IB Mathematics: Applications and Interpretation Quiz: Conditional Probability
20 questions · exam conditions
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Conditional ProbabilityQuestion 1 of 20

In a group of tourists, 60% speak English. Among the English speakers, 50% also speak French. Among those who do not speak English, 10% speak French.

What is the probability that a randomly chosen tourist speaks French, given that they do not speak English?

0.04
0.10
0.30
0.34
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Conditional Probability

Practice Conditional Probability 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 Conditional Probability, 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.

All questions

Question 1

In a group of tourists, 60% speak English. Among the English speakers, 50% also speak French. Among those who do not speak English, 10% speak French.

What is the probability that a randomly chosen tourist speaks French, given that they do not speak English?

  1. 0.04
  2. 0.10 (correct answer)
  3. 0.30
  4. 0.34
Explanation: This question tests careful reading of the prompt. Let E be the event that a tourist speaks English, and F be the event that a tourist speaks French. The question provides the information: P(E)=0.60P(E) = 0.60, P(F∣E)=0.50P(F|E) = 0.50, and P(F∣E′)=0.10P(F|E') = 0.10. The question asks for the probability that a tourist speaks French, given they do not speak English, which is exactly P(F∣E′)P(F|E'). The value is given directly in the text as 0.10.

Question 2

At a technology company, 80% of employees use a laptop (L) and 65% use a desktop computer (D). Every employee uses at least one of these devices.

Given that a randomly selected employee uses a laptop, what is the probability they also use a desktop computer?

  1. 0.450
  2. 0.563 (correct answer)
  3. 0.650
  4. 0.813
Explanation: Let L be the event that an employee uses a laptop and D be the event they use a desktop. We are given P(L)=0.80P(L) = 0.80 and P(D)=0.65P(D) = 0.65. Since every employee uses at least one device, P(L∪D)=1P(L \cup D) = 1. We can find the intersection using the addition rule: P(L∪D)=P(L)+P(D)−P(L∩D)P(L \cup D) = P(L) + P(D) - P(L \cap D). So, 1=0.80+0.65−P(L∩D)1 = 0.80 + 0.65 - P(L \cap D), which gives P(L∩D)=1.45−1=0.45P(L \cap D) = 1.45 - 1 = 0.45. We need to find P(D∣L)P(D|L), the probability they use a desktop given they use a laptop. Using the conditional probability formula: P(D∣L)=P(D∩L)P(L)=0.450.80=0.5625P(D|L) = \frac{P(D \cap L)}{P(L)} = \frac{0.45}{0.80} = 0.5625. Rounded to three significant figures, this is 0.563.

Question 3

For two events, A and B, it is known that P(A)=0.4P(A) = 0.4, P(B)=0.6P(B) = 0.6, and P(A∪B)=0.76P(A \cup B) = 0.76.

Which of the following statements about events A and B is true?

  1. A and B are mutually exclusive.
  2. P(A∣B)=P(A)/P(B)P(A|B) = P(A) / P(B)
  3. A and B are dependent.
  4. A and B are independent. (correct answer)
Explanation: When you encounter probability questions involving two events with given probabilities, you're typically being tested on the fundamental relationships between independence, dependence, and mutual exclusivity. To determine the relationship between events A and B, let's first find P(A∩B)P(A \cap B) using 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). Substituting the given values: 0.76=0.4+0.6−P(A∩B)0.76 = 0.4 + 0.6 - P(A \cap B), which gives us P(A∩B)=0.24P(A \cap B) = 0.24. Now we can test for independence. Two events are independent if P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B). Let's check: P(A)×P(B)=0.4×0.6=0.24P(A) \times P(B) = 0.4 \times 0.6 = 0.24. Since this equals our calculated P(A∩B)=0.24P(A \cap B) = 0.24, events A and B are independent, making D correct. Option A is wrong because mutually exclusive events cannot occur simultaneously, meaning P(A∩B)=0P(A \cap B) = 0. Since we found P(A∩B)=0.24≠0P(A \cap B) = 0.24 \neq 0, they're not mutually exclusive. Option B presents an incorrect formula. The correct conditional probability formula is P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}, not P(A)P(B)\frac{P(A)}{P(B)}. Option C is incorrect because we've already established independence. If events are independent, they cannot be dependent (these are mutually exclusive concepts). Remember this key strategy: always calculate P(A∩B)P(A \cap B) first using the addition rule, then test whether P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) to determine independence. This systematic approach will help you avoid common traps on probability questions.

Question 4

In a survey, 40% of participants were male and 60% were female. Of the males, 70% supported a new policy. Of the females, 80% supported the new policy.

Given that a randomly selected participant supports the new policy, what is the probability that the participant is male?

  1. 0.280
  2. 0.368 (correct answer)
  3. 0.480
  4. 0.760
Explanation: Let M be the event the participant is male, F be female, and S be supports the policy. We have P(M)=0.4P(M)=0.4, P(F)=0.6P(F)=0.6, P(S∣M)=0.7P(S|M)=0.7, and P(S∣F)=0.8P(S|F)=0.8. We want to find P(M∣S)P(M|S). First, find the overall probability of supporting the policy, P(S)P(S): P(S)=P(S∣M)P(M)+P(S∣F)P(F)=(0.7)(0.4)+(0.8)(0.6)=0.28+0.48=0.76P(S) = P(S|M)P(M) + P(S|F)P(F) = (0.7)(0.4) + (0.8)(0.6) = 0.28 + 0.48 = 0.76. Now, use the Bayes' theorem formula for P(M∣S)P(M|S): P(M∣S)=P(S∣M)P(M)P(S)=0.280.76≈0.368P(M|S) = \frac{P(S|M)P(M)}{P(S)} = \frac{0.28}{0.76} \approx 0.368.

Question 5

A car rental company has two types of cars: electric and gasoline. 60% of the cars are gasoline. 30% of the electric cars are SUVs. 50% of the gasoline cars are SUVs.

What is the probability that a car is an SUV, given that it is electric?

  1. 0.12
  2. 0.30 (correct answer)
  3. 0.42
  4. 0.50
Explanation: Let E be the event the car is electric, G be the event it is gasoline, and S be the event it is an SUV. We are given P(G)=0.6P(G) = 0.6, which means P(E)=1−0.6=0.4P(E) = 1 - 0.6 = 0.4. We are also given P(S∣E)=0.3P(S|E) = 0.3 and P(S∣G)=0.5P(S|G) = 0.5. The question asks for the probability that a car is an SUV, given that it is electric. This is a direct interpretation of the given information, P(S∣E)=0.30P(S|E) = 0.30.

Question 6

Two events X and Y are such that P(X)=0.5P(X) = 0.5, P(Y)=0.3P(Y) = 0.3, and P(X∣Y)=0.4P(X|Y) = 0.4.

Find the value of P(Y∣X)P(Y|X).

  1. 0.12
  2. 0.24 (correct answer)
  3. 0.40
  4. 0.60
Explanation: First, find P(X∩Y)P(X \cap Y) using the given conditional probability: P(X∣Y)=P(X∩Y)P(Y)P(X|Y) = \frac{P(X \cap Y)}{P(Y)}. Substituting values, 0.4=P(X∩Y)0.30.4 = \frac{P(X \cap Y)}{0.3}, so P(X∩Y)=0.4×0.3=0.12P(X \cap Y) = 0.4 \times 0.3 = 0.12. Now, we can find P(Y∣X)P(Y|X) using its formula: P(Y∣X)=P(X∩Y)P(X)=0.120.5=0.24P(Y|X) = \frac{P(X \cap Y)}{P(X)} = \frac{0.12}{0.5} = 0.24.

Question 7

At a university, 30% of students are in their first year. 25% of all students live on campus. It is also known that 10% of students are first-year students who live on campus.

Are the events 'is a first-year student' and 'lives on campus' independent? Why?

  1. Yes, because 0.30×0.25≠0.100.30 \times 0.25 \neq 0.10.
  2. No, because 0.30+0.25≠0.100.30 + 0.25 \neq 0.10.
  3. Yes, because 0.10/0.30≈0.330.10 / 0.30 \approx 0.33 which is not 0.25.
  4. No, because 0.30×0.25=0.0750.30 \times 0.25 = 0.075 which is not 0.10. (correct answer)
Explanation: Let F be the event that a student is in their first year, and C be the event that a student lives on campus. We are given P(F)=0.30P(F) = 0.30, P(C)=0.25P(C) = 0.25, and P(F∩C)=0.10P(F \cap C) = 0.10. For the events to be independent, the condition P(F∩C)=P(F)×P(C)P(F \cap C) = P(F) \times P(C) must be met. Calculating the product: P(F)×P(C)=0.30×0.25=0.075P(F) \times P(C) = 0.30 \times 0.25 = 0.075. Since 0.075≠0.100.075 \neq 0.10, the events are not independent (they are dependent).

Question 8

In a survey of 300 adults, it was found that 180 people have a pet, 150 people have a garden, and 90 people have both a pet and a garden.

What is the probability that a person has a garden, given that they do not have a pet?

  1. 0.20
  2. 0.50 (correct answer)
  3. 0.60
  4. 0.75
Explanation: Let P be the event of having a pet and G be the event of having a garden. We need to find P(G∣P′)P(G|P'). The number of people who do not have a pet is 300−180=120300 - 180 = 120. The number of people who have a garden but not a pet is the total number with a garden minus those with both: 150−90=60150 - 90 = 60. Therefore, P(G∣P′)=Number with Garden and no PetTotal Number with no Pet=60120=0.50P(G|P') = \frac{\text{Number with Garden and no Pet}}{\text{Total Number with no Pet}} = \frac{60}{120} = 0.50.

Question 9

The probability of rain on any given day in a city is 0.2. If it rains, the probability that the local football team wins their match is 0.4. If it does not rain, the probability that they win is 0.7.

Given that the football team won their match, what is the probability that it rained?

  1. 0.080
  2. 0.125 (correct answer)
  3. 0.400
  4. 0.640
Explanation: Let R be the event that it rains, and W be the event the team wins. We are given P(R)=0.2P(R) = 0.2, P(W∣R)=0.4P(W|R) = 0.4, and P(W∣R′)=0.7P(W|R') = 0.7. We want to find P(R∣W)P(R|W). First, find the total probability of winning, P(W)P(W): P(W)=P(W∣R)P(R)+P(W∣R′)P(R′)=(0.4)(0.2)+(0.7)(1−0.2)=0.08+(0.7)(0.8)=0.08+0.56=0.64P(W) = P(W|R)P(R) + P(W|R')P(R') = (0.4)(0.2) + (0.7)(1-0.2) = 0.08 + (0.7)(0.8) = 0.08 + 0.56 = 0.64. Now, use the formula for P(R∣W)P(R|W): P(R∣W)=P(W∣R)P(R)P(W)=0.080.64=0.125P(R|W) = \frac{P(W|R)P(R)}{P(W)} = \frac{0.08}{0.64} = 0.125.

Question 10

At a technology company, 80% of employees use a laptop (L) and 65% use a desktop computer (D). Every employee uses at least one of these devices.

Given that a randomly selected employee uses a laptop, what is the probability they also use a desktop computer?

  1. 0.450
  2. 0.563 (correct answer)
  3. 0.650
  4. 0.813
Explanation: Let L be the event that an employee uses a laptop and D be the event they use a desktop. We are given P(L)=0.80P(L) = 0.80 and P(D)=0.65P(D) = 0.65. Since every employee uses at least one device, P(L∪D)=1P(L \cup D) = 1. We can find the intersection using the addition rule: P(L∪D)=P(L)+P(D)−P(L∩D)P(L \cup D) = P(L) + P(D) - P(L \cap D). So, 1=0.80+0.65−P(L∩D)1 = 0.80 + 0.65 - P(L \cap D), which gives P(L∩D)=1.45−1=0.45P(L \cap D) = 1.45 - 1 = 0.45. We need to find P(D∣L)P(D|L), the probability they use a desktop given they use a laptop. Using the conditional probability formula: P(D∣L)=P(D∩L)P(L)=0.450.80=0.5625P(D|L) = \frac{P(D \cap L)}{P(L)} = \frac{0.45}{0.80} = 0.5625. Rounded to three significant figures, this is 0.563.

Question 11

In a survey of 300 adults, it was found that 180 people have a pet, 150 people have a garden, and 90 people have both a pet and a garden.

What is the probability that a person has a garden, given that they do not have a pet?

  1. 0.20
  2. 0.50 (correct answer)
  3. 0.60
  4. 0.75
Explanation: Let P be the event of having a pet and G be the event of having a garden. We need to find P(G∣P′)P(G|P'). The number of people who do not have a pet is 300−180=120300 - 180 = 120. The number of people who have a garden but not a pet is the total number with a garden minus those with both: 150−90=60150 - 90 = 60. Therefore, P(G∣P′)=Number with Garden and no PetTotal Number with no Pet=60120=0.50P(G|P') = \frac{\text{Number with Garden and no Pet}}{\text{Total Number with no Pet}} = \frac{60}{120} = 0.50.

Question 12

A factory has two machines, M1 and M2, that produce widgets. M1 produces 60% of the daily output, and M2 produces 40%. The defect rate for M1 is 5%, and the defect rate for M2 is 8%.

A widget is selected at random from the day's production and is found to be defective. What is the probability that it was produced by machine M2?

  1. 0.080
  2. 0.320
  3. 0.516 (correct answer)
  4. 0.620
Explanation: Let D be the event that a widget is defective. We want to find P(M2∣D)P(M2|D). We use Bayes' theorem: P(M2∣D)=P(D∣M2)P(M2)P(D)P(M2|D) = \frac{P(D|M2)P(M2)}{P(D)}. We are given P(M1)=0.6P(M1)=0.6, P(M2)=0.4P(M2)=0.4, P(D∣M1)=0.05P(D|M1)=0.05, and P(D∣M2)=0.08P(D|M2)=0.08. The total probability of a defect is P(D)=P(D∣M1)P(M1)+P(D∣M2)P(M2)=(0.05)(0.6)+(0.08)(0.4)=0.030+0.032=0.062P(D) = P(D|M1)P(M1) + P(D|M2)P(M2) = (0.05)(0.6) + (0.08)(0.4) = 0.030 + 0.032 = 0.062. The numerator is P(D∩M2)=P(D∣M2)P(M2)=(0.08)(0.4)=0.032P(D \cap M2) = P(D|M2)P(M2) = (0.08)(0.4) = 0.032. Therefore, P(M2∣D)=0.0320.062≈0.516P(M2|D) = \frac{0.032}{0.062} \approx 0.516.

Question 13

A bag contains 6 red marbles and 4 blue marbles. Two marbles are drawn from the bag without replacement.

What is the probability that the second marble drawn is red, given that the first marble drawn was also red?

  1. 0.500
  2. 0.556 (correct answer)
  3. 0.600
  4. 0.667
Explanation: Let R1 be the event that the first marble is red, and R2 be the event that the second marble is red. We want to find P(R2∣R1)P(R2|R1). After the first red marble is drawn, there are 9 marbles left in the bag. Of these 9 marbles, 5 are red and 4 are blue. Therefore, the probability of drawing a red marble second, given that the first was red, is 59≈0.556\frac{5}{9} \approx 0.556.

Question 14

In a certain town, 70% of households have broadband internet. Of the households with broadband, 40% also have a streaming TV service. Of the households without broadband, only 10% have a streaming TV service.

What is the probability that a randomly selected household has a streaming TV service?

  1. 0.28
  2. 0.31 (correct answer)
  3. 0.40
  4. 0.50
Explanation: Let B be the event of having broadband and S be the event of having a streaming service. We are given P(B)=0.70P(B) = 0.70, P(S∣B)=0.40P(S|B) = 0.40, and P(S∣B′)=0.10P(S|B') = 0.10. The probability of not having broadband is P(B′)=1−0.70=0.30P(B') = 1 - 0.70 = 0.30. We can find P(S)P(S) using the law of total probability: P(S)=P(S∣B)P(B)+P(S∣B′)P(B′)P(S) = P(S|B)P(B) + P(S|B')P(B'). So, P(S)=(0.40)(0.70)+(0.10)(0.30)=0.28+0.03=0.31P(S) = (0.40)(0.70) + (0.10)(0.30) = 0.28 + 0.03 = 0.31.

Question 15

Two events X and Y are such that P(X)=0.5P(X) = 0.5, P(Y)=0.3P(Y) = 0.3, and P(X∣Y)=0.4P(X|Y) = 0.4.

Find the value of P(Y∣X)P(Y|X).

  1. 0.12
  2. 0.24 (correct answer)
  3. 0.40
  4. 0.60
Explanation: First, find P(X∩Y)P(X \cap Y) using the given conditional probability: P(X∣Y)=P(X∩Y)P(Y)P(X|Y) = \frac{P(X \cap Y)}{P(Y)}. Substituting values, 0.4=P(X∩Y)0.30.4 = \frac{P(X \cap Y)}{0.3}, so P(X∩Y)=0.4×0.3=0.12P(X \cap Y) = 0.4 \times 0.3 = 0.12. Now, we can find P(Y∣X)P(Y|X) using its formula: P(Y∣X)=P(X∩Y)P(X)=0.120.5=0.24P(Y|X) = \frac{P(X \cap Y)}{P(X)} = \frac{0.12}{0.5} = 0.24.

Question 16

The probability of rain on any given day in a city is 0.2. If it rains, the probability that the local football team wins their match is 0.4. If it does not rain, the probability that they win is 0.7.

Given that the football team won their match, what is the probability that it rained?

  1. 0.080
  2. 0.125 (correct answer)
  3. 0.400
  4. 0.640
Explanation: Let R be the event that it rains, and W be the event the team wins. We are given P(R)=0.2P(R) = 0.2, P(W∣R)=0.4P(W|R) = 0.4, and P(W∣R′)=0.7P(W|R') = 0.7. We want to find P(R∣W)P(R|W). First, find the total probability of winning, P(W)P(W): P(W)=P(W∣R)P(R)+P(W∣R′)P(R′)=(0.4)(0.2)+(0.7)(1−0.2)=0.08+(0.7)(0.8)=0.08+0.56=0.64P(W) = P(W|R)P(R) + P(W|R')P(R') = (0.4)(0.2) + (0.7)(1-0.2) = 0.08 + (0.7)(0.8) = 0.08 + 0.56 = 0.64. Now, use the formula for P(R∣W)P(R|W): P(R∣W)=P(W∣R)P(R)P(W)=0.080.64=0.125P(R|W) = \frac{P(W|R)P(R)}{P(W)} = \frac{0.08}{0.64} = 0.125.

Question 17

A quality control inspector checks items for two types of defects, Defect A and Defect B. The probability of an item having Defect A is 0.12. Given that an item has Defect A, the probability it also has Defect B is 0.25.

What is the probability that a randomly selected item has both Defect A and Defect B?

  1. 0.03 (correct answer)
  2. 0.12
  3. 0.25
  4. 0.37
Explanation: Let A be the event of having Defect A and B be the event of having Defect B. We are given P(A)=0.12P(A) = 0.12 and P(B∣A)=0.25P(B|A) = 0.25. We need to find the probability of having both defects, which is P(A∩B)P(A \cap B). We can rearrange the conditional probability formula: P(B∣A)=P(A∩B)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)}. Therefore, P(A∩B)=P(B∣A)×P(A)=0.25×0.12=0.03P(A \cap B) = P(B|A) \times P(A) = 0.25 \times 0.12 = 0.03.

Question 18

Events A and B are independent. The probability of event A occurring is P(A)=0.8P(A) = 0.8 and the probability of event B occurring is P(B)=0.5P(B) = 0.5.

What is the probability of A occurring given that B has occurred, P(A∣B)P(A|B)?

  1. 0.40
  2. 0.50
  3. 0.80 (correct answer)
  4. 0.90
Explanation: By definition, two events A and B are independent if the occurrence of one does not affect the probability of the other. This means P(A∣B)=P(A)P(A|B) = P(A) and P(B∣A)=P(B)P(B|A) = P(B). Since events A and B are stated to be independent and P(A)=0.8P(A) = 0.8, it follows directly that P(A∣B)=0.8P(A|B) = 0.8.

Question 19

For two events C and D, P(C)=0.5P(C)=0.5 and P(C∩D)=0.2P(C \cap D) = 0.2.

Find P(D′∣C)P(D'|C), the probability that event D does not occur, given that event C has occurred.

  1. 0.4
  2. 0.5
  3. 0.6 (correct answer)
  4. 0.8
Explanation: We can first find P(D∣C)P(D|C) using the conditional probability formula: P(D∣C)=P(C∩D)P(C)=0.20.5=0.4P(D|C) = \frac{P(C \cap D)}{P(C)} = \frac{0.2}{0.5} = 0.4. The event D′∣CD'|C is the complement of the event D∣CD|C. Therefore, P(D′∣C)=1−P(D∣C)=1−0.4=0.6P(D'|C) = 1 - P(D|C) = 1 - 0.4 = 0.6.

Question 20

A car rental company has two types of cars: electric and gasoline. 60% of the cars are gasoline. 30% of the electric cars are SUVs. 50% of the gasoline cars are SUVs.

What is the probability that a car is an SUV, given that it is electric?

  1. 0.12
  2. 0.30 (correct answer)
  3. 0.42
  4. 0.50
Explanation: Let E be the event the car is electric, G be the event it is gasoline, and S be the event it is an SUV. We are given P(G)=0.6P(G) = 0.6, which means P(E)=1−0.6=0.4P(E) = 1 - 0.6 = 0.4. We are also given P(S∣E)=0.3P(S|E) = 0.3 and P(S∣G)=0.5P(S|G) = 0.5. The question asks for the probability that a car is an SUV, given that it is electric. This is a direct interpretation of the given information, P(S∣E)=0.30P(S|E) = 0.30.