Basic Probability — Practice Set
Bank code: STA-PRB · Section: statistics · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
Probability is favourable outcomes over total outcomes: get the total right first, read 'at least one' as 1 minus none, and subtract the overlap before adding two events.
Every question below is live in the question bank under the ID shown — the sheet and the portal are the same questions. Attempt a level with the clock running, then check Part C.
Part A — Questions
Quantitative Comparison — the four choices are always the same, so they are not reprinted: (A) Column A is greater · (B) Column B is greater · (C) The two quantities are equal · (D) The relationship cannot be determined from the information given
Level 1 · Easy — 5 questions · ~4 min
warm-up — these must be automatic
Q1 · STA-PRB-001 · MCQ · 40s
A bag contains 4 red marbles, 3 blue marbles, and 3 green marbles, and no others. One marble is drawn at random. What is the probability that it is red?
(A) 1/10
(B) 3/10
(C) 2/5
(D) 3/5
(E) 2/3
Q2 · STA-PRB-005 · MCQ · 35s
The probability that it will rain tomorrow is 0.35. What is the probability that it will NOT rain tomorrow?
(A) 0.15
(B) 0.35
(C) 0.65
(D) 0.70
(E) 1.35
Q3 · STA-PRB-009 · Numeric Entry · 45s
In a class of 30 students, 12 play soccer and the remaining 18 do not. One student is chosen at random. What is the probability that the student plays soccer? Enter your answer as a fraction in lowest terms.
Numeric entry — write the number.
Q4 · STA-PRB-012 · MCQ · 40s
A fair six-sided die is rolled once. What is the probability of rolling a 3 or a 5?
(A) 1/36
(B) 1/6
(C) 1/3
(D) 1/2
(E) 2/3
Q5 · STA-PRB-015 · QC · 50s
A fair six-sided die, with faces numbered 1 through 6, is rolled once.
Column A: P(the result is a prime number) Column B: P(the result is a composite number)
Level 2 · Medium — 8 questions · ~10 min
two or three steps, one planted trap each
Q6 · STA-PRB-018 · Numeric Entry · 75s
In a group of 50 students, 20 study French, 25 study Spanish, and 10 study both languages. One student is chosen at random. What is the probability that the student studies neither French nor Spanish? Enter your answer as a fraction in lowest terms.
Numeric entry — write the number.
Q7 · STA-PRB-019 · MCQ · 80s
One card is drawn at random from a standard 52-card deck. What is the probability that the card is a heart or a face card (jack, queen, or king)?
(A) 1/4
(B) 3/13
(C) 11/26
(D) 6/13
(E) 25/52
Q8 · STA-PRB-021 · MCQ · 80s
The table shows the favourite subject of each of the 200 students in a survey. Each student named exactly one subject.
Subject | Number of students
Math | 60
Science | 80
English | 40
History | 20
One student is chosen at random from those who did NOT name Science. What is the probability that this student named Math?
(A) 3/10
(B) 3/7
(C) 1/2
(D) 3/5
(E) 7/10
Q9 · STA-PRB-024 · QC · 75s
Events A and B are mutually exclusive, with P(A) = 0.4 and P(B) = 0.5.
Column A: P(A or B) Column B: P(A) + P(B) - P(A and B)
Q10 · STA-PRB-025 · Select all that apply · 85s
A fair six-sided die is rolled once. For which of the following events is the probability STRICTLY greater than 1/3? Select all that apply.
(A) Rolling a number less than 3
(B) Rolling an even number
(C) Rolling a number greater than 2
(D) Rolling a 1 or a 6
(E) Rolling a prime number
Q11 · STA-PRB-026 · MCQ · 80s
A square has side length 10. A circle of radius 3 is drawn with its centre at the centre of the square, so the circle lies entirely inside the square. A point is chosen at random inside the square, with every point equally likely. What is the probability that the point lies inside the circle?
(A) 3/10
(B) 9/100
(C) 3pi/20
(D) 9pi/100
(E) 9pi/25
Q12 · STA-PRB-033 · QC · 70s
An integer is chosen at random from the integers 1 through 20, inclusive.
Column A: P(the integer is a multiple of 3) Column B: P(the integer is a multiple of 4)
Q13 · STA-PRB-061 · MCQ · 65s
The odds in favour of a certain horse winning a race are 2 to 5. What is the probability that the horse wins the race?
(A) 2/5
(B) 2/7
(C) 2/3
(D) 5/7
(E) 5/2
Level 3 · Hard — 12 questions · ~19 min
where 162+ is won or lost
Q14 · STA-PRB-017 · MCQ · 90s
Events A and B are mutually exclusive, and P(A) = 3 x P(B). If P(A or B) = 0.6, what is P(not A)?
(A) 0.15
(B) 0.40
(C) 0.45
(D) 0.55
(E) 0.85
Q15 · STA-PRB-029 · QC · 95s
One card is drawn at random from a standard 52-card deck.
Column A: P(the card is an ace or a spade) Column B: P(the card is a king or a red card)
Q16 · STA-PRB-030 · Select all that apply · 100s
Events A and B satisfy P(A) = 0.5, P(B) = 0.4, and P(A and B) = 0.2. Which of the following statements are true? Select all that apply.
(A) P(A or B) = 0.7
(B) P(neither A nor B) = 0.1
(C) P(A occurs but B does not) = 0.3
(D) P(exactly one of A and B occurs) = 0.7
(E) P(not B) = 0.6
Q17 · STA-PRB-035 · Select all that apply · 95s
A fair coin is flipped 3 times. For which of the following events is the probability exactly 1/8? Select all that apply.
(A) Getting exactly 3 heads
(B) Getting exactly 2 heads
(C) Getting heads, then tails, then heads, in that exact order
(D) Getting at least 2 tails
(E) Getting heads on the first flip and tails on each of the other two flips
Q18 · STA-PRB-038 · MCQ · 90s
A fair coin is flipped 4 times. What is the probability of getting at least 2 heads?
(A) 5/16
(B) 3/8
(C) 5/8
(D) 11/16
(E) 15/16
Q19 · STA-PRB-041 · Numeric Entry · 100s
Two fair six-sided dice are rolled once. What is the probability that the sum of the two numbers showing is a prime number? Enter your answer as a fraction in lowest terms.
Numeric entry — write the number.
Q20 · STA-PRB-043 · Select all that apply · 100s
An integer is selected at random from the integers 1 through 30, inclusive. For which of the following events is the probability exactly 1/5? Select all that apply.
(A) The integer is divisible by 5
(B) The integer is a perfect square
(C) The integer is a multiple of 3 and less than 19
(D) The integer is a prime number no greater than 13
(E) The integer is both even and a multiple of 3
Q21 · STA-PRB-045 · Numeric Entry · 95s
A rectangular dartboard is 40 cm wide and 60 cm tall. Centred on the board is a circular target of diameter 20 cm. A dart is thrown and lands at a random point of the board, with every point equally likely. What is the probability that it lands inside the circular target? Enter your answer as a decimal rounded to 4 decimal places.
Numeric entry — write the number.
Q22 · STA-PRB-047 · MCQ · 90s
A bag contains only red, blue and yellow marbles. The probability of drawing a red marble at random is 3/8. The bag contains 6 red marbles and 5 blue marbles. What is the probability of drawing a marble that is neither red nor blue?
(A) 5/16
(B) 3/8
(C) 1/2
(D) 5/8
(E) 11/16
Q23 · STA-PRB-051 · QC · 100s
A bag contains 5 red and 5 blue balls. Three balls are drawn at random, all at once and without replacement.
Column A: P(at least one of the three balls is red) Column B: 11/12
Q24 · STA-PRB-052 · MCQ · 105s
Three non-overlapping circles, each of radius r, lie entirely inside a circle of radius 6. A point is chosen at random inside the large circle, with every point equally likely, and the probability that the point lies OUTSIDE all three small circles is 3/4. What is r?
(A) 1/2
(B) 1
(C) sqrt(3)
(D) 3
(E) 3 x sqrt(3)
Q25 · STA-PRB-053 · Numeric Entry · 105s
A class has 10 students, two of whom are named Ana and Ben. A committee of 3 students is chosen at random from the class, with every possible 3-student committee equally likely. What is the probability that Ana is on the committee and Ben is not? Enter your answer as a fraction in lowest terms.
Numeric entry — write the number.
Level 4 · Extreme — 5 questions · ~10 min
165+ — expect to need the insight, not the grind
Q26 · STA-PRB-042 · QC · 120s
A jar contains 4 red, 3 blue and 3 green marbles and no others. Two marbles are drawn at random at the same time, without replacement.
Column A: P(the two marbles are the same colour) Column B: 1/4
Q27 · STA-PRB-048 · Select all that apply · 125s
A jar contains r red, b blue and g green balls, where r, b and g are each at least 1, and no other balls. One ball is drawn at random. Which of the following must be true for every such jar? Select all that apply.
(A) P(red) + P(blue) + P(green) = 1
(B) P(not green) = P(red) + P(blue)
(C) P(red) > 1/3 if and only if r > b and r > g
(D) If the number of red balls is doubled while b and g are unchanged, then P(red) doubles
(E) If one more red ball is added to the jar, P(red) increases
Q28 · STA-PRB-056 · Numeric Entry · 125s
An integer is chosen at random from the integers 1 through 210, inclusive, with every integer equally likely. What is the probability that the chosen integer is divisible by none of 2, 3, 5 and 7? Enter your answer as a fraction in lowest terms.
Numeric entry — write the number.
Q29 · STA-PRB-058 · Numeric Entry · 125s
A square dartboard has side 10. Four quarter-circle regions of radius 3 are painted at the corners, each centred on a corner of the square, and one circular region of radius 2 is painted at the centre of the square. No two painted regions overlap. A dart lands at a random point of the board, with every point equally likely. What is the probability that it lands on an unpainted part of the board? Enter your answer as a decimal rounded to 4 decimal places.
Numeric entry — write the number.
Q30 · STA-PRB-060 · MCQ · 125s
In a room there are n people, where n >= 4, and each pair of people shakes hands exactly once. The handshakes occur one after another in a random order, with every order equally likely. In terms of n, what is the probability that the very first handshake is one that involves a particular person A?
(A) 1/n
(B) 1/C(n,2)
(C) 2/n
(D) 4/n
(E) 1/(n - 1)
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | STA-PRB-001 |
easy | MCQ | C |
| 2 | STA-PRB-005 |
easy | MCQ | C |
| 3 | STA-PRB-009 |
easy | Numeric Entry | 2/5 |
| 4 | STA-PRB-012 |
easy | MCQ | C |
| 5 | STA-PRB-015 |
easy | QC | A |
| 6 | STA-PRB-018 |
medium | Numeric Entry | 3/10 |
| 7 | STA-PRB-019 |
medium | MCQ | C |
| 8 | STA-PRB-021 |
medium | MCQ | C |
| 9 | STA-PRB-024 |
medium | QC | C |
| 10 | STA-PRB-025 |
medium | Select all that apply | B, C, E |
| 11 | STA-PRB-026 |
medium | MCQ | D |
| 12 | STA-PRB-033 |
medium | QC | A |
| 13 | STA-PRB-061 |
medium | MCQ | B |
| 14 | STA-PRB-017 |
hard | MCQ | D |
| 15 | STA-PRB-029 |
hard | QC | B |
| 16 | STA-PRB-030 |
hard | Select all that apply | A, C, E |
| 17 | STA-PRB-035 |
hard | Select all that apply | A, C, E |
| 18 | STA-PRB-038 |
hard | MCQ | D |
| 19 | STA-PRB-041 |
hard | Numeric Entry | 5/12 |
| 20 | STA-PRB-043 |
hard | Select all that apply | A, C, D |
| 21 | STA-PRB-045 |
hard | Numeric Entry | 0.1309 |
| 22 | STA-PRB-047 |
hard | MCQ | A |
| 23 | STA-PRB-051 |
hard | QC | C |
| 24 | STA-PRB-052 |
hard | MCQ | C |
| 25 | STA-PRB-053 |
hard | Numeric Entry | 7/30 |
| 26 | STA-PRB-042 |
extreme hard | QC | A |
| 27 | STA-PRB-048 |
extreme hard | Select all that apply | A, B, E |
| 28 | STA-PRB-056 |
extreme hard | Numeric Entry | 8/35 |
| 29 | STA-PRB-058 |
extreme hard | Numeric Entry | 0.5916 |
| 30 | STA-PRB-060 |
extreme hard | MCQ | C |
Part C — Worked Solutions
Q1 · STA-PRB-001 — Answer: C
Step 1 — count the total. 4 + 3 + 3 = 10 marbles, each equally likely to be drawn. Step 2 — count the favourable outcomes. There are 4 red marbles. Step 3 — divide. P(red) = 4/10 = 2/5. Check: P(red) + P(blue) + P(green) = 4/10 + 3/10 + 3/10 = 1, as the three colours cover the whole bag. Correct. Answer: 2/5.
Trap. Dividing by the 6 marbles that are NOT red: 4/6 = 2/3, choice E. The denominator of a probability is the entire sample space, never the leftovers. Choice D, 3/5, is P(not red); choice B, 3/10, is P(blue); choice A, 1/10, is the probability of one specific named marble.
Q2 · STA-PRB-005 — Answer: C
Step 1 — rain and no rain are complements: exactly one of the two must happen, so their probabilities add to 1. Step 2 — P(not rain) = 1 - P(rain) = 1 - 0.35 = 0.65. Check: 0.35 + 0.65 = 1.00. Correct. Answer: 0.65.
Trap. Subtracting from 0.5 instead of from 1: 0.5 - 0.35 = 0.15, choice A. The complement rule always subtracts from 1, because all outcomes together carry probability 1. Choice E, 1.35, comes from adding 1 + 0.35 and can be rejected on sight: no probability exceeds 1. Choice D, 0.70, comes from rounding 0.35 to 0.30 first.
Q3 · STA-PRB-009 — Answer: 2/5
Step 1 — the total is 30 students, each equally likely to be chosen. Step 2 — the favourable count is the 12 soccer players. Step 3 — P(soccer) = 12/30. Divide numerator and denominator by 6: 12/30 = 2/5. Check: the other 18 give 18/30 = 3/5, and 2/5 + 3/5 = 1. Correct. Answer: 2/5.
Trap. Writing 12/18 = 2/3, using the students who do NOT play as the denominator. The denominator is the whole class, 30. The second slip is leaving the answer as 12/30: the GRE expects a fraction in lowest terms.
Q4 · STA-PRB-012 — Answer: C
Step 1 — the sample space is {1, 2, 3, 4, 5, 6}: six equally likely outcomes. Step 2 — rolling a 3 and rolling a 5 cannot both happen on one roll, so the events are mutually exclusive and their probabilities simply add. P(3 or 5) = P(3) + P(5) = 1/6 + 1/6 = 2/6 = 1/3. Step 3 — the direct count agrees: 2 favourable outcomes out of 6, so 2/6 = 1/3. Answer: 1/3.
Trap. Multiplying instead of adding: (1/6) x (1/6) = 1/36, choice A. Multiplication answers 'and' across separate rolls; 'or' on a single roll adds. Choice D, 1/2, comes from reading 'a 3 or a 5' as the range 3 to 5 and counting 3, 4, 5; choice E, 2/3, from reading it as 'at least 3' and counting 3, 4, 5, 6.
Q5 · STA-PRB-015 — Answer: A
Step 1 — the primes on a die are 2, 3 and 5: three outcomes, so P(prime) = 3/6 = 1/2. Step 2 — the composites are 4 and 6: two outcomes, so P(composite) = 2/6 = 1/3. Step 3 — 1 is neither prime nor composite, so it belongs to neither group. The counts 3 + 2 + 1 = 6 account for the whole die. Step 4 — compare over a common denominator: 1/2 = 3/6 and 1/3 = 2/6, so Column A is greater. Answer: Column A is greater.
Trap. Counting 1 as composite gives P(composite) = 3/6 = 1/2 and the wrong answer C. Counting 1 as prime gives P(prime) = 4/6, which still points at A but for the wrong reason. The number 1 is neither prime nor composite and simply sits out of both lists.
Q6 · STA-PRB-018 — Answer: 3/10
Step 1 — the two groups overlap, so the union needs inclusion-exclusion: French or Spanish = 20 + 25 - 10 = 35 students. Step 2 — 'neither' is the complement of that union: 50 - 35 = 15 students. Step 3 — P(neither) = 15/50 = 3/10. Check the four regions: French only 10, Spanish only 15, both 10, neither 15, and 10 + 15 + 10 + 15 = 50. Correct. Answer: 3/10.
Trap. Adding 20 + 25 = 45 without removing the 10 who take both, which makes the union 45 and 'neither' 5/50 = 1/10. Students in the overlap are counted once in each list, so the overlap must be subtracted exactly once before the complement is taken.
Q7 · STA-PRB-019 — Answer: C
Step 1 — P(heart) = 13/52, and P(face card) = 12/52, since each of the 4 suits contributes a jack, a queen and a king. Step 2 — the two events overlap: the jack, queen and king of hearts appear in both lists, so P(heart and face) = 3/52. Step 3 — inclusion-exclusion: P(heart or face) = 13/52 + 12/52 - 3/52 = 22/52 = 11/26. Check by counting cards directly: 13 hearts plus the 9 face cards in the other three suits = 22 cards out of 52. Correct. Answer: 11/26.
Trap. Adding 13/52 + 12/52 = 25/52, choice E, and forgetting the overlap. 'Or' adds cleanly only for mutually exclusive events; hearts and face cards share three cards, which would otherwise be counted twice. Choice D, 6/13 = 24/52, subtracts just one overlapping card instead of three; choice B, 3/13 = 12/52, is P(face card) alone; choice A, 1/4 = 13/52, is P(heart) alone.
Q8 · STA-PRB-021 — Answer: C
Step 1 — the sample space has shrunk. Only students who did not name Science are eligible: 200 - 80 = 120 students. Step 2 — the favourable count is unchanged: 60 students named Math. Step 3 — P = 60/120 = 1/2. Check: among those 120 students, Math 60 + English 40 + History 20 = 120, and 60 is exactly half of them. Correct. Answer: 1/2.
Trap. Using 200 as the denominator: 60/200 = 3/10, choice A. The phrase 'chosen from those who did not name Science' redefines the total. Choice E, 7/10, is that same 3/10 with a reflex complement taken because the word NOT appears in the stem; the NOT restricts the pool, it does not negate the answer. Choice B, 3/7 = 60/140, removes the Math students from the total instead of the Science students; choice D, 3/5 = 120/200, is P(not Science), the answer to a different question.
Q9 · STA-PRB-024 — Answer: C
Step 1 — mutually exclusive means A and B cannot both occur, so P(A and B) = 0. Step 2 — Column B = 0.4 + 0.5 - 0 = 0.9. Step 3 — for mutually exclusive events the probabilities add directly, so Column A = P(A or B) = 0.4 + 0.5 = 0.9. Step 4 — both columns equal 0.9. The deeper point: Column B is the general addition rule, which holds for ANY two events, and Column A is the left side of that same identity. The two columns would be equal even if the numbers were different. Answer: the two quantities are equal.
Trap. Computing P(A and B) as P(A) x P(B) = 0.4 x 0.5 = 0.2, which drops Column B to 0.7 and points at A. Multiplying is the test for INDEPENDENT events; mutually exclusive is the opposite situation, in which the overlap is 0. Two events with positive probabilities can never be both mutually exclusive and independent.
Q10 · STA-PRB-025 — Answer: B, C, E
Each outcome has probability 1/6, so an event clears 1/3 = 2/6 only when it contains at least 3 outcomes. A - numbers less than 3 are 1 and 2: 2/6 = 1/3 exactly, which is not STRICTLY greater. FALSE. B - even numbers are 2, 4, 6: 3/6 = 1/2 > 1/3. TRUE. C - numbers greater than 2 are 3, 4, 5, 6: 4/6 = 2/3 > 1/3. TRUE. D - a 1 or a 6 is 2 outcomes: 2/6 = 1/3 exactly, not strictly greater. FALSE. E - primes on a die are 2, 3, 5: 3/6 = 1/2 > 1/3. TRUE. Answer: B, C, E.
Trap. Including A and D, whose probabilities are exactly 1/3 - the word 'strictly' rules out equality. The other slip is dropping E by forgetting that 2 is prime; counting only 3 and 5 gives 2/6 = 1/3 and pushes E into the false list.
Q11 · STA-PRB-026 — Answer: D
Step 1 — geometric probability is a ratio of AREAS, never of lengths. Step 2 — area of the square = 10^2 = 100. Step 3 — area of the circle = pi x 3^2 = 9pi. Step 4 — P = 9pi/100, which is about 0.283. Check the size: a circle of diameter 6 sitting inside a square of side 10 should cover well under half the square, and 0.283 is sensible. Answer: 9pi/100.
Trap. Choice A, 3/10, is the radius divided by the side - a length ratio, which is never the answer in geometric probability. Choice C, 3pi/20, is the circumference 6pi over the perimeter 40. Choice E, 9pi/25 = pi x 6^2/100, uses the diameter 6 as the radius and evaluates to about 1.13, greater than 1 and therefore impossible for any probability. Choice B, 9/100, drops the pi.
Q12 · STA-PRB-033 — Answer: A
Both columns have the same denominator, 20, so the comparison is really a count. Step 1 — multiples of 3 up to 20: 3, 6, 9, 12, 15, 18. That is 6 numbers, because 20/3 = 6.67 and only whole multiples count. P = 6/20 = 3/10. Step 2 — multiples of 4 up to 20: 4, 8, 12, 16, 20. That is 5 numbers, since 20/4 = 5 exactly. P = 5/20 = 1/4. Step 3 — 6/20 > 5/20, so Column A is greater. Answer: Column A is greater.
Trap. Rounding 20/3 = 6.67 up to 7 and reporting 7/20 for Column A, or assuming the smaller divisor must produce the smaller count and choosing B. Count multiples by taking the floor of 20/d, and note the boundary: 20 itself is a multiple of 4, while the next multiple of 3 after 18 is 21, outside the range.
Q13 · STA-PRB-061 — Answer: B
Step 1 — odds in favour of a to b mean the outcomes split into a favourable parts and b unfavourable parts, for a + b parts in all. Step 2 — here a = 2 and b = 5, so there are 2 + 5 = 7 equal parts. Step 3 — P(win) = a/(a + b) = 2/7. Check: P(lose) = 5/7, and 2/7 + 5/7 = 1. Correct. Answer: 2/7.
Trap. Reading the odds 2 to 5 straight off as the fraction 2/5, choice A - that would be the probability only if the odds were 2 to 3. Odds compare favourable to UNFAVOURABLE; probability compares favourable to the TOTAL. Choice C, 2/3, uses a/(b - a) = 2/3 instead of a/(a + b), subtracting the parts instead of adding them. Choice D, 5/7, is the probability the horse loses, and choice E, 5/2, is greater than 1 and cannot be a probability at all.
Q14 · STA-PRB-017 — Answer: D
Step 1 — for mutually exclusive events the probabilities add: P(A or B) = P(A) + P(B) = 0.6. Step 2 — substitute P(A) = 3 x P(B): 3P(B) + P(B) = 0.6, so 4P(B) = 0.6 and P(B) = 0.15. Step 3 — P(A) = 3 x 0.15 = 0.45. Step 4 — the question asks for the complement: P(not A) = 1 - 0.45 = 0.55. Check: 0.45 + 0.15 = 0.60, matching the given union. Correct. Answer: 0.55.
Trap. Stopping at P(A) = 0.45, choice C, instead of taking the complement the question actually asked for. Choice A, 0.15, is P(B); choice B, 0.40, is P(neither A nor B) = 1 - 0.6; choice E, 0.85, is 1 - P(B), the complement of the wrong event.
Q15 · STA-PRB-029 — Answer: B
Both events overlap, so both columns need inclusion-exclusion, and the overlaps are not the same size. Column A: P(ace) = 4/52 and P(spade) = 13/52. Exactly one card, the ace of spades, is in both lists, so the overlap is 1/52. P(ace or spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13. Column B: P(king) = 4/52 and P(red) = 26/52. Two cards, the king of hearts and the king of diamonds, are in both lists, so the overlap is 2/52. P(king or red) = 4/52 + 26/52 - 2/52 = 28/52 = 7/13. 4/13 < 7/13, so Column B is greater. Check by counting cards: Column A covers the 13 spades plus the 3 aces in other suits = 16 cards; Column B covers the 26 red cards plus the 2 black kings = 28 cards. Correct. Answer: Column B is greater.
Trap. Treating 'red card' as 'heart'. That turns Column B into 4/52 + 13/52 - 1/52 = 16/52, identical to Column A, and points at C; a standard deck has 26 red cards, the diamonds as well as the hearts. Reading Column B's 'or' as 'and' is the other way to go wrong: P(the card is a red king) = 2/52 is far below Column A's 16/52 and points at A.
Q16 · STA-PRB-030 — Answer: A, C, E
Build the four regions from the given numbers first. P(A or B) = P(A) + P(B) - P(A and B) = 0.5 + 0.4 - 0.2 = 0.7. A only = 0.5 - 0.2 = 0.3. B only = 0.4 - 0.2 = 0.2. Both = 0.2. Neither = 1 - 0.7 = 0.3. Check: 0.3 + 0.2 + 0.2 + 0.3 = 1.0. A - the union is 0.7. TRUE. B - 'neither' is the complement of the union: 1 - 0.7 = 0.3, not 0.1. FALSE. C - A but not B = 0.5 - 0.2 = 0.3. TRUE. D - exactly one = (A only) + (B only) = 0.3 + 0.2 = 0.5, not 0.7. The value 0.7 is 'at least one'. FALSE. E - P(not B) = 1 - 0.4 = 0.6. TRUE. Answer: A, C, E.
Trap. B is the value 1 - P(A) - P(B) = 0.1, which subtracts the 0.2 overlap twice and never adds it back. D confuses 'exactly one' with 'at least one': the union 0.7 includes the 0.2 who get both, so exactly one is 0.7 - 0.2 = 0.5.
Q17 · STA-PRB-035 — Answer: A, C, E
Three flips give 2^3 = 8 equally likely ordered outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Any single fully specified sequence therefore has probability 1/8. A - exactly 3 heads is the one sequence HHH. P = 1/8. TRUE. B - exactly 2 heads occurs as HHT, HTH or THH: 3 sequences, so P = 3/8. FALSE. C - H, T, H is one specified sequence. P = 1/8. TRUE. D - at least 2 tails covers HTT, THT, TTH and TTT: 4 sequences, so P = 4/8 = 1/2. FALSE. E - heads then tails then tails is the single sequence HTT. P = 1/8. TRUE. Answer: A, C, E.
Trap. Marking B because 'exactly 2 heads' sounds as specific as HHH. It is not: the order is unspecified, so it collects 3 of the 8 sequences and carries 3/8. Only a fully specified sequence is worth 1/8. D is the same error in reverse - 'at least 2 tails' sweeps up half the sample space.
Q18 · STA-PRB-038 — Answer: D
Step 1 — the sample space has 2^4 = 16 equally likely sequences. Step 2 — 'at least 2' is fastest through its complement, 'at most 1 head'. 0 heads: TTTT, 1 sequence. 1 head: HTTT, THTT, TTHT, TTTH, 4 sequences. So P(at most 1 head) = 5/16. Step 3 — P(at least 2 heads) = 1 - 5/16 = 11/16. Check by counting directly: exactly 2 heads occurs 6 ways, exactly 3 occurs 4 ways, exactly 4 occurs 1 way, and 6 + 4 + 1 = 11 out of 16. Correct. Answer: 11/16.
Trap. Choice B, 3/8 = 6/16, answers 'exactly 2 heads' instead of 'at least 2'. Choice C, 5/8 = 10/16, adds exactly 2 (6 ways) and exactly 3 (4 ways) but forgets the single all-heads sequence. Choice A, 5/16, is the complement itself - P(at most 1 head) - handed in without the final subtraction from 1. Choice E, 15/16, is 1 - 1/16, which answers 'at least 1 head'.
Q19 · STA-PRB-041 — Answer: 5/12
Step 1 — the two dice are distinguishable, so the sample space is 6 x 6 = 36 equally likely ordered pairs. Step 2 — the sums run from 2 to 12, and the primes among those are 2, 3, 5, 7 and 11. Step 3 — count the ways for each prime sum: sum 2: (1,1) = 1 way sum 3: (1,2), (2,1) = 2 ways sum 5: (1,4), (2,3), (3,2), (4,1) = 4 ways sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways sum 11: (5,6), (6,5) = 2 ways Step 4 — total favourable = 1 + 2 + 4 + 6 + 2 = 15, so P = 15/36 = 5/12. Check: the non-prime sums 4, 6, 8, 9, 10, 12 have counts 3 + 5 + 5 + 4 + 3 + 1 = 21, and 15 + 21 = 36. Correct. Answer: 5/12.
Trap. Treating (2,3) and (3,2) as one outcome, which collapses the sample space to 21 unordered pairs and gives 8/21. The dice are separate objects, so both orders count and the denominator is 36. The second slip is calling 9 prime (9 = 3 x 3) or calling 1 prime, either of which inflates the count.
Q20 · STA-PRB-043 — Answer: A, C, D
The denominator is 30, and 1/5 = 6/30, so an event qualifies exactly when it contains 6 of the integers. A - multiples of 5: 5, 10, 15, 20, 25, 30. That is 6, so P = 6/30 = 1/5. TRUE. B - perfect squares: 1, 4, 9, 16, 25. That is 5, so P = 5/30 = 1/6. FALSE. C - multiples of 3 below 19: 3, 6, 9, 12, 15, 18. That is 6, so P = 1/5. TRUE. D - primes at most 13: 2, 3, 5, 7, 11, 13. That is 6, so P = 1/5. TRUE. E - even and a multiple of 3 means a multiple of 6: 6, 12, 18, 24, 30. That is 5, so P = 1/6. FALSE. Answer: A, C, D.
Trap. B and E both contain 5 values rather than 6, and both feel like 6. In B the next perfect square, 36, is past 30; in E it is easy to slide an extra multiple of 6 into the list. In D, 1 is not prime, so the list starts at 2 and ends at 13 with exactly 6 members - counting 1 as prime would give 7 and put D in the false column.
Q21 · STA-PRB-045 — Answer: 0.1309
Step 1 — geometric probability is the ratio of areas. Step 2 — area of the board = 40 x 60 = 2400 square cm. Step 3 — the radius is half the diameter: 20/2 = 10 cm, so the target's area = pi x 10^2 = 100pi square cm. Step 4 — P = 100pi/2400 = pi/24. Step 5 — pi/24 = 3.14159/24 = 0.130899..., which rounds to 0.1309. Check the size: 100pi is about 314 square cm out of 2400, roughly one eighth of the board. Correct. Answer: 0.1309.
Trap. Using the diameter 20 as the radius: pi x 20^2 = 400pi gives 400pi/2400 = pi/6 = 0.5236, four times too large, because area scales with the square of the radius. Halve the diameter before squaring. Handing in the exact form pi/24 instead of the four-place decimal the question asked for also scores zero.
Q22 · STA-PRB-047 — Answer: A
Step 1 — recover the total from the red probability. If n is the total, 6/n = 3/8, so 3n = 48 and n = 16. Step 2 — count the yellow marbles: 16 - 6 red - 5 blue = 5 yellow. Step 3 — 'neither red nor blue' means yellow, so P = 5/16. Check: 6/16 + 5/16 + 5/16 = 16/16 = 1. Correct. Answer: 5/16.
Trap. Choice C, 1/2, divides the 5 yellow marbles by the 10 marbles that are not red instead of by all 16 - the 5 blue marbles are still in the bag and still in the sample space. Choice D, 5/8, is P(not red) = 10/16, which stops before removing the blue marbles; choice E, 11/16, is its complement 1 - 5/16, the probability of drawing red or blue; choice B, 3/8, simply repeats the given P(red).
Q23 · STA-PRB-051 — Answer: C
Step 1 — 'at least one' calls for the complement: P(at least one red) = 1 - P(no red) = 1 - P(all three blue). Step 2 — count the sample space: C(10,3) = (10 x 9 x 8)/(3 x 2 x 1) = 120 selections. Step 3 — count the all-blue selections: C(5,3) = (5 x 4 x 3)/(3 x 2 x 1) = 10. Step 4 — P(all blue) = 10/120 = 1/12, so P(at least one red) = 1 - 1/12 = 11/12. Column A = 11/12 = Column B. Check by the sequential route: (5/10)(4/9)(3/8) = 60/720 = 1/12, the same all-blue probability. Correct. Answer: the two quantities are equal.
Trap. Treating the draws as if each ball were replaced: (5/10)^3 = 1/8, which gives 1 - 1/8 = 7/8 and points at B. Without replacement the pool shrinks on every draw, so the denominators fall 10, 9, 8. Enumerating 'at least one red' as one red plus two reds plus three reds also works but costs three counts instead of one.
Q24 · STA-PRB-052 — Answer: C
Step 1 — take the complement first: P(the point lies inside one of the small circles) = 1 - 3/4 = 1/4. Step 2 — geometric probability is a ratio of AREAS. The large circle has area pi x 6^2 = 36pi, and the three small circles do not overlap, so together they cover 3 x pi x r^2 = 3pi r^2. Step 3 — set up the equation: 3pi r^2 / 36pi = 1/4. The pi cancels, leaving 3r^2/36 = 1/4, that is r^2/12 = 1/4. Step 4 — solve: r^2 = 3, and r > 0, so r = sqrt(3), about 1.73. Check: each small circle has area pi x 3 = 3pi, the three together 9pi, and 9pi/36pi = 1/4, so 3/4 of the large circle is left over. Correct, and three circles of radius 1.73 fit easily inside a circle of radius 6. Answer: sqrt(3).
Trap. Choice D, 3, comes from either of two slips that land on the same number: setting r^2/36 = 1/4, which forgets that THREE circles are covered, or setting 3r^2/36 = 3/4, which never takes the complement. Choice E, 3 x sqrt(3), makes both slips at once (r^2/36 = 3/4) and gives circles of radius 5.2 that cannot fit inside a circle of radius 6. Choice A, 1/2, works with lengths instead of areas: 3r/6 = 1/4. Choice B, 1, replaces the three circles by one circle of radius 3r: 9r^2/36 = 1/4.
Q25 · STA-PRB-053 — Answer: 7/30
Step 1 — count the sample space: C(10,3) = (10 x 9 x 8)/(3 x 2 x 1) = 120 committees. Step 2 — count the favourable committees. Ana occupies one seat. Ben is barred, so the other 2 seats are filled from the remaining 8 students: C(8,2) = (8 x 7)/2 = 28. Step 3 — P = 28/120 = 7/30. Check by subtraction: committees containing Ana number C(9,2) = 36, and those containing both Ana and Ben number C(8,1) = 8, so Ana-without-Ben = 36 - 8 = 28. Correct. Answer: 7/30.
Trap. Answering 3/10, which is P(Ana is on the committee) and ignores the requirement that Ben be off it. The other slip is C(8,1) instead of C(8,2): once Ana is seated and Ben is excluded, TWO seats remain to be filled, giving 8/120 = 1/15 - which is in fact the probability that both Ana and Ben are on the committee.
Q26 · STA-PRB-042 — Answer: A
Step 1 — count the sample space. Two marbles chosen from 10 give C(10,2) = (10 x 9)/2 = 45 equally likely pairs. Step 2 — count the same-colour pairs one colour at a time: red: C(4,2) = (4 x 3)/2 = 6 blue: C(3,2) = 3 green: C(3,2) = 3 Total same-colour pairs = 6 + 3 + 3 = 12. Step 3 — P(same colour) = 12/45 = 4/15. Step 4 — compare 4/15 with 1/4 over the common denominator 60: 4/15 = 16/60 and 1/4 = 15/60. 16/60 > 15/60, so Column A is greater - but only by 1/60. Check the complement: P(different colours) = 1 - 4/15 = 11/15, and the direct count (4x3 + 4x3 + 3x3)/45 = (12 + 12 + 9)/45 = 33/45 = 11/15. Correct. Answer: Column A is greater.
Trap. Mixing ordered and unordered counting: using 10 x 10 = 100 as the total gives (4x3 + 3x2 + 3x2)/100 = 24/100 = 6/25 = 0.24, which is BELOW 1/4 and points at B. Keep both counts in the same currency - 12 unordered same-colour pairs out of C(10,2) = 45. The other failure is eyeballing 4/15 as 'about a quarter' and answering C; the two differ by exactly 1/60.
Q27 · STA-PRB-048 — Answer: A, B, E
Write n = r + b + g for the total number of balls. A - the three colours are mutually exclusive and cover every ball, so r/n + b/n + g/n = n/n = 1. TRUE. B - P(not green) = 1 - g/n = (r + b)/n = r/n + b/n = P(red) + P(blue). TRUE. C - one half of the biconditional holds: r > b and r > g give 2r > b + g, hence 3r > r + b + g and P(red) > 1/3. The other half fails. Take r = 4, b = 1, g = 4: P(red) = 4/9 > 1/3, yet r is not greater than g. A statement joined by 'if and only if' needs both directions. FALSE. D - the new probability is 2r/(2r + b + g), while double the old one is 2r/(r + b + g). The new denominator is larger by r, which is at least 1, so the new probability is strictly LESS than double. With r = 3, b = 2, g = 1: P moves from 3/6 = 1/2 to 6/9 = 2/3, not to 1. FALSE. E - compare (r + 1)/(n + 1) with r/n. Cross-multiplying gives r(n + 1) = rn + r against n(r + 1) = rn + n, so the contest is r against n. Since b and g are at least 1, n > r, and therefore (r + 1)/(n + 1) > r/n. TRUE. Answer: A, B, E.
Trap. D is the seductive one: adding red balls raises the total as well as the favourable count, so a probability never scales like a raw count. C rewards a student who tests only the direction that works; 'if and only if' demands both, and the reverse direction breaks as soon as red merely ties for the largest group. E is true only because b and g are at least 1 - in a jar of red balls alone P(red) = 1 and another red changes nothing.
Q28 · STA-PRB-056 — Answer: 8/35
Step 1 — the sample space is the 210 equally likely integers, and 210 = 2 x 3 x 5 x 7. Because 210 is a multiple of each prime and of every product of them, each divisibility test cuts the count by an exact factor, with no rounding. Step 2 — remove the multiples of 2: a fraction 1/2 of the integers survive, 210 x 1/2 = 105. Step 3 — of those, a fraction 2/3 are not multiples of 3: 105 x 2/3 = 70. Then 4/5 are not multiples of 5: 70 x 4/5 = 56. Then 6/7 are not multiples of 7: 56 x 6/7 = 48. Step 4 — P = 48/210 = 8/35. Check the long way with inclusion-exclusion on the multiples: singles 105 + 70 + 42 + 30 = 247; pairs 35 + 21 + 15 + 14 + 10 + 6 = 101; triples 7 + 5 + 3 + 2 = 17; all four, 1. So the count divisible by at least one is 247 - 101 + 17 - 1 = 162, leaving 210 - 162 = 48. Correct. Answer: 8/35.
Trap. Adding the four divisibility probabilities and taking the complement: 1 - (1/2 + 1/3 + 1/5 + 1/7) = 1 - 247/210, which is negative. The four events overlap heavily - 6, 10, 14 and so on are counted twice or more - so they cannot simply be added. A second slip is dropping a prime: screening only for 2, 3 and 5 leaves 210 x (1/2)(2/3)(4/5) = 56 integers and the wrong answer 56/210 = 4/15.
Q29 · STA-PRB-058 — Answer: 0.5916
Step 1 — area of the board = 10^2 = 100. Step 2 — each corner region is a QUARTER of a circle of radius 3, so its area is (1/4) x pi x 3^2 = 9pi/4. The four together give 4 x 9pi/4 = 9pi. Step 3 — the central circle has area pi x 2^2 = 4pi. Step 4 — total painted area = 9pi + 4pi = 13pi = 40.8407. Step 5 — P(painted) = 13pi/100 = 0.408407, so P(unpainted) = 1 - 0.408407 = 0.591593, which rounds to 0.5916. Check the geometry: each corner region reaches only 3 units along its two edges, and the centre of the board is sqrt(5^2 + 5^2) = 7.07 units from any corner, more than 3 + 2 = 5, so nothing overlaps and the painted areas may simply be added. Answer: 0.5916.
Trap. Treating the corner regions as whole circles: 4 x 9pi + 4pi = 40pi = 125.7 square units, more paint than the board has area, and 1 - 1.257 comes out negative - a probability outside 0 to 1 is proof of a counting error. Only a quarter of each corner circle sits on the board. Dropping the central circle instead gives 1 - 9pi/100 = 0.7173.
Q30 · STA-PRB-060 — Answer: C
Step 1 — identify the sample space. The random object is a handshake, not a person: every handshake is equally likely to be the first one, and the number of handshakes is C(n,2) = n(n - 1)/2. Step 2 — count the favourable handshakes. Person A shakes hands with each of the other n - 1 people, so n - 1 handshakes involve A. Step 3 — divide: P = (n - 1) / [n(n - 1)/2] = (n - 1) x 2 / [n(n - 1)] = 2/n. The factor (n - 1) cancels, so the answer depends only on n. Check with n = 4 and people A, B, C, D: the handshakes are AB, AC, AD, BC, BD, CD - six in all, of which three involve A, giving 3/6 = 1/2, and 2/n = 2/4 = 1/2. Correct. Answer: 2/n.
Trap. Choice A, 1/n, answers the different question 'pick one of the n people at random'; because a handshake has two participants, A is twice as likely to appear in a random handshake as a random single person is to be A. Choice B, 1/C(n,2), is the probability that one NAMED handshake such as AB comes first. Choice D, 4/n, counts A's handshakes as ordered pairs - AB and BA as two - giving 2(n - 1) favourable outcomes over the unordered C(n,2) denominator; at n = 4 it equals 1, claiming A is certain to be in the first handshake, which flags the mismatch. Choice E, 1/(n - 1), shrinks the sample space to A's own n - 1 handshakes and asks which of THOSE is first, throwing away every handshake that does not involve A.