Conditional Probability
Conditional probability is the probability of one event happening given that another event has already happened. The SAT tests this almost exclusively through two-way tables — a grid showing how a population breaks down across two categorical variables. The trap that catches most students: misidentifying the right denominator. Conditional probability problems are won or lost in a single decision: which row or column of the table do you divide by?
Probability refresher
The probability of an event is the number of outcomes that match the event, divided by the total number of possible outcomes:
For example, if 20 students out of 100 are juniors, the probability that a randomly selected student is a junior is 20 / 100 = 20%.
Conditional probability narrows that calculation to a specific subset of the population. Instead of "probability of being a junior," you might ask: "given that a student plays a sport, what's the probability they're a junior?"
The conditional probability formula
Conditional probability is written P(A | B), read as "probability of A, given B." It means: "given that B has already happened, what is the probability of A?"
In practical terms with a two-way table, this simplifies to:
The denominator is determined by what's GIVEN. The phrase "given that..." or "if X is true..." tells you what subset of the population you're working with. The denominator is the total in that subset, NOT the overall total.
"Given that a student plays a sport, what's the probability..." → denominator = total students who play a sport.
"Among female students, what's the probability..." → denominator = total female students.
The SAT writes wrong-answer choices that use the wrong denominator. If you nail the denominator, you nail the question.
Reading two-way tables
A two-way table shows how a population breaks down across two categorical variables — for example, gender vs. grade level, or smoker vs. test result. The cells show how many people fall into each combination, with row totals and column totals on the edges.
Example: a school of 200 students broken down by grade and whether they play a sport.
| Plays sport | Doesn't play | Total | |
|---|---|---|---|
| Junior | 40 | 30 | 70 |
| Senior | 50 | 80 | 130 |
| Total | 90 | 110 | 200 |
From this table you can read off any conditional probability by identifying the right cell and the right total.
Using the table above, what is the probability that a student plays a sport, given that they're a junior?
- Identify the GIVEN. "Given that they're a junior" — we're conditioning on juniors. Use the JUNIOR row only.
- Find the denominator: total juniors = 70.
- Find the numerator: juniors who play a sport = 40.
- Compute:
P(plays sport | junior) = 40 / 70 = 4/7 ≈ 0.571
Using the same table, what is the probability that a student is a junior, given that they play a sport?
- Identify the GIVEN. "Given that they play a sport" — we're conditioning on sport-players. Use the PLAYS SPORT column only.
- Find the denominator: total sport-players = 90.
- Find the numerator: juniors who play a sport = 40.
- Compute:
P(junior | plays sport) = 40 / 90 = 4/9 ≈ 0.444
The two examples above use the SAME table and the SAME shared cell (40 juniors who play a sport), but give DIFFERENT answers (4/7 vs. 4/9). That's because the denominators are different.
P(plays sport | junior) = 40/70 — denominator is "all juniors"
P(junior | plays sport) = 40/90 — denominator is "all sport-players"
The SAT loves to write problems that swap which condition is being given to test whether students actually understand which population they're working with.
When the data is given as percentages
Sometimes the SAT gives you the table with percentages instead of counts. The same logic applies — identify the given, use the appropriate row or column total — but watch for whether percentages are calculated relative to row totals, column totals, or the grand total.
The table below shows the percentage of registered voters in a city, broken down by party and age group. The percentages are calculated relative to the GRAND TOTAL (all registered voters).
| Under 35 | 35 and over | Total | |
|---|---|---|---|
| Party A | 15% | 25% | 40% |
| Party B | 20% | 40% | 60% |
| Total | 35% | 65% | 100% |
What percent of voters under 35 belong to Party A?
- Identify the GIVEN. "Voters under 35" — we're conditioning on under-35 voters. Use the UNDER 35 column.
- Find the denominator: percentage of voters under 35 = 35%.
- Find the numerator: percentage who are under 35 AND in Party A = 15%.
- Compute:
P(Party A | under 35) = 15 / 35 = 3/7 ≈ 0.429 = 42.9%
Multi-step conditional probability
The hardest SAT conditional probability problems chain two conditions or use a single condition to filter a multi-row table. Strategy: shrink the population in stages.
A medical clinic tests patients for a disease. The two-way table shows the result of 1,000 patients:
| Tested positive | Tested negative | Total | |
|---|---|---|---|
| Has disease | 95 | 5 | 100 |
| Doesn't have disease | 45 | 855 | 900 |
| Total | 140 | 860 | 1,000 |
If a patient tested positive, what is the probability that they actually have the disease?
- Identify the GIVEN. "Tested positive" — use the TESTED POSITIVE column only.
- Find the denominator: total tested positive = 140.
- Find the numerator: tested positive AND has disease = 95.
- Compute:
P(has disease | tested positive) = 95 / 140 ≈ 0.679 = 67.9%
This result is surprising — even with a strong test (95% sensitivity), a positive result still only correctly identifies the disease about two-thirds of the time. This is because the disease is rare in the population (only 100 of 1,000), so false positives among the 900 healthy people swamp the true positives. This concept appears on hard SAT problems and in real medical decision-making.
Sample SAT-style problems
A class of 30 students consists of 18 boys and 12 girls. Of the boys, 12 have a pet; of the girls, 9 have a pet. If a student is selected at random and is a girl, what is the probability that she has a pet?
- "Given that the student is a girl" — denominator is total girls = 12.
- Numerator: girls who have a pet = 9.
- Compute:
P(pet | girl) = 9 / 12 = 3/4 = 0.75
The table below shows survey responses from 500 customers about whether they preferred Product A or Product B, broken down by age group:
| Product A | Product B | Total | |
|---|---|---|---|
| Under 30 | 120 | 80 | 200 |
| 30 and over | 90 | 210 | 300 |
| Total | 210 | 290 | 500 |
If a respondent preferred Product B, what is the probability that they were 30 or over?
- "Given that they preferred Product B" — denominator is total Product B preferences = 290.
- Numerator: Product B AND 30 or over = 210.
- Compute:
P(30+ | Product B) = 210 / 290 = 21/29 ≈ 0.724
Using the same Product A/B table from Sample 2, what is the probability that a respondent preferred Product B, given that they were 30 or over?
- Direction has reversed. "Given that they were 30 or over" — denominator is total 30+ = 300.
- Numerator: 30 or over AND Product B = 210 (same shared cell as before).
- Compute:
P(Product B | 30+) = 210 / 300 = 7/10 = 0.7
Notice this is different from Sample 2 (21/29 ≈ 0.724). Same shared cell (210), different denominator (290 vs. 300), different answer. This is exactly the kind of trap the SAT writes.
In a survey of 800 voters, 60% identified as Party X and 40% as Party Y. Among Party X voters, 45% supported a new tax proposal. Among Party Y voters, 70% supported it. If a randomly selected voter supports the tax proposal, what is the probability they belong to Party Y?
- Build the implied two-way table by computing actual counts:
Party X total: 0.60 × 800 = 480 Party Y total: 0.40 × 800 = 320 Party X who support: 0.45 × 480 = 216 Party Y who support: 0.70 × 320 = 224 Total who support: 216 + 224 = 440
- Build the table:
Supports Doesn't Total Party X 216 264 480 Party Y 224 96 320 Total 440 360 800 - "Given they support the proposal" — denominator is total supporters = 440.
- Numerator: Party Y AND supports = 224.
P(Party Y | supports) = 224 / 440 = 28/55 ≈ 0.509
1. Wrong denominator. The most common mistake. Always pause and ask: "What population am I working within, given the condition?" Then use that subset's total as the denominator.
2. Confusing P(A | B) with P(B | A). These are usually different numbers. The SAT writes wrong-answer choices that compute the OTHER direction. Read the question carefully — "given that" tells you what's after the bar.
3. Using percentages without checking the base. When a table gives percentages, ask: percentages of what? If they're row percentages (each row sums to 100%), they're already conditional probabilities. If they're column percentages or grand-total percentages, you may need to compute differently. Always check.