Digital SAT Math · Problem-Solving & Data Analysis
Conditional probability & two-way tables
A two-way table is a reading problem dressed as arithmetic. The phrase given that is not flavour text — it is an order to change the denominator. Once the conditioning group is fixed, the rest is one division: the intersection cell over that row or column total.
On the test
| Domain | Problem-Solving and Data Analysis (score report) |
| CB skill | Probability and conditional probability (table items) |
| What it looks like | A completed or nearly completed two-way table, or a short paragraph of counts that could fill one, plus a “given that / among those who” question |
| Often asked | “Given that…, what is the probability…?”, “Of the people who…, what fraction…?”, “Which of the following is closest to…?” |
| Format | Multiple choice and student-produced response |
| Calculator | Allowed throughout; useless until the right cell and the right total are already chosen |
Recognition cues: given that, among those who, of the students who, if it is known that, provided that, two-way or contingency table, row and column totals.
Pattern recognition
Almost every item on this page is the same move with different labels.
- Row-conditioned — the restriction lives in a row. Denominator = that row’s total.
- Column-conditioned — the restriction lives in a column. Denominator = that column’s total.
- Complement inside a restriction — “given B, probability of not A.” Same denominator; numerator is the other cell in that row or column (or the total minus the A cell).
- Fill-then-read — one or two interior cells are blank; restore them from the margins, then do the same conditional read.
Method
- Underline the conditioning group. Find the phrase after given that, among, of those who, or if it is known that. That group is .
- Set the denominator to ‘s row total or column total — the margin for that group alone.
- Set the numerator to the intersection cell: both the asked category and .
- Reduce or convert only after the fraction is built. Match the asked form — fraction, decimal, percent, or a count.
- Re-read the last line. If the stem asked for a count (“how many…”) and you built a probability, stop and multiply back; if it asked for a probability and you reported a cell, divide.
| Words in the stem | What they fix |
|---|---|
| given that, among those who, of the… who | denominator = that group’s total |
| and, both, who also | numerator is the intersection cell |
| not, neither, did not | numerator is the complement inside the same restriction |
| what is the probability, what fraction | answer is a proportion, not a raw count |
| how many of the… | answer is a count inside the restricted group |
Worked example 1 — column-conditioned read
Stem. A campus café recorded drink orders one morning.
| Hot drink | Cold drink | Total | |
|---|---|---|---|
| Student | 48 | 32 | 80 |
| Faculty | 27 | 13 | 40 |
| Total | 75 | 45 | 120 |
A person is selected at random from those who ordered a hot drink. What is the probability the person is a student?
Step 1 — condition. From those who ordered a hot drink → the hot-drink column. Denominator = 75.
Step 2 — intersection. Student and hot drink = 48.
Step 3 — divide.
Check. The hot-drink column is 48 + 27 = 75, so the fraction of that column that is student is . Under 1, as a probability must be. Answer: .
Trap watch. uses the grand total and answers a different question: the joint probability of student and hot, not the conditional. is the reverse order .
Worked example 2 — complement inside a restriction
Stem. A repair shop logged 200 service tickets.
| Warranty | No warranty | Total | |
|---|---|---|---|
| Fixed same day | 70 | 50 | 120 |
| Needed parts | 30 | 50 | 80 |
| Total | 100 | 100 | 200 |
Given that a ticket was covered by warranty, what is the probability the device needed parts?
Step 1 — condition. Warranty column. Denominator = 100.
Step 2 — asked event. “Needed parts” inside that column is the cell 30 (not the 70 same-day cell).
Step 3 — divide.
Check. Complement inside the same column: , and . Answer: .
Trap watch. is the grand-total slip. is the reverse, . Reporting 30 answers a count when the stem asked for a probability.
Worked example 3 — fill the missing cells, then condition
Stem. A media survey of 150 teens is only partly filled in.
| Streams music | Does not stream music | Total | |
|---|---|---|---|
| Streams video | 72 | ? | 96 |
| Does not stream video | ? | 36 | 54 |
| Total | 90 | 60 | 150 |
Given that a randomly selected teen streams music, what is the probability the teen also streams video?
Step 1 — restore the table. Video and not-music: . Not-video and music: . Check the no-music column: , which matches the printed margin.
Step 2 — condition on music. Music-column total = 90. Intersection with streams video = 72.
Step 3 — divide.
Check. Reverse order uses the video total: , a different answer — so the order mattered. Answer: .
Trap watch. Leaving the blanks unfilled and grabbing a printed total that is not the music margin is how the wrong-margin trap lands. is again the grand total.
Practice
Answer before you open the explanation. Two items are student-produced response (type the number, no choices), matching the real test, and at least four stems carry a full two-way table because that is how the skill actually appears. Every wrong choice below is a named trap — reverse order, grand total, wrong margin, or count-for-proportion.
Question 1 Warm-up
A bakery made 80 muffins: 50 blueberry and 30 chocolate. Of the blueberry muffins, 20 have sprinkles, and of the chocolate muffins, 12 have sprinkles. A muffin is selected at random from those that have sprinkles. What is the probability that the selected muffin is blueberry?
Show the answer Choice B
Why it is right
The condition restricts the sample to muffins with sprinkles. There are 20 + 12 = 32 sprinkled muffins, and 20 of them are blueberry, so the probability is 20/32 = 5/8. The grand total of 80 muffins is never used once the condition is applied. Check: five-eighths of the sprinkled muffins should be blueberry, and 5/8 of 32 is 20.
Why each other choice fails
- Choice A
- Uses the grand total as the denominator: 20/80 = 1/4. That is the joint probability of blueberry and sprinkles among all muffins, not the conditional probability among sprinkled muffins.
- Choice C
- Computes the reverse conditional P(sprinkles | blueberry) = 20/50 = 2/5. Same numerator, but the denominator is the blueberry count instead of the sprinkles count.
- Choice D
- Reports the intersection count 20 instead of the probability. The stem asked for a probability, so the 20 still has to be divided by the 32 sprinkled muffins.
Question 2 Standard
A company surveyed 120 employees about work location and schedule preference. | | Flexible hours | Fixed hours | Total | |---|---|---|---| | Remote | 36 | 24 | 60 | | On-site | 18 | 42 | 60 | | Total | 54 | 66 | 120 | Given that a randomly selected employee works remote, what is the probability the employee prefers flexible hours?
Show the answer Choice A
Why it is right
The phrase given that the employee works remote restricts attention to the Remote row. That row totals 60 employees, and 36 of them prefer flexible hours, so P(flexible | remote) = 36/60 = 3/5. The grand total 120 and the flexible-hours column total 54 are not the conditioning group.
Why each other choice fails
- Choice B
- Divides by the grand total: 36/120 = 3/10. That answers the joint probability of remote and flexible among all employees, ignoring the given-that restriction.
- Choice C
- Uses the reverse conditional P(remote | flexible) = 36/54 = 2/3. The numerator is the same intersection cell, but the denominator is the flexible-hours total instead of the remote total.
- Choice D
- Computes the unconditional probability of preferring flexible hours, 54/120 = 9/20, which never uses the remote restriction at all.
Question 3 Standard
Eighty science-fair projects were classified by subject and by whether they were solo or team projects. | | Biology | Chemistry | Physics | Total | |---|---|---|---|---| | Solo | 14 | 10 | 8 | 32 | | Team | 16 | 12 | 20 | 48 | | Total | 30 | 22 | 28 | 80 | A project is selected at random from the chemistry projects. What is the probability that it is a team project?
Show the answer Choice C
Why it is right
Chemistry is the conditioning column, with total 22. Of those 22 chemistry projects, 12 are team projects, so P(team | chemistry) = 12/22 = 6/11. Only the chemistry column is in play; the team-row total 48 and the grand total 80 are the wrong denominators.
Why each other choice fails
- Choice A
- Computes the reverse conditional P(chemistry | team) = 12/48 = 1/4. Same intersection cell, denominator switched to the team-row total.
- Choice B
- Divides by the grand total: 12/80 = 3/20. That is the joint share of team chemistry among all projects, not the conditional among chemistry projects.
- Choice D
- Uses a wrong margin: 16/48 = 1/3 is the share of team projects that are biology, not chemistry. The cell and the total both drifted off the chemistry column.
Question 4 Standard
Of 200 tickets sold for a play, 80 were for the evening show and the rest were for the matinee. Of the evening tickets, 50 were adult tickets. Of the matinee tickets, 90 were adult tickets. Given that a randomly selected ticket is an adult ticket, what is the probability it is for the evening show?
Show the answer Choice D
Why it is right
Adult tickets total 50 + 90 = 140. Among those 140 adult tickets, 50 are evening tickets, so P(evening | adult) = 50/140 = 5/14. The condition is adult, not evening, so the evening total of 80 is not the denominator. Check by reducing: divide numerator and denominator by 10 to get 5/14, which is less than one half, matching that fewer than half of the adult tickets are evening tickets.
Why each other choice fails
- Choice A
- Computes the reverse conditional P(adult | evening) = 50/80 = 5/8. The restriction in the stem is on adult tickets, not on evening tickets.
- Choice B
- Divides by the grand total of tickets: 50/200 = 1/4. That ignores the given-that adult restriction entirely.
- Choice C
- Reports the unconditional probability of an evening ticket, 80/200 = 2/5, which does not use the adult counts at all.
Question 5 Standard
A clinic recorded 120 visits on one day by time of day and by patient type. | | New patient | Returning | Total | |---|---|---|---| | Morning | 25 | 35 | 60 | | Afternoon | 15 | 45 | 60 | | Total | 40 | 80 | 120 | Given that a randomly selected visit is from a returning patient, what is the probability the visit was in the afternoon?
Show the answer Choice A
Why it is right
Returning patients form the conditioning column, totaling 80 visits. Of those 80, 45 were afternoon visits, so P(afternoon | returning) = 45/80 = 9/16. The afternoon row total of 60 would be the denominator only if the condition were afternoon instead of returning. Reduce by dividing numerator and denominator by 5: 9/16. Most returning visits are afternoon, so a value just above one half is the right neighbourhood.
Why each other choice fails
- Choice B
- Uses the reverse conditional P(returning | afternoon) = 45/60 = 3/4. Same intersection, wrong margin — the afternoon row instead of the returning column.
- Choice C
- Divides by the grand total: 45/120 = 3/8. That is the joint probability of afternoon and returning among all visits.
- Choice D
- Reports the unconditional share of returning patients, 80/120 = 2/3, which never applies the afternoon question.
Question 6 Harder
A library recorded 250 checkouts classified by patron type and by genre. | | Fiction | Nonfiction | Total | |---|---|---|---| | Adult | 70 | 50 | 120 | | Student | 90 | 40 | 130 | | Total | 160 | 90 | 250 | Given that a checkout is fiction, what is the probability the patron is a student?
Show the answer Choice B
Why it is right
Fiction is the conditioning column, totaling 160 checkouts. Of those, 90 are student checkouts, so P(student | fiction) = 90/160 = 9/16. Reduce before matching choices: divide numerator and denominator by 10. The student-row total 130 would be correct only for the reverse question P(fiction | student), and the grand total 250 answers a joint probability, not this conditional.
Why each other choice fails
- Choice A
- Uses the grand total: 90/250 = 9/25. That is the joint share of student fiction among all checkouts, not the conditional among fiction checkouts.
- Choice C
- Computes the reverse conditional P(fiction | student) = 90/130 = 9/13. The student row total is the wrong denominator for a fiction condition.
- Choice D
- Reports the unconditional probability of fiction, 160/250 = 16/25, which does not involve student counts.
Question 7 Harder Student-produced response
A summer camp recorded activity choices for 80 campers by age group. | | Kayaking | Hiking | Total | |---|---|---|---| | Age 14 or under | 22 | 18 | 40 | | Age 15 or over | 14 | 26 | 40 | | Total | 36 | 44 | 80 | Given that a randomly selected camper chose kayaking, what is the probability the camper is age 14 or under? (Enter a fraction or decimal.)
Show the answer 11/18
Why it is right
Kayaking is the conditioning column, totaling 36 campers. Of those 36, 22 are age 14 or under, so the probability is 22/36 = 11/18. As a decimal, 11/18 ≈ 0.611. The age-row total 40 would be correct only for the reverse condition.
Answers students type instead
- 11/40
- Divides by the grand total of campers: 22/80 = 11/40. The condition is kayaking, so the denominator must be the kayaking total 36, not 80.
- 11/20
- Computes the reverse conditional P(kayaking | age 14 or under) = 22/40 = 11/20. Same intersection cell, denominator switched to the age-row total.
- 9/20
- Reports the unconditional probability of kayaking, 36/80 = 9/20, which never uses the age restriction asked for in the numerator.
Question 8 Harder
A restaurant tracked 100 diners by whether they ordered an appetizer and whether they ordered dessert. | | Appetizer | No appetizer | Total | |---|---|---|---| | Dessert | 28 | 12 | 40 | | No dessert | 32 | 28 | 60 | | Total | 60 | 40 | 100 | Given that a diner ordered an appetizer, what is the probability the diner did not order dessert?
Show the answer Choice C
Why it is right
The condition is ordered an appetizer, so the denominator is the appetizer column total 60. Inside that column, 32 diners ordered no dessert, so P(no dessert | appetizer) = 32/60 = 8/15. Equivalently, 1 − 28/60 = 1 − 7/15 = 8/15.
Why each other choice fails
- Choice A
- Computes P(dessert | appetizer) = 28/60 = 7/15, the complement of the asked event. The stem asked for no dessert, not dessert.
- Choice B
- Divides by the grand total: 32/100 = 8/25. That is the joint share of appetizer-and-no-dessert among all diners.
- Choice D
- Uses a wrong margin: 28/40 = 7/10 is P(dessert | no appetizer), mixing the dessert cell with the no-appetizer column.
Question 9 Harder
A factory inspected 200 units from two production lines. | | Pass | Fail | Total | |---|---|---|---| | Line A | 85 | 15 | 100 | | Line B | 70 | 30 | 100 | | Total | 155 | 45 | 200 | Given that a randomly selected unit failed inspection, what is the probability it came from Line A?
Show the answer Choice D
Why it is right
Failed units form the conditioning column, totaling 45. Of those 45 failures, 15 are from Line A, so P(Line A | fail) = 15/45 = 1/3. The answer is a proportion of the failure group, not a raw count of failures from Line A.
Why each other choice fails
- Choice A
- Computes the reverse conditional P(fail | Line A) = 15/100 = 3/20. The condition is failure, so the denominator is 45, not the Line A total.
- Choice B
- Divides by the grand total: 15/200 = 3/40. That is the joint share of Line A failures among all units.
- Choice C
- Reports the intersection count 15 instead of the probability 15/45. The stem asked for a probability, so the count still needs the failure total underneath.
Question 10 Hardest
A media survey of 150 teens is only partly filled in. | | Streams music | Does not stream music | Total | |---|---|---|---| | Streams video | 72 | ? | 96 | | Does not stream video | ? | 36 | 54 | | Total | 90 | 60 | 150 | Given that a randomly selected teen streams music, what is the probability the teen also streams video?
Show the answer Choice A
Why it is right
Restore the music column first: streams video and music is already 72, and the music total is 90, so not-video music is 90 − 72 = 18. Conditioning on music uses denominator 90 and numerator 72, so P(video | music) = 72/90 = 4/5. The video-row total 96 is the reverse denominator.
Why each other choice fails
- Choice B
- Uses the reverse conditional P(music | video) = 72/96 = 3/4. Same intersection, but the condition in the stem is streams music, not streams video.
- Choice C
- Divides by the grand total: 72/150 = 12/25. That ignores the given-that music restriction.
- Choice D
- Reports the unconditional probability of streaming video, 96/150 = 16/25, which never uses the music column.
Question 11 Hardest Student-produced response
A charity auction sold 100 items classified by category and by sales channel. | | Online | In-person | Total | |---|---|---|---| | Art | 30 | 20 | 50 | | Jewelry | 10 | 40 | 50 | | Total | 40 | 60 | 100 | Given that an item was sold in person, what is the probability it was jewelry? (Enter a fraction or decimal.)
Show the answer 2/3
Why it is right
In-person sales form the conditioning column, totaling 60 items. Of those 60, 40 are jewelry, so P(jewelry | in-person) = 40/60 = 2/3. As an unreduced fraction that is 40/60, which the grid also accepts. The jewelry-row total of 50 is the reverse denominator and would give 4/5, a different answer. Check: two thirds of the in-person sales should be jewelry, and two thirds of 60 is 40.
Answers students type instead
- 4/5
- Computes the reverse conditional P(in-person | jewelry) = 40/50 = 4/5. Same intersection cell, denominator switched to the jewelry total.
- 2/5
- Divides by the grand total: 40/100 = 2/5. That is the joint share of in-person jewelry among all items.
- 1/2
- Reports the unconditional probability of jewelry, 50/100 = 1/2, which never applies the in-person restriction.
Question 12 Hardest
A hospital recorded vaccination status for 100 staff members by job category. | | Vaccinated | Not vaccinated | Total | |---|---|---|---| | Clinical staff | 48 | 12 | 60 | | Admin staff | 27 | 13 | 40 | | Total | 75 | 25 | 100 | A staff member is selected at random from those who are not vaccinated. What is the probability the person is clinical staff?
Show the answer Choice C
Why it is right
The condition is not vaccinated, so the denominator is that column's total of 25 staff. Of those 25 unvaccinated staff, 12 are clinical, so P(clinical | not vaccinated) = 12/25. The clinical-row total of 60 would answer the reverse question P(not vaccinated | clinical). The vaccinated column would answer a different condition entirely: 48/75 = 16/25, which is a live wrong choice on this item.
Why each other choice fails
- Choice A
- Divides by the grand total: 12/100 = 3/25. That is the joint share of unvaccinated clinical staff among all staff.
- Choice B
- Computes the reverse conditional P(not vaccinated | clinical) = 12/60 = 1/5. Same intersection, wrong margin — the clinical row instead of the not-vaccinated column.
- Choice D
- Uses the wrong conditioning group: 48/75 = 16/25 is P(clinical | vaccinated), the complementary column rather than the not-vaccinated column the stem named.
Common mistakes
- Grand total as denominator — dividing the intersection by everyone in the table instead of by the conditioning group. The most common miss on the whole skill.
- Reverse order computed as — same numerator, wrong margin. The two answers match only by accident.
- Wrong margin — using the other category’s row or column total (the group the stem did not restrict to).
- Count reported when a proportion was asked (or the reverse) — leaving the answer as the cell value, or multiplying a correct probability back into a count nobody wanted.
- Unconditional probability — answering or from a margin over the grand total when the stem clearly conditioned.
- Ignoring a blank cell — computing from an incomplete table without restoring the intersection from the margins first.
- Complement applied to the wrong total — computing with the grand total instead of inside the restricted group.
- Adding cells that straddle two restrictions — summing a row cell and a column cell that are not the same intersection.
- Percent of the wrong base — converting a correct fraction of the restricted group into a percent of the whole sample.
FAQ
Is this the same as the probability skill? Overlapping idea, different spine. Unconditional “bag of marbles” and simple compound events belong on the probability page. This page is for given that language and two-way tables — the place where the denominator actually changes.
Do I need the formula ? Only if you think in probabilities already. On the Digital SAT the counts are sitting in the table, so the working form is . Same math, fewer symbols.
What if the table is missing a cell? Row total minus the known cell in that row, or column total minus the known cell in that column. Do that before any probability. A blank is not a reason to switch to the grand total.
How do I enter a probability in a grid-in?
Enter the number the last line asks for — a reduced fraction like 3/5, a decimal like 0.4, or a percent without the % sign if the stem said “by what percent.” Equivalent forms of the same value are accepted; a raw cell count is not equivalent to a probability.
Can a conditional probability equal the unconditional one? Yes, when the two events are independent — but the test almost never asks you to prove independence. Treat equality of and as a possible numerical coincidence, not as a shortcut that lets you skip reading the condition.