IB Mathematics: Applications and Interpretation Quiz: Probability Fundamentals
20 questions · exam conditions
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Probability FundamentalsQuestion 1 of 20

Two fair six-sided dice are rolled. Given that the sum of the numbers is less than 6, what is the probability that at least one die shows a 1?

0.40
0.50
0.60
0.70
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Probability Fundamentals

Practice Probability Fundamentals in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Probability Fundamentals, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Two fair six-sided dice are rolled. Given that the sum of the numbers is less than 6, what is the probability that at least one die shows a 1?

  1. 0.40
  2. 0.50
  3. 0.60
  4. 0.70 (correct answer)
Explanation: Let A be the event that at least one die shows a 1. Let B be the event that the sum is less than 6. We want to find P(A∣B)=n(A∩B)n(B)P(A|B) = \frac{n(A \cap B)}{n(B)}. First, let's list the outcomes for event B (sum < 6): Sum=2: (1,1). Sum=3: (1,2), (2,1). Sum=4: (1,3), (3,1), (2,2). Sum=5: (1,4), (4,1), (2,3), (3,2). Total outcomes for B: n(B)=1+2+3+4=10n(B) = 1+2+3+4 = 10. Now, let's find the outcomes that are in both A and B (sum < 6 AND at least one 1): (1,1), (1,2), (2,1), (1,3), (3,1), (1,4), (4,1). Total outcomes for A∩BA \cap B: n(A∩B)=7n(A \cap B) = 7. Therefore, the probability is P(A∣B)=710=0.70P(A|B) = \frac{7}{10} = 0.70.

Question 2

A company has 200 employees. 120 employees drive to work. Of the employees who drive to work, 75 are in a management position. Of the employees who do not drive to work, 25 are in a management position. An employee is selected at random. What is the probability that the employee is in a management position, given that they do not drive to work?

  1. 0.125
  2. 0.250
  3. 0.313 (correct answer)
  4. 0.500
Explanation: Let M be the event that an employee is in a management position, and D be the event they drive to work. We want to find P(M∣D′)P(M|D'). The number of employees who do not drive to work is n(D′)=200−120=80n(D') = 200 - 120 = 80. The number of employees who are managers and do not drive to work is n(M∩D′)=25n(M \cap D') = 25. The conditional probability is calculated as P(M∣D′)=n(M∩D′)n(D′)=2580=0.3125P(M|D') = \frac{n(M \cap D')}{n(D')} = \frac{25}{80} = 0.3125. So the probability is approximately 0.313.

Question 3

A bag contains 20 balls numbered 1 to 20. A ball is drawn at random. Let A be the event that the number is a multiple of 3, and B be the event that the number is a multiple of 5. Calculate P(A∪B)P(A \cup B).

  1. 0.05
  2. 0.45 (correct answer)
  3. 0.50
  4. 0.55
Explanation: The sample space has 20 outcomes. The multiples of 3 are {3,6,9,12,15,18}\{3, 6, 9, 12, 15, 18\}, so n(A)=6n(A)=6 and P(A)=6/20P(A) = 6/20. The multiples of 5 are {5,10,15,20}\{5, 10, 15, 20\}, so n(B)=4n(B)=4 and P(B)=4/20P(B) = 4/20. The events are not mutually exclusive because 15 is a multiple of both 3 and 5. The intersection is A∩B={15}A \cap B = \{15\}, so P(A∩B)=1/20P(A \cap B) = 1/20. Using the addition rule: P(A∪B)=P(A)+P(B)−P(A∩B)=620+420−120=920=0.45P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{6}{20} + \frac{4}{20} - \frac{1}{20} = \frac{9}{20} = 0.45.

Question 4

For students at a university, let H be the event that a student is taking a humanities course and S be the event that a student is taking a science course. It is known that P(H)=0.6P(H) = 0.6, P(S)=0.5P(S) = 0.5, and P(S∣H)=0.4P(S|H) = 0.4. Find the probability that a randomly selected student is taking a humanities course but not a science course.

  1. 0.24
  2. 0.26
  3. 0.30
  4. 0.36 (correct answer)
Explanation: We need to find P(H∩S′)P(H \cap S'). The formula for this is P(H)−P(H∩S)P(H) - P(H \cap S). First, we must find P(H∩S)P(H \cap S) using the conditional probability formula: P(S∣H)=P(S∩H)P(H)P(S|H) = \frac{P(S \cap H)}{P(H)}. Rearranging gives P(S∩H)=P(S∣H)×P(H)=0.4×0.6=0.24P(S \cap H) = P(S|H) \times P(H) = 0.4 \times 0.6 = 0.24. Now we can find the required probability: P(H∩S′)=P(H)−P(H∩S)=0.6−0.24=0.36P(H \cap S') = P(H) - P(H \cap S) = 0.6 - 0.24 = 0.36.

Question 5

At a coffee shop, the probability that a customer buys a coffee is 0.75, the probability that they buy a pastry is 0.40, and the probability that they buy both is 0.25. What is the probability that a randomly selected customer buys neither a coffee nor a pastry?

  1. 0.10 (correct answer)
  2. 0.15
  3. 0.75
  4. 0.90
Explanation: Let C be the event of buying a coffee and P be the event of buying a pastry. We are given P(C)=0.75P(C) = 0.75, P(P)=0.40P(P) = 0.40, and P(C∩P)=0.25P(C \cap P) = 0.25. We want to find the probability of buying neither, which is P(C′∩P′)P(C' \cap P'). Using De Morgan's laws, this is equivalent to P((C∪P)′)P((C \cup P)'), which is 1−P(C∪P)1 - P(C \cup P). First, we find P(C∪P)P(C \cup P) using the addition rule: P(C∪P)=P(C)+P(P)−P(C∩P)=0.75+0.40−0.25=0.90P(C \cup P) = P(C) + P(P) - P(C \cap P) = 0.75 + 0.40 - 0.25 = 0.90. Therefore, the probability of buying neither is 1−0.90=0.101 - 0.90 = 0.10.

Question 6

Two production lines, A and B, produce microchips. The probability that a chip from line A is defective is 0.04. The probability that a chip from line B is defective is 0.07. The two events are independent. If one chip is selected from each line, what is the probability that at least one of the chips is not defective?

  1. 0.0028
  2. 0.8900
  3. 0.8928
  4. 0.9972 (correct answer)
Explanation: The event 'at least one chip is not defective' is the complement of the event 'both chips are defective'. Let DAD_A be the event a chip from A is defective, and DBD_B be the event a chip from B is defective. We are given P(DA)=0.04P(D_A) = 0.04 and P(DB)=0.07P(D_B) = 0.07. Since the events are independent, the probability that both are defective is P(DA∩DB)=P(DA)×P(DB)=0.04×0.07=0.0028P(D_A \cap D_B) = P(D_A) \times P(D_B) = 0.04 \times 0.07 = 0.0028. The probability of the complement event (at least one is not defective) is 1−0.0028=0.99721 - 0.0028 = 0.9972.

Question 7

A box contains 5 red balls and 3 green balls. A ball is drawn and its color is noted. It is not replaced. A second ball is then drawn. Given that the second ball drawn is red, what is the probability that the first ball drawn was also red?

  1. 0.500
  2. 0.571 (correct answer)
  3. 0.625
  4. 0.714
Explanation: Let R1 be the event the first ball is red, and R2 be the event the second ball is red. We want to find P(R1∣R2)P(R1|R2). Using the formula for conditional probability, P(R1∣R2)=P(R1∩R2)P(R2)P(R1|R2) = \frac{P(R1 \cap R2)}{P(R2)}. The numerator is the probability of drawing two red balls in a row: P(R1∩R2)=58×47=2056P(R1 \cap R2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}. The denominator, P(R2)P(R2), can be found using the law of total probability: P(R2)=P(R2∣R1)P(R1)+P(R2∣G1)P(G1)=(47)(58)+(57)(38)=2056+1556=3556P(R2) = P(R2|R1)P(R1) + P(R2|G1)P(G1) = (\frac{4}{7})(\frac{5}{8}) + (\frac{5}{7})(\frac{3}{8}) = \frac{20}{56} + \frac{15}{56} = \frac{35}{56}. Therefore, P(R1∣R2)=20/5635/56=2035=47≈0.571P(R1|R2) = \frac{20/56}{35/56} = \frac{20}{35} = \frac{4}{7} \approx 0.571.

Question 8

At a library, 70% of checked-out items are books (B) and 30% are other media (B'). Of the books checked out, 40% are fiction (F). Of the other media, 10% are categorized as fiction (F). Given that a randomly selected item is categorized as fiction, what is the probability that it is not a book?

  1. 0.030
  2. 0.097 (correct answer)
  3. 0.100
  4. 0.310
Explanation: We are looking for P(B′∣F)P(B'|F). Using the conditional probability formula, P(B′∣F)=P(B′∩F)P(F)P(B'|F) = \frac{P(B' \cap F)}{P(F)}. We can find the terms using a tree diagram. P(B′∩F)=P(F∣B′)×P(B′)=0.10×0.30=0.03P(B' \cap F) = P(F|B') \times P(B') = 0.10 \times 0.30 = 0.03. To find P(F)P(F), we need to consider both paths that lead to a fiction item: P(F)=P(F∩B)+P(F∩B′)=(P(F∣B)×P(B))+0.03=(0.40×0.70)+0.03=0.28+0.03=0.31P(F) = P(F \cap B) + P(F \cap B') = (P(F|B) \times P(B)) + 0.03 = (0.40 \times 0.70) + 0.03 = 0.28 + 0.03 = 0.31. Therefore, P(B′∣F)=0.030.31≈0.09677P(B'|F) = \frac{0.03}{0.31} \approx 0.09677, which is 0.097 to three decimal places.

Question 9

A player plays a game where they first flip a biased coin with P(Heads)=0.4P(\text{Heads})=0.4. If it's heads, they roll a standard six-sided die. If it's tails, they roll a four-sided die with faces numbered 1, 2, 3, 4. The player wins if they roll a number greater than 3. What is the probability that the player wins the game?

  1. 0.125
  2. 0.350 (correct answer)
  3. 0.375
  4. 0.750
Explanation: We can model this with a tree diagram. There are two paths to winning. Path 1: Flip Heads AND roll > 3 on a 6-sided die. The numbers greater than 3 are {4,5,6}\{4, 5, 6\}, so the probability is 3/6=0.53/6 = 0.5. The probability of this path is 0.4×0.5=0.200.4 \times 0.5 = 0.20. Path 2: Flip Tails AND roll > 3 on a 4-sided die. The probability of tails is 1−0.4=0.61-0.4=0.6. The only number greater than 3 is {4}\{4\}, so the probability is 1/4=0.251/4 = 0.25. The probability of this path is 0.6×0.25=0.150.6 \times 0.25 = 0.15. The total probability of winning is the sum of the probabilities of these two paths: 0.20+0.15=0.350.20 + 0.15 = 0.35.

Question 10

A company has 200 employees. 120 employees drive to work. Of the employees who drive to work, 75 are in a management position. Of the employees who do not drive to work, 25 are in a management position. An employee is selected at random. What is the probability that the employee is in a management position, given that they do not drive to work?

  1. 0.125
  2. 0.250
  3. 0.313 (correct answer)
  4. 0.500
Explanation: Let M be the event that an employee is in a management position, and D be the event they drive to work. We want to find P(M∣D′)P(M|D'). The number of employees who do not drive to work is n(D′)=200−120=80n(D') = 200 - 120 = 80. The number of employees who are managers and do not drive to work is n(M∩D′)=25n(M \cap D') = 25. The conditional probability is calculated as P(M∣D′)=n(M∩D′)n(D′)=2580=0.3125P(M|D') = \frac{n(M \cap D')}{n(D')} = \frac{25}{80} = 0.3125. So the probability is approximately 0.313.

Question 11

A box contains 8 dark chocolates and 4 milk chocolates. Sarah picks one chocolate at random, eats it, and then picks a second one. What is the probability that she picks two different types of chocolate?

  1. 0.242
  2. 0.444
  3. 0.485 (correct answer)
  4. 0.515
Explanation: There are two ways to pick two different types: (Dark then Milk) or (Milk then Dark). The total number of chocolates is 12. P(Dark then Milk)=812×411=32132P(\text{Dark then Milk}) = \frac{8}{12} \times \frac{4}{11} = \frac{32}{132}. P(Milk then Dark)=412×811=32132P(\text{Milk then Dark}) = \frac{4}{12} \times \frac{8}{11} = \frac{32}{132}. The total probability is the sum of these two probabilities: 32132+32132=64132≈0.485\frac{32}{132} + \frac{32}{132} = \frac{64}{132} \approx 0.485.

Question 12

A player plays a game where they first flip a biased coin with P(Heads)=0.4P(\text{Heads})=0.4. If it's heads, they roll a standard six-sided die. If it's tails, they roll a four-sided die with faces numbered 1, 2, 3, 4. The player wins if they roll a number greater than 3. What is the probability that the player wins the game?

  1. 0.125
  2. 0.350 (correct answer)
  3. 0.375
  4. 0.750
Explanation: We can model this with a tree diagram. There are two paths to winning. Path 1: Flip Heads AND roll > 3 on a 6-sided die. The numbers greater than 3 are {4,5,6}\{4, 5, 6\}, so the probability is 3/6=0.53/6 = 0.5. The probability of this path is 0.4×0.5=0.200.4 \times 0.5 = 0.20. Path 2: Flip Tails AND roll > 3 on a 4-sided die. The probability of tails is 1−0.4=0.61-0.4=0.6. The only number greater than 3 is {4}\{4\}, so the probability is 1/4=0.251/4 = 0.25. The probability of this path is 0.6×0.25=0.150.6 \times 0.25 = 0.15. The total probability of winning is the sum of the probabilities of these two paths: 0.20+0.15=0.350.20 + 0.15 = 0.35.

Question 13

In a survey of 80 students, 30 students said they play a musical instrument (M), and 45 said they play a team sport (S). 15 students said they do both. How many students do exactly one of these activities?

  1. 20
  2. 45 (correct answer)
  3. 60
  4. 75
Explanation: We want to find the number of students who play an instrument but not a sport, plus the number of students who play a sport but not an instrument. Number who play only a musical instrument = n(M)−n(M∩S)=30−15=15n(M) - n(M \cap S) = 30 - 15 = 15. Number who play only a team sport = n(S)−n(M∩S)=45−15=30n(S) - n(M \cap S) = 45 - 15 = 30. The total number of students who do exactly one activity is 15+30=4515 + 30 = 45.

Question 14

The probability of snow on a given day is 0.2. If it snows, the probability a bus is late is 0.6. If it does not snow, the probability a bus is late is 0.1. What is the probability that on a randomly chosen day, it snows and the bus is not late?

  1. 0.08 (correct answer)
  2. 0.12
  3. 0.20
  4. 0.72
Explanation: Let S be the event that it snows, and L be the event that the bus is late. We want to find P(S∩L′)P(S \cap L'). We are given P(S)=0.2P(S) = 0.2 and P(L∣S)=0.6P(L|S) = 0.6. The probability that the bus is not late, given that it snows, is P(L′∣S)=1−P(L∣S)=1−0.6=0.4P(L'|S) = 1 - P(L|S) = 1 - 0.6 = 0.4. The probability of the intersection is P(S∩L′)=P(L′∣S)×P(S)=0.4×0.2=0.08P(S \cap L') = P(L'|S) \times P(S) = 0.4 \times 0.2 = 0.08.

Question 15

A manufacturer finds that 15% of its products have a cosmetic flaw, 8% have a functional flaw, and 3% have both. A product is selected at random. What is the probability that it has a functional flaw but not a cosmetic flaw?

  1. 0.03
  2. 0.05 (correct answer)
  3. 0.12
  4. 0.20
Explanation: Let C be the event that a product has a cosmetic flaw, and F be the event that it has a functional flaw. We are given P(C)=0.15P(C) = 0.15, P(F)=0.08P(F) = 0.08, and P(C∩F)=0.03P(C \cap F) = 0.03. We want to find the probability of a functional flaw but not a cosmetic flaw, which is P(F∩C′)P(F \cap C'). This is calculated as P(F)−P(F∩C)P(F) - P(F \cap C). So, P(F∩C′)=0.08−0.03=0.05P(F \cap C') = 0.08 - 0.03 = 0.05.

Question 16

A box contains 8 dark chocolates and 4 milk chocolates. Sarah picks one chocolate at random, eats it, and then picks a second one. What is the probability that she picks two different types of chocolate?

  1. 0.242
  2. 0.444
  3. 0.485 (correct answer)
  4. 0.515
Explanation: There are two ways to pick two different types: (Dark then Milk) or (Milk then Dark). The total number of chocolates is 12. P(Dark then Milk)=812×411=32132P(\text{Dark then Milk}) = \frac{8}{12} \times \frac{4}{11} = \frac{32}{132}. P(Milk then Dark)=412×811=32132P(\text{Milk then Dark}) = \frac{4}{12} \times \frac{8}{11} = \frac{32}{132}. The total probability is the sum of these two probabilities: 32132+32132=64132≈0.485\frac{32}{132} + \frac{32}{132} = \frac{64}{132} \approx 0.485.

Question 17

The probability of rain on Monday is 0.3. The probability of rain on Tuesday is 0.6. The probability of rain on at least one of these two days is 0.72. Let M be the event that it rains on Monday and T be the event that it rains on Tuesday. Which statement is correct?

  1. The events M and T are independent. (correct answer)
  2. The events M and T are not independent.
  3. The events M and T are mutually exclusive.
  4. P(M∣T)=0.25P(M|T) = 0.25.
Explanation: When you encounter probability questions involving two events and phrases like "at least one," you're typically dealing with concepts of independence and the addition rule for probability. To determine if events M and T are independent, you need to check if P(M∩T)=P(M)×P(T)P(M \cap T) = P(M) \times P(T). First, find P(M∩T)P(M \cap T) using the addition rule: P(M∪T)=P(M)+P(T)−P(M∩T)P(M \cup T) = P(M) + P(T) - P(M \cap T). Substituting the given values: 0.72=0.3+0.6−P(M∩T)0.72 = 0.3 + 0.6 - P(M \cap T), so P(M∩T)=0.18P(M \cap T) = 0.18. Now test for independence: P(M)×P(T)=0.3×0.6=0.18P(M) \times P(T) = 0.3 \times 0.6 = 0.18. Since P(M∩T)=0.18P(M \cap T) = 0.18 equals P(M)×P(T)=0.18P(M) \times P(T) = 0.18, the events are independent. Option B is incorrect because we just proved the events are independent. Option C is wrong because mutually exclusive events cannot occur simultaneously (P(M∩T)=0P(M \cap T) = 0), but here P(M∩T)=0.18≠0P(M \cap T) = 0.18 \neq 0. Option D is incorrect because P(M∣T)=P(M∩T)P(T)=0.180.6=0.3P(M|T) = \frac{P(M \cap T)}{P(T)} = \frac{0.18}{0.6} = 0.3, not 0.25. Therefore, A is correct. Study tip: Always use the addition rule P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) to find the intersection probability first, then test independence with P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B). Remember that mutually exclusive events have zero intersection probability and cannot be independent unless one has zero probability.

Question 18

In a certain population, 2% of people have a particular disease. A test for this disease is 95% accurate for people who have it (true positive) and 90% accurate for people who do not have it (true negative). A person is selected at random and tests positive. What is the probability that this person actually has the disease, correct to three significant figures?

  1. 0.019
  2. 0.117
  3. 0.162 (correct answer)
  4. 0.950
Explanation: Let D be the event of having the disease and + be the event of testing positive. We want to find P(D∣+)P(D|+). We can use a tree diagram or conditional probability formula. P(D)=0.02P(D) = 0.02, P(D′)=0.98P(D')=0.98, P(+∣D)=0.95P(+|D)=0.95, and P(−∣D′)=0.90P(-|D')=0.90, which implies P(+∣D′)=1−0.90=0.10P(+|D')=1-0.90=0.10. The probability of testing positive is P(+)=P(+∣D)P(D)+P(+∣D′)P(D′)=(0.95)(0.02)+(0.10)(0.98)=0.019+0.098=0.117P(+) = P(+|D)P(D) + P(+|D')P(D') = (0.95)(0.02) + (0.10)(0.98) = 0.019 + 0.098 = 0.117. The probability of having the disease and testing positive is P(D∩+)=P(+∣D)P(D)=0.019P(D \cap +) = P(+|D)P(D) = 0.019. Thus, P(D∣+)=P(D∩+)P(+)=0.0190.117≈0.16239P(D|+) = \frac{P(D \cap +)}{P(+)} = \frac{0.019}{0.117} \approx 0.16239, which is 0.162 to 3 s.f.

Question 19

In a sports center with 100 members, 40 play Tennis (T), 55 play Squash (S), and 35 play Badminton (B). 15 play Tennis and Squash, 12 play Squash and Badminton, 10 play Tennis and Badminton, and 7 play all three. What is the probability that a randomly selected member plays exactly one sport?

  1. 0.16
  2. 0.56
  3. 0.77 (correct answer)
  4. 0.93
Explanation: We can use a Venn diagram to solve this. First find the number of members in each region. n(all three)=7n(\text{all three}) = 7. n(T and S only)=15−7=8n(\text{T and S only}) = 15 - 7 = 8. n(S and B only)=12−7=5n(\text{S and B only}) = 12 - 7 = 5. n(T and B only)=10−7=3n(\text{T and B only}) = 10 - 7 = 3. Now find the number who play exactly one sport: n(T only)=40−8−3−7=22n(\text{T only}) = 40 - 8 - 3 - 7 = 22. n(S only)=55−8−5−7=35n(\text{S only}) = 55 - 8 - 5 - 7 = 35. n(B only)=35−3−5−7=20n(\text{B only}) = 35 - 3 - 5 - 7 = 20. The total number of members playing exactly one sport is 22+35+20=7722 + 35 + 20 = 77. The probability is 77100=0.77\frac{77}{100} = 0.77.

Question 20

A security system has two independent alarms, A and B. The probability that alarm A fails is 0.15. The probability that alarm B fails is 0.05. What is the probability that at least one alarm functions correctly?

  1. 0.0075
  2. 0.8075
  3. 0.9925 (correct answer)
  4. 1.0000
Explanation: The event 'at least one alarm functions correctly' is the complement of the event 'both alarms fail'. Let F_A be the event A fails and F_B be the event B fails. We are given P(FA)=0.15P(F_A) = 0.15 and P(FB)=0.05P(F_B) = 0.05. Since the alarms are independent, the probability that both fail is P(FA∩FB)=P(FA)×P(FB)=0.15×0.05=0.0075P(F_A \cap F_B) = P(F_A) \times P(F_B) = 0.15 \times 0.05 = 0.0075. The probability that at least one functions is 1−P(both fail)=1−0.0075=0.99251 - P(\text{both fail}) = 1 - 0.0075 = 0.9925.