Let the Condition Narrow the Table
In Reading the Given Condition in a Sentence, you practiced identifying the reference group named by a condition. A two-way table makes that group visible: it is one row or one column. To find \(P(A\mid B)\), focus on the row or column for \(B\), then find what proportion of that group also belongs to \(A\).
The key is to use two counts from the table. The numerator is the count in the cell where the target and condition overlap. The denominator is the total for the condition group. The grand total is not the denominator for a conditional probability; it would describe a selection from the full table rather than from the restricted group.
This is the count-based version of the conditional probability formula from What Conditional Probability Means. A practical way to use it is to first circle or name the condition total, then identify the cell within that total that also satisfies the target. This order helps prevent reversing the events or accidentally using the grand total.
A Routine for Reading a Two-Way Table
Rows and columns can represent either event. The layout does not change the meaning of the probability: whichever direction the table uses, the condition determines the reference group. Use this routine before doing the division.
For \(P(A\mid B)\), \(A\) is the event whose probability you want, and \(B\) is the group you are restricting attention to.
Locate the row or column containing all people in \(B\). Its marginal total is the denominator.
Within the condition group, locate the cell that also meets the target \(A\). Its count is the numerator.
Divide the overlap count by the condition total. Describe the result as a proportion of the condition group, in context.
A useful visual check is to imagine temporarily setting aside every row or column outside the condition group. The remaining group is the new reference group. The cell count for the target is then compared with the total number of people in that group—not with the number of people in the whole table.
Worked Example: Flexible Schedules Among Transit Riders
Worked Example: Flexible Schedules Among Transit Riders
A fictional survey records the usual commute and work-schedule flexibility of 300 people. Let \(T\) be the event that a randomly selected person uses public transit, and let \(F\) be the event that the person has a flexible schedule. The counts are shown below.
| Usual commute | Flexible schedule | Not flexible | Total |
|---|---|---|---|
| Public transit | 72 | 48 | 120 |
| Walk or bike | 36 | 24 | 60 |
| Drive | 54 | 66 | 120 |
| Total | 162 | 138 | 300 |
State: The question asks for the probability of having a flexible schedule, given that the person uses public transit. In notation, this is \(P(F\mid T)\).
Plan: The condition is public transit, so restrict attention to the public-transit row. There are 120 transit riders. Of those, 72 have flexible schedules. The row total is the denominator, and the overlapping cell is the numerator.
Do:
The division is \(72\div120=0.60\). As a check using the full-table probabilities, \(P(F\cap T)=72/300=0.24\) and \(P(T)=120/300=0.40\). Their ratio is \(0.24/0.40=0.60\), the same result.
Conclude: Among the people in this survey who use public transit, 60% have a flexible work schedule. This describes a randomly selected person from the transit-rider group in the table, not from all 300 people.
Notice how the condition changes the denominator. If the question instead asked for the probability of using public transit, given a flexible schedule, the condition would be \(F\). The flexible-schedule column total, 162, would be the denominator, and the same overlap count, 72, would be the numerator. That probability is \(72/162\approx0.4444\), not 0.60. The cell is unchanged, but the reference group is different.
Worked Example: Home Type Given No Garden
Worked Example: Home Type Given No Garden
In a fictional survey of 300 residents, each person reports whether their home is an apartment or a house, and whether they garden at home. Let \(H\) mean that a resident lives in a house, and let \(N\) mean that the resident does not garden at home.
| Home type | Gardens | Does not garden | Total |
|---|---|---|---|
| Apartment | 18 | 72 | 90 |
| House | 126 | 84 | 210 |
| Total | 144 | 156 | 300 |
Find the probability that a randomly selected resident lives in a house, given that the resident does not garden at home.
State: The target is living in a house, and the condition is not gardening. The requested probability is \(P(H\mid N)\).
Plan: The condition “does not garden” identifies the second column. Its total, 156, is the reference-group count. Within that column, 84 residents live in houses.
Do:
The fraction simplifies because both 84 and 156 are divisible by 12. The decimal is rounded to four places. As a check with probabilities from all 300 residents, \(P(H\cap N)=84/300=0.28\) and \(P(N)=156/300=0.52\), so \(0.28/0.52\approx0.5385\).
Conclude: Among the residents who do not garden at home, about 53.85% live in houses.
A common misreading would divide 84 by 210, the total number of residents living in houses. That would answer a different question: the probability of not gardening, given that the resident lives in a house, \(P(N\mid H)\). The condition is not “house”; it is “does not garden,” so the no-garden column supplies the denominator.
Worked Example: Print Newsletter Readers in an Age Group
Worked Example: Print Newsletter Readers in an Age Group
A fictional community survey of 300 people records each respondent's age group and preferred newsletter format. Let \(O\) mean that a respondent is 40 or older, and let \(P\) mean that the respondent prefers a print newsletter.
| Age group | Digital | Total | |
|---|---|---|---|
| 18–39 | 96 | 24 | 120 |
| 40 or older | 72 | 108 | 180 |
| Total | 168 | 132 | 300 |
Find the probability that a randomly selected respondent is 40 or older, given that the respondent prefers print.
State: The target is being 40 or older. The condition is preferring print, so the requested probability is \(P(O\mid P)\).
Plan: Restrict attention to the print column. The column total is 132. The cell where “40 or older” and “print” overlap contains 108 respondents. The condition total is positive, so this conditional probability is defined.
Do:
The count totals are consistent: the age-group row totals are \(120+180=300\), and the format column totals are \(168+132=300\). The print column contains \(24+108=132\) respondents, confirming the denominator used. The simplified fraction \(9/11\) is approximately 0.8182.
Conclude: Among respondents who prefer a print newsletter, about 81.82% are 40 or older.
If the question were “What is the probability that a respondent prefers print, given that the respondent is 40 or older?” the condition would instead select the “40 or older” row. The answer would be \(108/180=0.60\). Conditional probabilities can differ when the target and condition switch because the reference group—and therefore the denominator—changes.
Common Mistakes and AP Exam Tips
- Using the grand total automatically. The grand total is appropriate for a probability based on a selection from everyone represented in the table. For a conditional probability, use the total for the condition group.
- Choosing the wrong margin. A row total is not always the denominator, and neither is a column total always the denominator. Find the event after the bar in \(P(A\mid B)\); whichever row or column represents \(B\) supplies the denominator.
- Using the target's total. For \(P(A\mid B)\), the denominator is the total for \(B\), not the total for \(A\). Reversing them answers \(P(B\mid A)\), which can have a different value.
- Putting a marginal count in the numerator. The numerator must satisfy both the target and the condition. It is the count in their intersection cell, not a row or column total.
- Leaving out the reference group in the interpretation. A full-credit interpretation identifies the group named by the condition and describes the result as a proportion of that group, in context.
- Reporting a count when asked for a probability. A cell such as 108 is a number of respondents. The conditional probability is the cell count divided by the condition-group total.
A reliable written response makes the choices visible: write the conditional probability, identify the condition total as the denominator, and point to the overlap cell as the numerator. Then interpret the result using “among” or “of those who” to name the reference group clearly.
Check Your Understanding
For each question, identify the condition group before choosing the denominator.
- A table records whether 80 hikers brought a map and whether they used a marked trail. If 52 brought a map and used a marked trail, and 65 brought a map, what counts would you use to find the probability of using a marked trail, given that a hiker brought a map?
- In a table of 300 residents, 48 live in apartments and grow vegetables, while 72 live in apartments and do not grow vegetables. Find the probability that an apartment resident grows vegetables. What is the denominator?
- A table shows 90 people prefer tea, including 30 who choose a reusable cup. Another 110 people prefer coffee, including 50 who choose a reusable cup. Find the probability of preferring tea, given that a person chooses a reusable cup.
- Explain why the cell count for people who prefer tea and choose a reusable cup can be used in two conditional probabilities, while the denominators may differ.
- For \(P(A\mid B)\), explain why the marginal total for \(B\), rather than the grand total, represents the relevant reference group.