25 hard SAT Probability & conditional probability questions
Real questions from the SAT Climb bank, all at the hard difficulty tier. Pick an answer before you open the explanation. Every question tells you why the answer is right and why each wrong choice is tempting.
Math · Problem-Solving and Data Analysis~2 per testHard tier
Uses the grand total when the question restricts to a subgroup.
Swaps numerator and denominator.
Misreads which row or column the condition names.
Question 1Hard
A box contains 120 marbles. Each marble is either red or blue and either large or small. There are 50 red marbles, of which 18 are large. There are 35 large marbles in total. What is the probability that a randomly selected marble is blue and small?
Show the answer and explanation
Why B is right
First construct the implicit two-way table. Red marbles total 50, blue marbles total 70. Large marbles total 35 (18 red large, so 17 blue large). Small marbles total 85 (32 red small, 53 blue small). Blue and small = 53 out of 120 total marbles, so the probability is 53/120.
Why the others are wrong
AThis incorrectly uses only the blue small marble count divided by the total, but uses 67 which is the count of blue marbles that are not large red (a marginal confusion).
CThis adds the count of blue marbles (70) and small marbles (85) instead of finding their intersection, then divides by 120.
DThis uses the correct numerator (53 blue and small) but divides by 85 (total small marbles) instead of the full sample space of 120.
Question 2Hard
The table summarizes the distribution of major and class year for 120 students at a university. One of these students will be selected at random. What is the probability of selecting a junior, given that the student is a mathematics major?
| | Sophomore | Junior | Senior | Total |
|-----------|-----------|--------|--------|-------|
| Biology | 15 | 12 | 13 | 40 |
| Mathematics| 8 | 18 | 14 | 40 |
| Physics | 17 | 10 | 13 | 40 |
| Total | 40 | 40 | 40 | 120 |
Show the answer and explanation
Why C is right
The probability of selecting a junior given that the student is a mathematics major is the number of mathematics majors who are juniors divided by the total number of mathematics majors. From the table, there are 18 juniors who are mathematics majors and 40 total mathematics majors, so the probability is 18/40 = 9/20.
Why the others are wrong
AA student who computes the unconditional P(junior)=40/120 instead of conditioning on math majors gets 1/3.
BThis incorrectly uses the total number of students as the denominator instead of only mathematics majors.
DThis incorrectly adds probabilities instead of computing the conditional probability.
Question 3Hard
A company has 320 employees. Of these employees, 192 work in the sales department. Of the employees who work in the sales department, 48 have been with the company for more than 10 years.
What fraction of the employees who work in the sales department have been with the company for more than 10 years?
Show the answer and explanation
Why A is right
The question asks for the fraction of sales department employees who have been with the company for more than 10 years. There are 192 employees in the sales department, and 48 of them have been with the company for more than 10 years. The fraction is 48/192 = 1/4.
Why the others are wrong
BThis represents 48/320, which uses the total number of employees as the denominator instead of restricting to sales department employees only.
CThis represents 192/320, the fraction of all employees who work in the sales department, not the fraction asked.
DThis represents the sales employees with 10 years or less (144/192 = 3/4) or inverts the relationship between the two groups.
Question 4Hard
A spinner is divided into 8 equal sections numbered 1 through 8. The spinner is spun twice. What is the probability that at least one of the two spins results in a number greater than 5?
Show the answer and explanation
Why D is right
The probability that at least one spin results in a number greater than 5 equals 1 minus the probability that both spins result in numbers 5 or less. There are 5 numbers (1, 2, 3, 4, 5) that are not greater than 5. The probability that the first spin is 5 or less is 5/8, and the probability that the second spin is also 5 or less is 5/8. Thus P(both ≤5) = 8585=25/64. Therefore P(at least one >5) = 1−25/64=39/64.
Why the others are wrong
AThis is the probability of one spin resulting in a number greater than 5, not accounting for two spins.
BThis results from incorrectly adding probabilities: 3/8 + 3/8, which does not account for the overlap using inclusion-exclusion.
CThis uses an incorrect denominator or fails to properly apply the complement rule for the compound event.
Question 5Hard
A bag contains 15 red marbles, 20 blue marbles, and 25 green marbles. A marble is selected at random from the bag and not replaced. Then a second marble is selected at random from the bag. What is the probability that both marbles selected are red?
Show the answer and explanation
Why B is right
The probability that the first marble is red is 15/60 = 1/4. Given that the first marble is red, there are now 14 red marbles out of 59 total marbles remaining. The probability that the second marble is also red is 14/59. The probability that both marbles are red is (1/4)(14/59) = 14/236 = 7/118.
Why the others are wrong
AThis is only the probability of selecting one red marble on the first draw, not the probability of both marbles being red.
CThis uses the wrong denominator by failing to reduce the total from 60 to 59 after the first marble is removed.
DThis incorrectly adds the probabilities (1/4 + 1/4) instead of multiplying them for independent selections without replacement.
Question 6Hard
In a survey of 200 employees, 120 employees work in sales and 90 employees have been with the company for more than 5 years. If 60 employees both work in sales and have been with the company for more than 5 years, what is the probability that a randomly selected employee either works in sales or has been with the company for more than 5 years?
Show the answer and explanation
Why B is right
Using the inclusion-exclusion principle, the number of employees who either work in sales or have been with the company for more than 5 years is 120 + 90 - 60 = 150. The probability is 150/200 = 3/4.
Why the others are wrong
AThis incorrectly computes (120 + 90)/(200 + 200) by doubling the denominator.
CThis is the probability of working in sales only, not accounting for the union of both events.
DThis uses an incorrect denominator of 100 instead of 200.
Question 7Hard
A bag contains red marbles and blue marbles. The probability of selecting a red marble at random is 73. If there are 12 red marbles in the bag, what is the probability of selecting a blue marble at random?
Show the answer and explanation
Why B is right
Since the probability of selecting a red marble is 3/7, and there are 12 red marbles, the total number of marbles is 12 ÷ (3/7) = 28. The number of blue marbles is 28 - 12 = 16. The probability of selecting a blue marble is 16/28 = 4/7.
Why the others are wrong
AThis is the probability of selecting a red marble, not a blue marble.
CThis incorrectly uses 12 red marbles as the denominator instead of the total number of marbles.
DThis incorrectly combines the numerator and denominator from the given probability without proper calculation.
Question 8Hard
The table summarizes the results of a quality control test for 150 items produced by two machines. One item will be selected at random. What is the probability that the item is defective, given that it was produced by Machine B?
| | Defective | Not Defective | Total |
|-----------|-----------|---------------|-------|
| Machine A | 5 | 70 | 75 |
| Machine B | 9 | 66 | 75 |
| Total | 14 | 136 | 150 |
Show the answer and explanation
Why C is right
The probability that an item is defective given that it was produced by Machine B is the number of defective items from Machine B divided by the total number of items from Machine B. From the table, 9 items from Machine B are defective and 75 items total were produced by Machine B, so the probability is 9/75 = 3/25.
Why the others are wrong
AThis incorrectly uses the total number of all items as the denominator instead of only items from Machine B.
BThis is the marginal probability of an item being defective, not conditional on being from Machine B.
DThis incorrectly adds probabilities without proper calculation.
Question 9Hard
A school has 240 students. Of these students, 144 participate in at least one sport. Of the students who participate in at least one sport, 96 play basketball.
What fraction of the students who participate in at least one sport play basketball?
Show the answer and explanation
Why D is right
The question asks for the fraction of students who participate in at least one sport that play basketball. There are 144 students who participate in at least one sport, and 96 of them play basketball. The fraction is 96/144 = 2/3.
Why the others are wrong
AThis represents the non-basketball athletes (48/144 = 1/3) or uses incorrect grouping of the data.
BThis represents 96/240, which uses all students as the denominator instead of restricting to students who participate in at least one sport.
CThis represents 144/240, the fraction of all students who participate in at least one sport, not the fraction asked.
Question 10Hard
A bag contains 60 marbles: 15 red, 20 blue, and 25 green. A marble is randomly selected from the bag and not replaced. Then a second marble is randomly selected. What is the probability that both marbles selected are blue?
Show the answer and explanation
Why B is right
The probability that the first marble is blue is 20/60 = 1/3. After removing one blue marble, there are 19 blue marbles remaining out of 59 total marbles. The probability that the second marble is blue given the first was blue is 19/59. The probability that both are blue is (20/60) × (19/59) = 380/3540 = 19/177.
Why the others are wrong
AThis is the marginal probability of selecting one blue marble (20/60 = 1/3), not the conditional probability of selecting two blue marbles in succession without replacement.
CThis incorrectly uses the original total of 60 for the denominator of the second selection instead of accounting for the fact that one marble has already been removed, making the denominator 59.
DThis adds the probabilities (20/60 + 20/60 = 40/120) instead of multiplying them to find the probability of both events occurring.
Question 11Hard
A box contains 8 white cards, 12 black cards, and 10 gray cards. Two cards are selected at random from the box without replacement. What is the probability that the first card is white and the second card is black?
Show the answer and explanation
Why C is right
The probability that the first card is white is 8/30 = 4/15. After removing one white card, there are 29 cards remaining, of which 12 are black. The probability that the second card is black given the first is white is 12/29. The probability that both events occur is (4/15)(12/29) = 48/435 = 16/145.
Why the others are wrong
AThis is only the probability of drawing a white card on the first draw, not the probability of both events occurring.
BThis incorrectly uses 30 as the denominator for the second draw instead of 29, yielding (8/30)(12/30) = 4/75, which then gets miscalculated.
A box contains red marbles and blue marbles. The probability of randomly selecting a red marble is 127. If there are 35 red marbles in the box, what is the probability of selecting a blue marble?
Show the answer and explanation
Why B is right
Since the probability of selecting a red marble is 7/12, and there are 35 red marbles, the total number of marbles is 35 ÷ (7/12) = 60. The number of blue marbles is 60 - 35 = 25. The probability of selecting a blue marble is 25/60 = 5/12.
Why the others are wrong
AThis incorrectly uses the number of red marbles as the denominator instead of the total.
CThis is the probability of selecting a red marble, not a blue marble.
DThis inverts and combines the given values without proper calculation.
Question 13Hard
The table summarizes the distribution of employment status and education level for 150 adults in a survey. High School College Graduate School Total Employed 28 42 35 105 Unemployed 12 18 15 45 Total 40 60 50 150 One of these adults will be selected at random. What is the probability of selecting an employed adult, given that the adult has a College or Graduate School education?
Show the answer and explanation
Why C is right
The number of employed adults with College or Graduate School education is 42 + 35 = 77. The total number of adults with College or Graduate School education is 60 + 50 = 110. The conditional probability is 77/110 = 7/10.
Why the others are wrong
AThis is the unconditional probability of selecting an employed adult with College or Graduate School education from all 150 adults, not restricted to the given condition.
BThis uses the total number of employed adults as the denominator instead of the total number of adults with College or Graduate School education.
DThis incorrectly adds the total employed count to the denominator or uses a flawed calculation of the conditional probability.
Question 14Hard
A box contains 60 marbles. Of these marbles, 24 are red, 18 are blue, and the rest are green. Of the red marbles, 9 are large and the rest are small. Of the blue marbles, 6 are large and the rest are small. Of the green marbles, 3 are large and the rest are small. One marble will be selected at random from the box. What is the probability that the selected marble is large, given that it is red or blue?
Show the answer and explanation
Why B is right
There are 24 red marbles and 18 blue marbles, so 42 marbles are red or blue. Of the red marbles, 9 are large. Of the blue marbles, 6 are large. Therefore, 15 marbles are large and either red or blue. The probability is 15/42 = 5/14.
Why the others are wrong
AThis is the probability of selecting a large marble from all marbles (15/60), not conditional on the marble being red or blue.
CThis uses 60 as the denominator instead of 42, the number of red or blue marbles.
DThis incorrectly adds the number of red marbles (24) and large marbles (9) from red marbles, treating them as separate events.
Question 15Hard
A student takes a two-part quiz. The probability that the student answers the first question correctly is 0.6. If the student answers the first question correctly, the probability of answering the second question correctly is 0.9. If the student answers the first question incorrectly, the probability of answering the second question correctly is 0.3. What is the probability that the student answers both questions correctly?
Show the answer and explanation
Why B is right
For both questions to be correct, the student must answer the first correctly (probability 0.6) and then answer the second correctly given the first was correct (probability 0.9). The probability is 0.6 × 0.9 = 0.54.
Why the others are wrong
AThis incorrectly uses the conditional probability for when the first question is answered incorrectly: 0.6 × 0.7 would give a value near this, or uses 0.4 × 0.3 plus other paths incorrectly.
CThis is the product of 0.6 and 1.2, incorrectly adding the two conditional probabilities (0.9 + 0.3) instead of using only the relevant one.
DThis uses only the conditional probability of the second question given the first is correct, ignoring the probability of the first question being correct.
Question 16Hard
A company has 300 employees. Of these, 180 have a college degree, 150 have more than 5 years of experience, and 90 have both a college degree and more than 5 years of experience. If an employee with more than 5 years of experience is selected at random, what is the probability that the employee has a college degree?
Show the answer and explanation
Why B is right
Given that an employee has more than 5 years of experience, the probability that the employee has a college degree is the number of employees with both qualifications divided by the number of employees with more than 5 years of experience: 90/150 = 3/5.
Why the others are wrong
AThis uses the marginal probability 90/300 of randomly selecting an employee with both qualifications from all employees, not conditioning on experience.
CThis uses the wrong denominator by dividing by the number of employees with a college degree instead of those with more than 5 years of experience.
DThis incorrectly adds 90/300 + 150/300 instead of computing the conditional probability.
Question 17Hard
A deck contains cards numbered 1 through 50. The probability of drawing a card with a number that is a multiple of 8 is 50k. What is the probability of drawing a card with a number that is NOT a multiple of 8?
Show the answer and explanation
Why A is right
The multiples of 8 from 1 to 50 are 8, 16, 24, 32, 40, and 48, for a total of 6 cards. Therefore k=6. The number of cards that are NOT multiples of 8 is 50 - 6 = 44, so the probability is 44/50.
Why the others are wrong
BThis is the probability of drawing a card that IS a multiple of 8, not the complement.
CThis incorrectly uses 44 as the denominator instead of the total number of cards.
DThis incorrectly divides the total by 2 without proper reasoning.
Question 18Hard
The table summarizes the results of a quality control inspection of 300 manufactured parts. Pass Fail Total Machine X 85 15 100 Machine Y 110 40 150 Machine Z 45 5 50 Total 240 60 300 One of these parts will be selected at random. What is the probability of selecting a part that passed inspection, given that it was produced by Machine X or Machine Y?
Show the answer and explanation
Why C is right
The number of parts that passed inspection from Machine X or Machine Y is 85 + 110 = 195. The total number of parts produced by Machine X or Machine Y is 100 + 150 = 250. The conditional probability is 195/250 = 39/50.
Why the others are wrong
AThis is the unconditional probability of selecting a passing part from Machine X or Y out of all 300 parts, not restricted to the given condition.
BThis uses the total number of passing parts as the denominator instead of the total number of parts from Machine X or Y.
DThis incorrectly uses the total number of passing parts in the numerator or applies a flawed simplification.
Question 19Hard
A conference has 320 attendees classified by profession and registration type. Of these attendees, 120 are engineers, 100 are scientists, and 100 are educators. Of the engineers, 80 registered early and 40 registered late. Of the scientists, 60 registered early and 40 registered late. Of the educators, 50 registered early and 50 registered late. One attendee will be selected at random. What is the probability that the selected attendee is a scientist, given that the attendee registered early?
Show the answer and explanation
Why B is right
Given that the attendee registered early, we consider only the early registrants. There are 80 + 60 + 50 = 190 early registrants. Of these, 60 are scientists. The probability is 60/190 = 6/19.
Why the others are wrong
AThis uses 320 as the denominator instead of 190, the number of early registrants.
CThis is the unconditional probability of being a scientist (100/320), not conditional on registering early.
DThis treats the three professions as equally likely among early registrants without using the actual counts.
Question 20Hard
A jar contains 120 marbles: 45 red, 35 blue, and 40 green. A marble is selected at random from the jar and not replaced. Then a second marble is selected at random. What is the probability that both marbles selected are red?
Show the answer and explanation
Why B is right
The probability that the first marble is red is 45/120. After one red marble is removed, 44 red marbles remain out of 119 total marbles. The probability that the second marble is also red is 44/119. Multiplying these probabilities gives (45/120)(44/119) = 1980/14280 = 33/238.
Why the others are wrong
AThis results from adding the probabilities instead of multiplying: 45/120 + 44/119, which is incorrect for independent sequential events.
CThis is the probability of selecting one red marble on the first draw, ignoring the condition that both marbles must be red.
DA student who treats the draws as with replacement computes (45/120)^2 = 9/64.
Question 21Hard
At a school fundraiser, 65 percent of the attendees purchased raffle tickets and 40 percent of the attendees purchased both raffle tickets and auction items. An attendee will be selected at random. What is the probability that the attendee purchased auction items, given that the attendee purchased raffle tickets?
Show the answer and explanation
Why A is right
The probability that an attendee purchased auction items given that they purchased raffle tickets is the proportion who purchased both divided by the proportion who purchased raffle tickets. This is 0.40/0.65 = 40/65 = 8/13.
Why the others are wrong
BThis is the unconditional probability that an attendee purchased both raffle tickets and auction items (40%), not the conditional probability.
CThis uses the wrong denominator by incorrectly computing the complement or using an arithmetic error in the conditional probability calculation.
DThis incorrectly adds the percentages (65% + 40% = 105% = 21/20), which exceeds 1 and is impossible for a probability.
Question 22Hard
The table summarizes the distribution of major and class year for 200 students at a university. Freshman Sophomore Junior Senior Total Engineering 18 22 15 25 80 Business 12 16 18 14 60 Liberal Arts 20 18 12 10 60 Total 50 56 45 49 200 One of these students will be selected at random. What is the probability of selecting an Engineering major, given that the student is a Junior or Senior?
Show the answer and explanation
Why A is right
The number of Engineering majors who are Juniors or Seniors is 15 + 25 = 40. The total number of students who are Juniors or Seniors is 45 + 49 = 94. The conditional probability is 40/94.
Why the others are wrong
BThis is the unconditional probability of selecting an Engineering major who is a Junior or Senior from all 200 students, not accounting for the given condition.
CThis incorrectly adds probabilities or uses the total number of Engineering majors in the numerator instead of only those who are Juniors or Seniors.
DThis uses the total number of Engineering majors as the denominator instead of the total number of Juniors and Seniors.
Question 23Hard
A library has 240 books classified by genre and availability. Of these books, 100 are fiction, 80 are nonfiction, and 60 are reference. Of the fiction books, 70 are available and 30 are checked out. Of the nonfiction books, 50 are available and 30 are checked out. Of the reference books, 40 are available and 20 are checked out. One book will be selected at random. What is the probability that the selected book is fiction, given that it is available?
Show the answer and explanation
Why A is right
Given that the book is available, we consider only the books that are available. There are 70 + 50 + 40 = 160 available books. Of these, 70 are fiction. The probability is 70/160 = 7/16.
Why the others are wrong
BThis is the unconditional probability of selecting a fiction book (100/240), not conditional on the book being available.
CThis uses 240 as the denominator instead of 160, the number of available books.
DThis treats the total number of fiction books as the probability without considering the condition.
Question 24Hard
A quality control inspector examines a batch of products in two stages. In the first stage, each product has a probability of 0.8 of passing inspection. If a product passes the first stage, it moves to the second stage, where it has a probability of 0.7 of passing. What is the probability that a randomly selected product passes both stages of inspection?
Show the answer and explanation
Why A is right
The probability that a product passes both stages is the product of the individual probabilities: 0.8 × 0.7 = 0.56. This represents the compound probability of two independent sequential events both occurring.
Why the others are wrong
BThis is the average of the two probabilities, (0.8 + 0.7)/2 = 0.75, which incorrectly treats the problem as finding a mean rather than a compound probability.
CThis gives only the probability of passing the first stage, ignoring the requirement that the product must also pass the second stage.
DThis incorrectly adds the two probabilities, 0.8 + 0.7 = 1.5, rather than multiplying them for compound probability.
Question 25Hard
A box contains tiles numbered 1 through 20. If two tiles are drawn without replacement, what is the probability that both tiles show even numbers?
Show the answer and explanation
Why A is right
There are 10 even numbers among the 20 tiles. The probability of drawing an even number first is 10/20 = 1/2. After drawing one even tile, 9 even tiles remain among 19 total tiles. The probability is (10/20) × (9/19) = 9/38.
Why the others are wrong
BThis represents 1/2 × 1/2, incorrectly assuming replacement or failing to adjust the denominator for the second draw.
CThis uses only the probability of drawing one even number (10/20), ignoring the requirement that both draws must be even.
DThis uses 10/20 for the first draw but incorrectly keeps 10 in the numerator for the second draw, failing to reduce the count of even tiles after the first draw.
These 25 are a sample, not a study plan
These 25 come from a bank of 17,599 questions across all 31 SAT skills and three difficulty tiers. Inside SAT Climb you get the rest of Probability & conditional probability, a diagnostic that finds which skills are actually costing you points, and a grid that shows what to drill next.
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