total unless the question asks for an unconditional probability.
Worked Example: Updating a Reservation Chance for a Subgroup
A fictional campus survey records whether students use a study room and whether they reserve it ahead of time. The table summarizes 240 students. Find the probability that a student reserved ahead, given that the student used a study room. Then compare it with the probability that a student used a study room, given that the student reserved ahead.
| Reserved ahead | Did not reserve | Total | |
|---|---|---|---|
| Used study room | 54 | 36 | 90 |
| Did not use study room | 30 | 120 | 150 |
| Total | 84 | 156 | 240 |
State and plan. Let \(R\) mean that a student reserved ahead, and let \(U\) mean that the student used a study room. For \(P(R\mid U)\), the condition \(U\) selects the 90 students in the “Used study room” row. Among them, 54 reserved ahead. For \(P(U\mid R)\), the condition \(R\) selects the 84 students in the “Reserved ahead” column.
Do. Calculate each conditional probability using its own condition-group total:
As a check on the first denominator, \(54+36=90\) students used a study room. For the second, \(54+30=84\) students reserved ahead. The two probabilities are not equal because the condition groups have different totals, even though both numerators refer to the 54 students who did both.
Conclude. In this survey, 60% of students who used a study room had reserved ahead. In contrast, about 64.29% of students who reserved ahead used a study room. The first percentage describes reservation behavior among room users; the second describes room use among students who reserved. Neither percentage should be substituted for the other, and the survey counts alone do not establish that reserving caused a student to use a study room.
Updating in the Reverse Direction
Some problems give information in one direction but ask for the probability in the other. For example, a test may be very likely to be positive for a person with a condition, but the question may ask for the chance a person has the condition given a positive result. Those are different conditional probabilities. To reverse the direction, account for all ways the given information could occur.
A probability tree can help. Each branch represents a conditional probability. To find the probability of a complete path, multiply the probabilities along that path. If an outcome can happen by multiple paths, add those path probabilities. This lets us find the probability of the new information first, then calculate the share of those cases that also have the event we want.
Worked Example: Interpreting a Positive Screening Result
Suppose a fictional screening model describes a population in which 2% of people have a particular condition. The model says that 90% of people with the condition test positive, and 5% of people without the condition also test positive. Find the model probability that a randomly selected person has the condition, given a positive result.
State. Let \(C\) be the event that a person has the condition and \(+\) be the event that the test is positive. We seek \(P(C\mid +)\), not \(P(+\mid C)\).
Plan. The supplied probabilities are \(P(C)=0.02\), \(P(+\mid C)=0.90\), and \(P(+\mid C^c)=0.05\). The condition has positive probability, and so does the positive-result group under this model. Use the tree-path multiplication rule to find the two ways a positive result can occur: a person has the condition and tests positive, or a person does not have it and tests positive.
Do. The probability of having the condition and testing positive is:
The probability of not having the condition is \(1-0.02=0.98\). The probability of not having the condition and testing positive is:
These are the only two ways to get a positive result, so:
Among all positive results, the fraction that also belong to people with the condition is:
As a check, the positive-result probability is also \(P(C)P(+\mid C)+P(C^c)P(+\mid C^c)\), or \(0.02(0.90)+0.98(0.05)=0.018+0.049=0.067\), the same denominator.
Conclude. According to this model, among people who test positive, the probability that a randomly selected person has the condition is about \(0.2687\), or 26.87%. This is not the same as the 90% chance of a positive result among people who have the condition. The interpretation depends on the model’s assumed prevalence and test-result probabilities, and it does not make a positive result a certainty or replace appropriate medical interpretation.
Common Mistakes and AP Exam Communication
Conditional probability is easy to misread because the same pair of events can produce two different questions. Before calculating, translate the vertical bar in \(P(A\mid B)\) into “given that \(B\) has occurred.” This identifies the reference group and therefore the denominator.
- Using the overall total as the denominator. For \(P(R\mid U)\) in the table example, the denominator is the 90 students who used a study room, not all 240 students. The condition determines the group.
- Reversing the condition and event. \(P(C\mid +)\) is not \(P(+\mid C)\). State in words which information is known and which event’s chance is being calculated.
- Adding instead of dividing. The conditional probability is a fraction within the condition group. Find the joint count or probability for the numerator and divide by the condition’s total probability or count.
- Calling an association a cause. A higher conditional chance in one group describes a difference in the model or data. It does not alone show that subgroup membership caused the outcome.
- Leaving out the model qualification. When probabilities depend on assumptions, say “according to this model.” A calculated value is not automatically a guarantee about every real case.
Key Takeaway
Additional information can refine a probability model by narrowing attention to a subgroup of outcomes. Conditional probability measures the chance of the event of interest within that condition group. The direction of the condition matters, and an updated probability remains a model-based chance rather than a guarantee or a causal explanation.
Check Your Understanding
Use the conditional-probability formula and explain each result in context.
- A model gives \(P(W)=0.40\) for windy conditions and \(P(D\text{ and }W)=0.12\) for a delayed train and windy conditions. Find \(P(D\mid W)\).
- In a two-way table, 36 of 60 students who joined a robotics club also attended a workshop. What is the conditional probability of workshop attendance given club membership? Identify the denominator group.
- Explain in words the difference between \(P(A\mid B)\) and \(P(B\mid A)\). Why are their denominators generally different?
- A screening model gives \(P(+\mid C)=0.85\). Does this value directly answer \(P(C\mid +)\)? Explain what additional information would be needed to calculate the latter.
- If \(P(B)=0\), why can the formula for \(P(A\mid B)\) not be used to describe a chance within the group \(B\)?