Make Your Reasoning Easy to Follow
A correct numerical answer is important, but a free-response solution should also make clear what probability you calculated and why your denominator fits the question. In Checking Answers by Reduced Sample Space, you used the condition to identify the relevant group of outcomes. Here, the goal is to communicate that reasoning clearly from start to finish.
A useful response names the target event and condition, shows how the condition sets the reference group, gives the calculation, and finishes with an interpretation in context. If you use the notation \(P(A\mid B)\), say what \(A\) and \(B\) mean. That lets a reader see that your symbols match the question rather than having to guess.
As in The Conditional Probability Formula, when \(P(B)>0\), the probability of \(A\) given \(B\) is \(P(A\cap B)/P(B)\). When working from equally likely outcomes or counts, the same structure appears as the number satisfying both events divided by the number satisfying the condition. You do not need to prove the formula in every response. You do need to show enough work to demonstrate that you used it correctly.
A Four-Part Response Routine
For a written solution, organize the reasoning into State, Plan, Do, and Conclude. You do not always need to label these steps, but using them as a checklist helps prevent missing notation, work, or context.
Name the events and write the requested probability. Make clear which event is the target and which event is the condition.
Identify the information you will use, such as a table count or the probabilities along relevant paths. Explain briefly why the denominator matches the condition.
Show the relevant numerator and denominator and carry out the calculation. Keep the work legible and include appropriate rounding if you give a decimal.
Answer the question in a complete sentence, naming the target event and the group specified by the condition.
A Plan can be one sentence. For example: “I will restrict attention to students who registered early, then find the proportion of that group who completed the workshop.” A Do step should make the calculation visible, not merely state a calculator result. The Conclude step translates the probability back into the situation.
When the problem gives a table of counts, use the count in the cell where both events occur for the numerator and the total for the condition group for the denominator. When the problem gives probabilities for different paths, the numerator is the probability of the target-and-condition path or paths; the denominator accounts for all paths that satisfy the condition. In either case, check that the condition group has positive probability or a positive count.
Worked Example: Workshop Completion by Registration Group
Worked Example: Workshop Completion by Registration Group
A school program records whether students completed a weekend workshop and whether they registered early or late. The invented counts are shown below. Find the probability that a randomly selected early registrant completed the workshop.
| Registration | Completed | Did not complete | Total |
|---|---|---|---|
| Early | 72 | 18 | 90 |
| Late | 48 | 42 | 90 |
| Total | 120 | 60 | 180 |
State: Let \(C\) be the event that a student completed the workshop and \(E\) the event that a student registered early. The requested probability is \(P(C\mid E)\).
Plan: The condition is early registration, so use the 90 early registrants as the reference group. Among them, 72 completed the workshop.
Do: Divide the count in both \(C\) and \(E\) by the total count in \(E\):
The numerator is the count of early registrants who completed the workshop. The denominator is all early registrants, including both those who completed and those who did not.
Conclude: Among students who registered early, the probability of completing the workshop is \(0.800\), or 80%. This describes the early-registration group, not all 180 students.
Notice that a well-written answer does not just say “72 divided by 90.” Naming the events and explaining the denominator makes the reference group unmistakable. The conclusion also preserves the condition by beginning “Among students who registered early.”
Worked Example: Residents Among Bike Commuters
Worked Example: Residents Among Bike Commuters
A community center conducts an invented survey of 300 people who use a nearby campus. Respondents are classified by whether they live in the surrounding neighborhood and whether they bike to campus. The results are summarized here.
| Group | Bikes to campus | Does not bike | Total |
|---|---|---|---|
| Neighborhood resident | 96 | 64 | 160 |
| Not a neighborhood resident | 42 | 98 | 140 |
| Total | 138 | 162 | 300 |
Find the probability that a randomly selected bike commuter is a neighborhood resident.
State: Let \(R\) mean that a person is a neighborhood resident and \(B\) mean that a person bikes to campus. The question asks for \(P(R\mid B)\), not \(P(B\mid R)\).
Plan: Since the condition is that the person bikes, restrict attention to the 138 bike commuters. Of those, 96 are neighborhood residents.
Do: Use the resident-and-bike count over the total number of bike commuters:
The decimal is rounded to four places. The denominator is 138 rather than 300 because the probability is conditional on being a bike commuter.
Conclude: Among the people surveyed who bike to campus, the probability that a randomly selected person is a neighborhood resident is approximately \(0.6957\), or 69.57%.
The wording of the conclusion matters. Saying “the probability a resident bikes to campus” would describe \(P(B\mid R)\), which uses the 160 residents as its reference group. The requested question instead starts with the 138 bike commuters and asks what fraction of that group are residents. As emphasized in Why \(P(A\mid B)\) Is Not \(P(B\mid A)\), reversing the condition changes the question.
Worked Example: Carrier Given a Delayed Delivery
Worked Example: Carrier Given a Delayed Delivery
An online shop uses Carrier A for 60% of its packages and Carrier B for the other 40%. In this invented scenario, 8% of Carrier A’s packages are delayed, while 15% of Carrier B’s packages are delayed. Given that a package is delayed, find the probability that Carrier A handled it.
State: Let \(A\) be the event that Carrier A handled the package, \(B\) the event that Carrier B handled it, and \(D\) the event that it was delayed. We want \(P(A\mid D)\).
Plan: The condition is a delayed package. The numerator is the probability that a package was both handled by A and delayed. The denominator is the probability of a delay by either carrier, since both carrier paths can lead to a delayed package.
Do: First find the probability of each delayed path by multiplying the carrier-selection probability by the delay probability for that carrier:
The two delayed paths are disjoint because a package is handled by one carrier or the other. Thus, the probability of a delay is \(0.048+0.060=0.108\). This is positive, so conditioning on a delay is defined. Now divide the A-and-delayed probability by the probability of any delay:
The result is rounded to four decimal places. The denominator includes delayed packages handled by both A and B; using only the Carrier A path would not represent all delayed packages.
Conclude: Given that a package is delayed, the probability that Carrier A handled it is approximately \(0.4444\), or 44.44%.
A frequent shortcut is to use 8%, the delay rate for Carrier A, as the answer. But 8% answers the different question \(P(D\mid A)\): the probability of a delay given that Carrier A handled the package. The requested probability conditions on a delay and asks which carrier handled the package.
Common Mistakes and Full-Credit Communication
Many conditional probability errors are communication errors that reveal a setup problem. Before finalizing a response, check that the notation, calculation, and conclusion all answer the same question.
- Reversing the condition. \(P(A\mid B)\) and \(P(B\mid A)\) are generally different. State the condition in words and make sure it matches the event after the bar.
- Using the grand total as the denominator. For a table question asking for a conditional probability, the denominator comes from the condition group, not automatically from the whole table.
- Giving a number without its setup. A decimal alone does not show which events were used. Write the relevant count fraction or probability ratio.
- Omitting the target or condition in the conclusion. “The answer is 0.70” does not interpret the result. State what the probability describes and among whom or under what condition.
- Rounding too early or inconsistently. Keep the exact fraction or unrounded values through the calculation, then round the reported decimal as appropriate.
- Leaving out a path that meets the condition. In a probability model, include every relevant way the condition can occur in its denominator.
A concise, full-credit style of response often sounds like this: “Let \(A\) represent [target] and \(B\) represent [condition]. Because the condition group contains [denominator count or probability], \(P(A\mid B)=[\text{numerator}]/[\text{denominator}]=[\text{result}]\). Thus, among [condition group], the probability of [target event] is [result].” Adapt the wording to the situation; do not fill in the brackets mechanically.
For table problems, explain what the denominator counts. For path problems, explain why all relevant paths appear in the condition probability. If the task asks for a value from a provided model, show the calculation rather than relying only on a calculator display. A brief but specific justification is stronger than a longer explanation that never identifies the reference group.
Put the Response Together
The State, Plan, Do, and Conclude routine is a communication tool, not a new probability rule. Use it to show that you understood which event is being predicted and which group defines the reference set. When counts are available, a clear fraction often makes that reference set especially visible. When probabilities are supplied for paths, show how the paths contribute to the numerator and denominator.
Before moving on, check your solution against three questions: Did I define the events? Does my denominator correspond to the condition? Does my final sentence interpret the probability in context? If all three answers are yes and the arithmetic is shown, your solution is much easier to evaluate and understand.
Check Your Understanding
For each prompt, focus on how you would communicate the solution, including event notation, the denominator, and a context-based conclusion.
- In a group of 120 garden volunteers, 45 brought reusable water bottles; 30 of those 45 also brought reusable food containers. Define events and write the conditional probability of bringing a reusable food container given that a volunteer brought a reusable water bottle. Show the count fraction and describe the denominator.
- A randomly selected passenger is known to have used the express bus. Explain which group supplies the denominator when finding the probability that the passenger paid with a transit card.
- A device is made by Line X with probability 0.7 and Line Y with probability 0.3. The probability of a warning light is 0.02 for Line X devices and 0.05 for Line Y devices. Set up, but do not calculate, the probability that a device came from Line Y given that its warning light is on. Identify all terms needed in the denominator.
- A student reports \(P(\text{late}\mid\text{rain})=0.25\) as “25% of late trips happened in rain.” Explain why that sentence does not match the conditional probability stated.
- Write one context-based conclusion for \(P(A\mid B)=0.42\), where \(A\) means a randomly selected library visitor borrowed a book and \(B\) means the visitor came after school.