Digital SAT Math · Problem-Solving & Data Analysis

Probability

On the Digital SAT, probability is almost never a formula hunt. It is a counting problem: name the group you are choosing from, count the outcomes that match what the stem asks for, and write the exact fraction. The whole difficulty is which total goes underneath.

On the test


Domain	Problem-Solving and Data Analysis (score report)
CB skill	Probability and conditional probability (unconditional / simple-compound items)
What it looks like	A jar, spinner, short survey, or two-way table; one random selection (sometimes two)
Often asked	“What is the probability that…?”, “Which of the following is closest to…?”, type-in a fraction
Format	Multiple choice and student-produced response
Calculator	Allowed throughout; usually unnecessary if you keep the fraction exact

Recognition cues: at random, probability that, selected from, or, and, not, a two-way table of counts, without replacement.

Pattern recognition

Almost every item on this page is one of these:

Single event — count favourable, count the reference group, divide.
Complement — the stem asks for not A; compute $1 - P(A)$ when that is faster.
Simple compound — A or B (watch overlap) or A and B (multiply only when independent).
Table read — pull a joint cell or a row/column total over the right denominator (no heavy “given that” ladder here).

Method

Name the reference group the stem restricts you to. Underline the phrase that sets the pool (from the jar, from the sophomores, one of the students who play a sport).
Count favourable outcomes inside that group only. Joint events need the cell that satisfies both conditions.
Write the fraction and keep it exact. Reduce only if the choices are reduced; never switch to a rounded decimal unless the stem asks for one.
Use the complement when “not A” is shorter: $P(\text{not }A) = 1 - P(A)$ .
For or and and, decide the rule before arithmetic. Mutually exclusive or → add. Overlapping or → add, then subtract the overlap once. Independent and → multiply. Without replacement → second factor uses the updated counts.

Words in the stem	What they force
at random / selected	every outcome in the reference group is equally likely
not, neither, fails to	complement: $1 - P(A)$ is often faster
or (can both happen)	add, then subtract the double-counted overlap
or (cannot both happen)	add only
and / both / in succession	multiply; update the total if without replacement
without replacement	second draw sees one fewer item
two-way table, no “given that”	joint cell or margin over the grand total

Worked example 1 — single event, exact fraction

Stem. A jar contains 9 red, 6 blue, and 5 green marbles. One marble is drawn at random. What is the probability that the marble is blue?

Step 1 — reference group. The whole jar: $9 + 6 + 5 = 20$ marbles.

Step 2 — favourable. Blue marbles: $6$ .

Step 3 — fraction.

P(\text{blue}) = \frac{6}{20} = \frac{3}{10}

Check. Numerator is a count of blue only; denominator is every marble; fraction is between 0 and 1. Answer: $3/10$ .

Trap watch. $\frac{6}{15}$ drops the green marbles from the total. $\frac{9}{20}$ answers red. $\frac{6}{9}$ compares blue to red only — a ratio, not a probability from the jar.

Worked example 2 — two-way table, joint over grand total

Stem. A club survey of 100 students recorded whether each takes art and whether each takes band:

	Band	No band	Total
Art	22	28	50
No art	18	32	50
Total	40	60	100

If one student is selected at random from the 100, what is the probability that the student takes both art and band?

Step 1 — reference group. All 100 surveyed students (no smaller pool named).

Step 2 — favourable. The joint cell art-and-band: $22$ .

Step 3 — fraction.

P(\text{art and band}) = \frac{22}{100} = \frac{11}{50}

Check. The joint cell is smaller than either margin ( $22 < 50$ and $22 < 40$ ), as a joint must be. Answer: $11/50$ .

Trap watch. $\frac{50}{100}$ is “takes art” (row total). $\frac{40}{100}$ is “takes band” (column total). $\frac{22}{50}$ is “takes band among art students” — a restricted denominator the stem never asked for. That last trap is the doorway into conditional probability; here the stem said from the 100.

Worked example 3 — complement, then overlapping or

Stem. An integer from 1 to 12 inclusive is chosen at random.

(a) What is $P(\text{not a multiple of }4)$ ?

(b) What is $P(\text{even or greater than }9)$ ?

(a) Complement. Multiples of 4 in the set: $4, 8, 12$ → three values. So

P(\text{multiple of }4) = \frac{3}{12} = \frac{1}{4}, \qquad P(\text{not}) = 1 - \frac{1}{4} = \frac{3}{4}.

Direct count of non-multiples: $1,2,3,5,6,7,9,10,11$ → nine values → $9/12 = 3/4$ . Same answer.

(b) Overlapping or. Even numbers: $2,4,6,8,10,12$ → 6. Greater than 9: $10,11,12$ → 3. Both (even and $>9$ ): $10,12$ → 2.

P(\text{even or }>9) = \frac{6 + 3 - 2}{12} = \frac{7}{12}.

Check. List the union: $\{2,4,6,8,10,11,12\}$ — seven numbers out of twelve. Answer: (a) $3/4$ ; (b) $7/12$ .

Trap watch. On (a), reporting $\frac{1}{4}$ is answering the event you were asked to exclude. On (b), $\frac{6+3}{12} = \frac{9}{12}$ double-counts 10 and 12. $\frac{6}{12}$ answers “even” only.

Practice

Answer before you open the explanation. Two items are student-produced response (type the fraction or decimal, no choices), matching the real test. At least two stems use two-way table data. Every wrong choice below is a specific named trap — when you miss one, log the trap name, not just the item number.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

A jar contains 8 red marbles, 6 blue marbles, and 2 green marbles. One marble is drawn at random. What is the probability that the marble is blue?

Show the answer Choice A

Why it is right

The reference group is the whole jar: 8 + 6 + 2 = 16 marbles. Exactly 6 of them are blue, so P(blue) = 6/16. Reducing by 2 gives 3/8. The numerator counts only blue marbles, and the denominator counts every marble that could have been drawn — both required pieces of a single-event probability.

Why each other choice fails

Choice B: Compares blue to red only: 6/8 = 3/4. That is a blue-to-red ratio, not a probability of drawing from the whole jar. The denominator must be every marble (16), not just the red count.
Choice C: Reports 8/16 = 1/2, which is the probability of drawing a red marble, not a blue one. The favourable colour was blue (6), not red (8).
Choice D: Uses 6/14 = 3/7 by removing the 2 green marbles from the total as if they were not in the jar. Every marble in the jar is a possible outcome.

Question 2 Standard

A fair spinner is divided into 8 equal sections labeled 1 through 8. The spinner is spun once. What is the probability that it lands on a number that is not a multiple of 3?

Show the answer Choice C

Why it is right

The multiples of 3 among 1 through 8 are 3 and 6 — two sections. So P(multiple of 3) = 2/8 = 1/4. The stem asks for the complement: P(not a multiple of 3) = 1 − 1/4 = 3/4. Direct count agrees: the six labels 1, 2, 4, 5, 7, and 8 give 6/8 = 3/4. Either route is exact; the complement is faster once you see the word not.

Why each other choice fails

Choice A: Answers P(is a multiple of 3) = 2/8 = 1/4 and never applies the complement. The stem asked for not a multiple of 3.
Choice B: Treats only 3 as a multiple of 3 and forgets that 6 is also a multiple, so the non-multiples are counted as 7 instead of 6, giving 7/8.
Choice D: Counts five non-multiples instead of six — a listing slip that usually drops one of 1, 2, 4, 5, 7, or 8 — and reports 5/8.

Question 3 Standard

A survey of 80 students recorded whether each is in the drama club and whether each is in the choir. The two-way table of counts is: drama and choir, 18; drama only, 24; choir only, 12; neither, 26. If one student is selected at random from the 80, what is the probability that the student is in the drama club?

Show the answer Choice B

Why it is right

Drama club membership is the sum of the two drama cells: 18 (drama and choir) + 24 (drama only) = 42 students. The reference group is all 80 surveyed students, so P(drama) = 42/80. Reducing by 2 gives 21/40. The stem asked for the drama margin over the grand total, not a joint cell and not a restricted subgroup.

Why each other choice fails

Choice A: Uses only the joint cell 18/80 = 9/40 — the probability of being in both drama and choir. The stem asked for drama (with or without choir), which needs both drama cells.
Choice C: Uses the choir column total 30/80 = 3/8. That is P(choir), not P(drama). The drama margin is 42, not 30.
Choice D: Uses 18/42 = 3/7, the share of drama students who are also in choir. That restricts the denominator to drama students, which the stem never did — it selected from all 80.

Question 4 Standard

A school has 200 students: 80 sophomores and 120 juniors. Of the sophomores, 24 take chemistry. Of the juniors, 48 take chemistry. One student is selected at random from the sophomores. What is the probability that the selected student takes chemistry?

Show the answer Choice D

Why it is right

The stem restricts the reference group to the sophomores before the draw: 80 students. Of those, 24 take chemistry. So P(takes chemistry) = 24/80 = 3/10. The junior counts and the school-wide total are real numbers in the stem, but they are not the pool being drawn from. Underline from the sophomores, then divide only inside that group.

Why each other choice fails

Choice A: Uses all chemistry students over the whole school: (24 + 48)/200 = 72/200 = 9/25. That is the school-wide chemistry rate, not the rate among sophomores only.
Choice B: Uses the junior chemistry rate 48/120 = 2/5. The stem selected from sophomores, not juniors.
Choice C: Puts the sophomore chemistry count over the school total: 24/200 = 3/25. The numerator is right for sophomores, but the denominator must also be the sophomore count (80), not 200.

Question 5 Standard

A bag contains 5 red marbles, 7 blue marbles, and 4 yellow marbles. One marble is drawn at random. What is the probability that the marble is red or yellow?

Show the answer Choice A

Why it is right

Red and yellow cannot both happen on a single draw, so the events are mutually exclusive. Favourable marbles: 5 + 4 = 9. Total marbles: 5 + 7 + 4 = 16. P(red or yellow) = 9/16. Adding the two counts is correct here because no marble is both red and yellow — there is no overlap to subtract.

Why each other choice fails

Choice B: Reports only P(red) = 5/16 and ignores the yellow marbles the stem included with or.
Choice C: Reports only P(yellow) = 4/16 = 1/4 and ignores the red marbles.
Choice D: Multiplies 5/16 × 4/16 = 5/64 as if the stem asked for red and yellow on one draw (impossible) or two independent draws. The connective was or, and a single draw is one event — add the mutually exclusive counts.

Question 6 Harder

An integer from 1 to 20 inclusive is selected at random. What is the probability that the integer is even or a multiple of 5?

Show the answer Choice B

Why it is right

Even integers from 1 to 20: ten values (2, 4, …, 20). Multiples of 5: four values (5, 10, 15, 20). Integers that are both even and a multiple of 5: two values (10 and 20). The union count is 10 + 4 − 2 = 12, so P = 12/20 = 3/5. Subtracting the overlap once is required because 10 and 20 would otherwise be counted twice.

Why each other choice fails

Choice A: Adds 10 + 4 = 14 without subtracting the two overlapping values, then reports 14/20 = 7/10. Classic double-count on a non-mutually-exclusive or.
Choice C: Reports only P(even) = 10/20 = 1/2 and drops the multiples of 5 that are not even (namely 5 and 15).
Choice D: Reports only P(multiple of 5) = 4/20 = 1/5 and drops the even integers that are not multiples of 5.

Question 7 Harder Student-produced response

A fair 12-sided die has faces numbered 1 through 12. The die is rolled once. What is the probability that the number rolled is greater than 9? Enter your answer as a fraction or decimal.

Your answer

Show the answer 1/4

Why it is right

Faces greater than 9 are 10, 11, and 12 — three faces out of twelve equally likely outcomes. P = 3/12 = 1/4. The inequality is strict (greater than 9), so 9 itself is not favourable. Equivalent unreduced form 3/12 and the terminating decimal 0.25 match the same value and are accepted.

Answers students type instead

1/3: Uses 3 favourable faces over a total of 9 (the faces up through 9), as if the die only had faces 1–9. The die has 12 faces.
1/2: Counts six faces (7 through 12) as if the stem said greater than 6, or includes 9 and miscounts the upper half.
9/12: Answers P(number ≤ 9) — the complement of what was asked — or confuses the threshold with the favourable count.

Question 8 Harder

A fair coin is flipped and a fair 6-sided die is rolled. What is the probability that the coin shows heads and the die shows an even number?

Show the answer Choice C

Why it is right

The coin flip and the die roll are independent: neither outcome changes the other. P(heads) = 1/2. Even faces on the die are 2, 4, and 6, so P(even) = 3/6 = 1/2. For both events together, multiply: (1/2) × (1/2) = 1/4. Sample-space check: 2 × 6 = 12 equally likely pairs, and the favourable pairs are (H,2), (H,4), (H,6) — three of twelve, which is again 1/4.

Why each other choice fails

Choice A: Computes P(heads or even) with a sloppy inclusion: 1/2 + 1/2 − 1/4 = 3/4, or simply adds without care. The stem asked for and, not or.
Choice B: Multiplies 1/2 by 1/6 as if only one even face existed (or as if the die had to show a specific face such as 2). Three faces are even, so the die factor is 1/2, not 1/6.
Choice D: Reports only one of the two factors — P(heads) or P(even) — and never multiplies for the joint event.

Question 9 Harder

A survey of 150 students recorded whether each plays a sport and whether each plays a musical instrument. The two-way table of counts is: sport and instrument, 28; sport and no instrument, 52; no sport and instrument, 22; no sport and no instrument, 48. One student is selected at random from those who play a sport. What is the probability that the selected student plays a musical instrument?

Show the answer Choice D

Why it is right

The stem restricts the pool to students who play a sport before the selection. Sport total: 28 + 52 = 80. Of those, 28 also play an instrument. P = 28/80 = 7/20. The grand total 150 and the instrument column total 50 are real table numbers, but they are not the reference group named by from those who play a sport.

Why each other choice fails

Choice A: Uses the joint cell over the grand total: 28/150 = 14/75. That would be P(sport and instrument) for a student drawn from everyone, not from the sport players only.
Choice B: Uses all instrument players over everyone: 50/150 = 1/3. That ignores both the sport restriction and the joint requirement.
Choice C: Uses 28/50 = 14/25 — instrument players who play a sport, over all instrument players. That reverses the restriction (conditioning on instrument instead of on sport).

Question 10 Hardest

A box contains 6 red tiles and 4 blue tiles. Two tiles are drawn at random one after another without replacement. What is the probability that both tiles are red?

Show the answer Choice A

Why it is right

Without replacement, the second draw depends on the first. P(first red) = 6/10. After one red is removed, 5 red remain among 9 tiles, so P(second red | first red) = 5/9. Multiply along the path: (6/10) × (5/9) = 30/90 = 1/3. The events are not independent, so the second factor must use the updated counts.

Why each other choice fails

Choice B: Multiplies (6/10) × (6/10) = 36/100 = 9/25 as if the first tile were replaced (or as if the draws were independent with the original counts). Without replacement changes both the red count and the total.
Choice C: Updates the red count to 5 but leaves the total at 10: (6/10) × (5/10) = 3/10. The total must also drop by one after the first draw, to 9.
Choice D: Reports only the second-draw factor 5/9 and never multiplies by P(first red) = 6/10. Both tiles red is a two-step path: multiply the factors, do not stop after the conditional second draw.

Question 11 Hardest Student-produced response

A survey of 120 customers recorded whether each prefers online ordering and whether each would recommend the store. The two-way table of counts is: online and yes, 45; online and no, 15; in-person and yes, 35; in-person and no, 25. One customer is selected at random from the 120. What is the probability that the customer prefers online ordering or would recommend the store (or both)? Enter your answer as a fraction.

Your answer

Show the answer 19/24

Why it is right

Online total: 45 + 15 = 60. Recommend-yes total: 45 + 35 = 80. Both (online and yes): 45. The union count is 60 + 80 − 45 = 95, so P = 95/120. Reducing by 5 gives 19/24. Subtracting the joint cell once prevents double-counting the 45 customers who satisfy both conditions. The reference group is all 120 because the stem selected from the full survey.

Answers students type instead

7/6: Adds 60 + 80 = 140 and divides by 120 without subtracting the overlap, producing 140/120 = 7/6 — a probability greater than 1 and a sure sign of double-counting.
3/8: Reports only the joint cell 45/120 = 3/8 (online and yes), which is the intersection, not the union the stem asked for with or.
1/2: Reports only P(online) = 60/120 = 1/2 and drops the in-person customers who would still recommend the store.

Question 12 Hardest

A box holds 20 raffle tickets, of which 8 are winning tickets. Two tickets are drawn at random one after another without replacement. What is the probability that both tickets are winning tickets?

Show the answer Choice C

Why it is right

P(first winning) = 8/20. After one winning ticket is removed, 7 winning tickets remain among 19 tickets, so P(second winning | first winning) = 7/19. Multiply: (8/20) × (7/19) = 56/380. Reducing by 4 gives 14/95. Without replacement forces the second factor to use 7 and 19, not the original 8 and 20.

Why each other choice fails

Choice A: Multiplies (8/20) × (8/20) = 64/400 = 4/25 as if each draw were independent with replacement. The stem says without replacement.
Choice B: Updates the winning count to 7 but leaves the total at 20: (8/20) × (7/20) = 56/400 = 7/50. The total must fall to 19 after the first draw.
Choice D: Adds the two path factors instead of multiplying: 8/20 + 7/19 = 152/380 + 140/380 = 292/380 = 73/95. Addition is the or rule; both tickets winning is and — multiply along the path.

Common mistakes

Grand total under a restricted pool — the stem already limited you to sophomores (or “students who play a sport”) and you still divided by the whole school.
Subgroup total when the stem did not restrict — using a row total as the denominator when the question was about a random student from everyone.
Double-counting or — adding two overlapping sets without subtracting the intersection once.
Multiplying non-independent events — treating a without-replacement second draw as if the counts never changed, or multiplying events that clearly depend on each other.
Adding when the stem said and — or multiplying when the stem said or. The connective is the rule.
Answering the complement by accident — computing $P(A)$ when the stem asked for not A, or the reverse.
Favourable count from the wrong cell — reading a margin when the stem asked for a joint, or a joint when it asked for a margin.
Leaving a probability greater than 1 — a sure sign the denominator is too small or an or was double-counted without later division by the right total.

FAQ

Do I need to memorize the addition and multiplication “rules”? You need the ideas, not the textbook names. Or means “in at least one of the sets” — count the union. And means “in both” — count the intersection, or multiply when draws are independent. Without replacement, update the counts before the second factor.

When is a probability allowed to be greater than 1? Never. If your fraction exceeds 1, the denominator is wrong or you added overlapping sets and forgot you were still counting people, not probabilities of separate experiments.

Fraction or decimal on an SPR? Enter what the stem asks for. Equivalent fractions are usually accepted (3/8 and 6/16 when both equal the key). A rounded decimal is safe only when the stem says “nearest…” or the value terminates cleanly (0.25 for 1/4).

Is this the same as conditional probability? Related, split on purpose. This page is single-event, complement, simple and/or, and table reads over the pool the stem names. When every item is “given that…, change the denominator,” use the conditional-probability-tables skill.

Desmos? Yes for an ugly product or a complement subtraction. No for deciding which total is the reference group — see the box above.

Editorial team · last reviewed 2026-08-12

Probability

On the test

Pattern recognition

Method

Worked example 1 — single event, exact fraction

Worked example 2 — two-way table, joint over grand total

Worked example 3 — complement, then overlapping or

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

Why it is right

Why each other choice fails

Question 3 Standard

Why it is right

Why each other choice fails

Question 4 Standard

Why it is right

Why each other choice fails

Question 5 Standard

Why it is right

Why each other choice fails

Question 6 Harder

Why it is right

Why each other choice fails

Question 7 Harder Student-produced response

Why it is right

Answers students type instead

Question 8 Harder

Why it is right

Why each other choice fails

Question 9 Harder

Why it is right

Why each other choice fails

Question 10 Hardest

Why it is right

Why each other choice fails

Question 11 Hardest Student-produced response

Why it is right

Answers students type instead

Question 12 Hardest

Why it is right

Why each other choice fails

Common mistakes

FAQ

Related skills