20 medium SAT Probability & conditional probability questions

Medium is where most scores are actually won and lost. These questions are not tricky for the sake of it, but every one of them has a wrong answer built to catch a specific shortcut.

Every question below is a real item from the SAT Climb bank, tagged medium by the same difficulty model the app uses to build your practice. Pick an answer before you open the explanation.

Math · Problem-Solving and Data Analysis~2 per testMedium tier
Question 1Medium

A box contains 120 marbles. Each marble is either glass or plastic, and each marble is either large or small. There are 48 glass marbles, of which 18 are large. There are 72 plastic marbles, of which 30 are large. If one marble is selected at random from the box, what is the probability of selecting a small plastic marble?

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Why A is right

There are 72 plastic marbles total, and 30 of them are large. Therefore, 72 minus 30 equals 42 plastic marbles that are small. The probability of selecting a small plastic marble is 42 out of 120 total marbles, or 42/120.

Why the others are wrong

  • BThis is the number of large plastic marbles, not small plastic marbles.
  • CThis uses the total number of plastic marbles without accounting for the requirement that the marble must also be small.
  • DThis incorrectly adds the number of small glass marbles (48 minus 18 equals 30) to the number of large plastic marbles (30) instead of finding the small plastic marbles.
Question 2Medium

TABLE: A survey of 200 customers at a store categorized by purchase type. Rows: 'Electronics' and 'Clothing'. Columns: 'Used Coupon' and 'No Coupon'. Electronics: Used Coupon 30, No Coupon 70, Total 100. Clothing: Used Coupon 45, No Coupon 55, Total 100. Total: Used Coupon 75, No Coupon 125, Total 200. If one customer is selected at random from those who purchased electronics, what is the probability that the customer used a coupon?

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Why B is right

The probability that a customer used a coupon given that they purchased electronics is the number of electronics customers who used a coupon divided by the total number of electronics customers. From the table, 30 electronics customers used a coupon out of 100 electronics customers total. Therefore, the probability is 30/100 = 0.30.

Why the others are wrong

  • AThis results from dividing the number of electronics customers who used a coupon by the total number of all customers (30/200 = 0.15) instead of by electronics customers only.
  • CThis results from using the overall coupon usage rate (75/200 = 0.375) without conditioning on the electronics purchase.
  • DThis results from using an incorrect calculation, possibly adding probabilities or using the wrong conditional setup.
Question 3Medium
A company has 150 employees. Of these employees, 90 work in the sales department. Of the employees who work in the sales department, 54 have been with the company for more than 5 years.

What fraction of the employees in the sales department have been with the company for more than 5 years?

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Why B is right

The question asks for the fraction of sales department employees who have been with the company for more than 5 years. There are 90 employees in the sales department, and 54 of them have been with the company for more than 5 years. The fraction is 54/90 = 3/5.

Why the others are wrong

  • AThis uses the total number of employees (150) as the denominator instead of the number in the sales department: 54/150 = 9/25.
  • CThis incorrectly uses 81 as the denominator (90 - 9 or another miscomputation): 54/81 = 2/3.
  • DThis uses the number of employees not in sales (60) in the calculation, or computes 90/(150-30) yielding a confused result of 3/4.
Question 4Medium

A jar contains 15 marbles: 5 are green, 6 are blue, and 4 are red. If two marbles are selected at random without replacement, what is the probability that both marbles are blue?

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Why A is right

The probability that the first marble is blue is 6/15. After one blue marble is selected, there are 5 blue marbles left out of 14 total marbles. The probability that the second marble is also blue is 5/14. The probability that both are blue is (6/15) × (5/14) = 30/210 = 1/7.

Why the others are wrong

  • BThis results from using only the marginal probability of selecting a blue marble on the first draw (6/15 = 2/5) without accounting for the second draw.
  • CThis results from adding the probabilities instead of multiplying them, such as 6/15 + 6/15 = 12/15, then incorrectly computing.
  • DThis results from using the wrong denominator by treating the draws as with replacement: (6/15) × (6/15) = 36/225.
Question 5Medium

The table summarizes the distribution of major and class year for 200 college students. TABLE: Rows are Sophomore/Junior/Senior, Columns are Engineering/Business/Total. Sophomore: 24 Engineering, 36 Business, 60 Total. Junior: 32 Engineering, 28 Business, 60 Total. Senior: 44 Engineering, 36 Business, 80 Total. Column Totals: 100 Engineering, 100 Business, 200 Total. If one of these students is selected at random, what is the probability of selecting a junior given that the student is an engineering major?

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Why A is right

This is a conditional probability question asking for P(junior | engineering major). There are 100 engineering majors total, and 32 of them are juniors. Therefore, the probability is 32/100 = 8/25.

Why the others are wrong

  • BThis is the marginal probability of being a junior (60/200 = 3/10), not the conditional probability given the student is an engineering major.
  • CThis results from adding the probability of being a junior (60/200) and the probability of being an engineering major (100/200) to get 160/200 = 4/5, rather than using conditional probability.
  • DThis uses 60 as the denominator (the total number of juniors) instead of 100 (the total number of engineering majors), giving 32/60 = 8/15.
Question 6Medium

A survey of 240 students asked whether they prefer tea or coffee. Of the students surveyed, 156 prefer coffee and 84 prefer tea. If one of these students is selected at random, what is the probability of selecting a student who prefers tea?

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Why A is right

The probability of selecting a student who prefers tea is the number of students who prefer tea divided by the total number of students. There are 84 students who prefer tea out of 240 total students, so the probability is 84/240, which simplifies to 7/20.

Why the others are wrong

  • BThis distractor uses the number of students who prefer coffee instead of tea.
  • CThis distractor uses the number of coffee-preferring students as the denominator instead of the total number of students.
  • DThis distractor incorrectly adds probabilities or uses an incorrect calculation method.
Question 7Medium

TABLE: A table shows the number of employees at a company by department and work mode. Rows: 'Sales' and 'Marketing'. Columns: 'Remote' and 'In-office'. Sales: Remote 18, In-office 32, Total 50. Marketing: Remote 27, In-office 33, Total 60. Total: Remote 45, In-office 65, Total 110. If one employee from the marketing department is selected at random, what is the probability that the employee works remotely?

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Why C is right

The probability that an employee works remotely given that they are in the marketing department is the number of remote marketing employees divided by the total number of marketing employees. From the table, 27 marketing employees work remotely out of 60 total marketing employees. Therefore, the probability is 27/60 = 0.45.

Why the others are wrong

  • AThis results from dividing the number of remote marketing employees by the total number of all employees (27/110 ≈ 0.245, rounded to 0.27 or similar error).
  • BThis results from using the overall remote work rate (45/110 ≈ 0.409, rounded to 0.41) without conditioning on the marketing department.
  • DThis results from using the total marketing employees divided by total employees (60/110 ≈ 0.545) or another computational error.
Question 8Medium
A school has 200 students enrolled in language classes. Of these students, 75 are enrolled in Spanish classes. Of the students enrolled in Spanish classes, 45 are also enrolled in an advanced literature course.

What fraction of the students enrolled in Spanish classes are also enrolled in an advanced literature course?

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Why D is right

The question asks for the fraction of Spanish class students who are also enrolled in an advanced literature course. There are 75 students enrolled in Spanish classes, and 45 of them are also enrolled in an advanced literature course. The fraction is 45/75 = 3/5.

Why the others are wrong

  • AThis uses an incorrect denominator such as 100 (a miscomputation): 45/300 simplifies incorrectly to 3/20.
  • BThis uses the total number of students (200) as the denominator instead of the number in Spanish classes: 45/200 = 9/40.
  • CThis uses the number of students not in Spanish classes (125) incorrectly: 50/125 = 2/5, or computes 80/200.
Question 9Medium

A bag contains 5 red marbles and 3 blue marbles. A marble is drawn at random, replaced, and then a second marble is drawn at random. What is the probability that both marbles drawn are red?

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Why C is right

The probability of drawing red on the first draw is 5/8. Since the marble is replaced, the probability remains 5/8 for the second draw. The compound probability is (5/8) × (5/8) = 25/64.

Why the others are wrong

  • AThis incorrectly uses a denominator of 16 instead of squaring the total count of 8 marbles.
  • BThis incorrectly treats the draws as dependent, using 7 remaining marbles for the second draw.
  • DThis gives the probability of drawing red on a single draw rather than on both draws.
Question 10Medium

A jar contains 18 red balls and 12 white balls. If one ball is selected at random, replaced, and then a second ball is selected at random, what is the probability that both balls selected are red?

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Why B is right

The total number of balls is 18 + 12 = 30. The probability of selecting a red ball on the first draw is 18/30 = 3/5. Since the ball is replaced, the probability of selecting a red ball on the second draw is also 3/5. The combined probability is (3/5) × (3/5) = 9/25.

Why the others are wrong

  • AThis is the probability of selecting one red ball (18/30 = 3/5), not both balls being red.
  • CThis results from adding 3/5 + 3/5 and then dividing incorrectly, or from another error in combining probabilities.
  • DThis treats the problem as without replacement, giving (18/30) × (17/29) = 306/870 = 51/145.
Question 11Medium

A library has 350 books classified by genre and format. Of these books, 140 are fiction and 210 are nonfiction. Among the fiction books, 84 are hardcover. Among the nonfiction books, 105 are hardcover. If one book is selected at random from the library, what is the probability of selecting a paperback fiction book?

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Why A is right

There are 140 fiction books total, and 84 of them are hardcover. Therefore, 140 minus 84 equals 56 fiction books that are paperback. The probability of selecting a paperback fiction book is 56 out of 350 total books, or 56/350.

Why the others are wrong

  • BThis is the number of hardcover fiction books, not paperback fiction books.
  • CThis uses the total number of fiction books without accounting for the requirement that the book must also be paperback.
  • DThis incorrectly adds the total number of paperback books (350 minus 189 equals 161) instead of finding only the paperback fiction books.
Question 12Medium

A survey of 180 students asked whether they prefer online classes, in-person classes, or both. The results are shown in the table.

Online OnlyIn-Person OnlyBothTotal
Freshmen15251050
Sophomores20301565
Juniors18321565
Total538740180
If one of the students who prefer both types of classes is selected at random, what is the probability of selecting a sophomore?

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Why B is right

The question asks for the conditional probability of selecting a sophomore given that the student prefers both types of classes. There are 40 students who prefer both types of classes total, and 15 of those are sophomores. Therefore, the probability is 15/40.

Why the others are wrong

  • AThis incorrectly uses the total number of sophomores (65) as the numerator and the total number of all students (180) as the denominator, ignoring the condition that the student must prefer both types of classes.
  • CThis uses the correct numerator (15 sophomores who prefer both) but incorrectly uses 65 (total sophomores) as the denominator instead of 40 (total students who prefer both).
  • DThis incorrectly adds the number of freshmen who prefer online only (15) and in-person only (25) instead of identifying the correct subset.
Question 13Medium
A survey of 80 students found that 32 students play a musical instrument and 48 students do not play a musical instrument. Of the students who play a musical instrument, 20 are in the school band.

What fraction of the students who play a musical instrument are in the school band?

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Why C is right

The question asks for the fraction of students who play a musical instrument that are also in the school band. There are 32 students who play a musical instrument, and 20 of them are in the school band. The fraction is 20/32 = 5/8.

Why the others are wrong

  • AThis uses the total number of students (80) as the denominator instead of the number who play a musical instrument: 20/80 = 1/4.
  • BThis uses the number of students who do not play a musical instrument (48) in the computation: 32/80 = 2/5.
  • DThis subtracts incorrectly or uses the complement: (32 - 20)/80 + 20/32 arrives at a confused calculation yielding 4/5.
Question 14Medium

A box contains 120 tokens. Each token is labeled with exactly one letter: A, B, C, or D. There are 36 tokens labeled A, 28 tokens labeled B, and 32 tokens labeled C. If one token is selected at random from the box, what is the probability of selecting a token labeled D?

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Why B is right

The total number of tokens is 120. The number labeled A, B, or C is 36 + 28 + 32 = 96. Therefore, the number labeled D is 120 - 96 = 24. The probability of selecting a token labeled D is 24/120 = 1/5.

Why the others are wrong

  • AThis assumes there are 30 tokens of each type, dividing 120 equally by 4, which ignores the given distribution.
  • CThis uses 96/120 = 4/5, which is the probability of selecting a token that is NOT labeled D.
  • DThis incorrectly adds probabilities 36/120 + 28/120 and simplifies to 2/5, which does not represent any meaningful event.
Question 15Medium

A box contains 15 green marbles, 24 red marbles, and 21 yellow marbles. If one marble is selected at random from the box, what is the probability of selecting a marble that is not yellow?

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Why B is right

The total number of marbles is 15 + 24 + 21 = 60. The number of marbles that are not yellow is 15 + 24 = 39. Therefore, the probability of selecting a marble that is not yellow is 39/60, which simplifies to 13/20.

Why the others are wrong

  • AThis distractor uses the number of yellow marbles divided by total marbles instead of the complement.
  • CThis distractor uses an incorrect denominator, possibly from counting only green and red marbles.
  • DThis distractor results from using the wrong combination of values, possibly dividing green plus red by yellow.
Question 16Medium

TABLE: A two-way table shows 150 library visitors categorized by age group and borrowing status. Rows: 'Adult' and 'Child'. Columns: 'Borrowed Books' and 'Did Not Borrow'. Adult: Borrowed Books 54, Did Not Borrow 36, Total 90. Child: Borrowed Books 42, Did Not Borrow 18, Total 60. Total: Borrowed Books 96, Did Not Borrow 54, Total 150. If one visitor who borrowed books is selected at random, what is the probability that the visitor is an adult?

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Why B is right

The probability that a visitor is an adult given that they borrowed books is the number of adults who borrowed books divided by the total number of visitors who borrowed books. From the table, 54 adults borrowed books out of 96 total visitors who borrowed books. Therefore, the probability is 54/96 = 0.5625, which rounds to 0.56.

Why the others are wrong

  • AThis results from dividing the number of adults who borrowed books by the total number of all visitors (54/150 = 0.36) instead of by those who borrowed books.
  • CThis results from using the proportion of adults among all visitors (90/150 = 0.60) without conditioning on borrowing status.
  • DThis results from using the overall borrowing rate (96/150 = 0.64) or another computational error.
Question 17Medium
A library has 240 books in its fiction section. Of these books, 144 are hardcover and 96 are paperback. Of the hardcover books, 108 were published after 2010.

What fraction of the hardcover books in the fiction section were published after 2010?

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Why C is right

The question asks for the fraction of hardcover books that were published after 2010. There are 144 hardcover books, and 108 of them were published after 2010. The fraction is 108/144 = 3/4.

Why the others are wrong

  • AThis uses the total number of books in the fiction section (240) as the denominator instead of the number of hardcover books: 108/240 = 9/20.
  • BThis uses the number of paperback books (96) in the calculation: 96/144 = 2/3.
  • DThis uses an incorrect denominator such as 120 (144 - 24): 108/120 = 9/10.
Question 18Medium

A deck contains 60 cards. Each card is marked with exactly one symbol: star, circle, or triangle. There are 18 cards with stars and 25 cards with circles. If one card is selected at random from the deck, what is the probability of selecting a card with a triangle?

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Why A is right

The number of cards with stars or circles is 18 + 25 = 43. Therefore, the number of cards with triangles is 60 - 43 = 17. The probability of selecting a card with a triangle is 17/60.

Why the others are wrong

  • BThis uses 43/60, which is the probability of selecting a card with a star or circle, not a triangle.
  • CThis uses 17/43, incorrectly using the count of non-triangle cards as the denominator.
  • DThis results from adding 18/60 + 25/60 and simplifying incorrectly, which does not represent any meaningful event.
Question 19Medium

The table shows the distribution of 240 employees at a company by department and employment status.

Full-TimePart-TimeTotal
Sales451560
Marketing502070
Operations603090
Human Resources15520
Total17070240
If one employee from the marketing department is selected at random, what is the probability of selecting a full-time employee?

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Why C is right

The question asks for the conditional probability of selecting a full-time employee given that the employee is from the marketing department. There are 70 employees in marketing total, and 50 of those are full-time. The probability is 50/70, which simplifies to 5/7.

Why the others are wrong

  • AThis uses the correct numerator (50 full-time marketing employees) but incorrectly uses 170 (total full-time employees) as the denominator instead of 70 (total marketing employees).
  • BThis incorrectly uses the total number of all employees (240) as the denominator, ignoring the condition that the employee must be from marketing.
  • DThis is the marginal probability of selecting any marketing employee from all employees, not the conditional probability of selecting a full-time employee from marketing.
Question 20Medium

A research study categorized 450 participants by age group and exercise frequency. The results are shown in the table.

RarelySometimesOftenTotal
18-30 years254580150
31-50 years306060150
51+ years355560150
Total90160200450
If one participant who exercises often is selected at random, what is the probability of selecting a participant who is 31-50 years old?

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Why C is right

The question asks for the conditional probability of selecting a participant aged 31-50 given that the participant exercises often. There are 200 participants who exercise often total, and 60 of those are aged 31-50. The probability is 60/200, which simplifies to 3/10.

Why the others are wrong

  • AThis incorrectly uses the total number of all participants (450) as the denominator, ignoring the condition that the participant must exercise often.
  • BThis uses the correct numerator (60 participants aged 31-50 who exercise often) but incorrectly uses 150 (total participants aged 31-50) as the denominator instead of 200 (total participants who exercise often).
  • DThis is the marginal probability of selecting any participant who rarely exercises, not the conditional probability requested.

What to do after medium

Medium is the tier that decides most scores. If these are landing, the hard set is where the remaining points are.

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