IB Mathematics: Analysis and Approaches Quiz: Conditional Probability

20 questions · exam conditions

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Conditional ProbabilityQuestion 1 of 20

Which of the following statements is always true for any two events A and B, where $P(B) > 0$ ?

If A and B are independent, then

P(A|B) = P(B|A)

P(A|B) = P(A)

, then

P(B|A) = P(B)

P(A|B) + P(A'|B) = 1

P(A) > P(B)

, then

P(A|B) > P(B|A)

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Conditional Probability

Practice Conditional Probability in IB Mathematics: Analysis and Approaches 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: Analysis and Approaches.

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

Which of the following statements is always true for any two events A and B, where $P(B) > 0$ ?

If A and B are independent, then $P(A|B) = P(B|A)$ .
If $P(A|B) = P(A)$ , then $P(B|A) = P(B)$ .
$P(A|B) + P(A'|B) = 1$ . (correct answer)
If $P(A) > P(B)$ , then $P(A|B) > P(B|A)$ .

Explanation: Let's analyze each option: A. If A and B are independent,

P(A|B) = P(A)

and

P(B|A) = P(B)

. The statement implies

P(A)=P(B)

, which is not always true for independent events. So A is false. B. If

P(A|B) = P(A)

, A and B are independent. This implies

P(B|A) = P(B)

, but only if

P(A)>0

. If

P(A)=0

, then

P(A|B)=0

, so the condition

0=0

holds, but

P(B|A)

is undefined. So B is not always true. C.

P(A|B) + P(A'|B) = \frac{P(A \cap B)}{P(B)} + \frac{P(A' \cap B)}{P(B)} = \frac{P(A \cap B) + P(A' \cap B)}{P(B)}

. The events

A \cap B

and

A' \cap B

are mutually exclusive and their union is B. Thus,

P(A \cap B) + P(A' \cap B) = P(B)

. The expression simplifies to

\frac{P(B)}{P(B)} = 1

. This is always true. D.

P(A|B) > P(B|A)

is equivalent to

\frac{P(A \cap B)}{P(B)} > \frac{P(A \cap B)}{P(A)}

. If

P(A \cap B) > 0

, this simplifies to

P(A) > P(B)

. However, if

P(A \cap B) = 0

, the inequality becomes

0 > 0

, which is false. So D is not always true.

Question 2

A bag contains 3 red and 7 blue marbles. Two marbles are drawn at random without replacement.

Given that the second marble drawn was blue, what is the probability that the first marble was red?

$3/10$
$1/3$ (correct answer)
$7/10$
$7/9$

Explanation: Let R1 be the event 'first marble is red' and B2 be the event 'second marble is blue'. We want to find

P(R1|B2)

. Using the formula for conditional probability,

P(R1|B2) = \frac{P(R1 \cap B2)}{P(B2)}

. First, find the probability of the intersection,

P(R1 \cap B2)

. This is the probability of drawing red then blue:

P(R1 \cap B2) = P(R1) \times P(B2|R1) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90} = \frac{7}{30}

. Next, find the overall probability of the second marble being blue,

P(B2)

, using the law of total probability:

P(B2) = P(B2|R1)P(R1) + P(B2|B1)P(B1)

P(B2) = (\frac{7}{9})(\frac{3}{10}) + (\frac{6}{9})(\frac{7}{10}) = \frac{21}{90} + \frac{42}{90} = \frac{63}{90} = \frac{7}{10}

. Finally,

P(R1|B2) = \frac{7/30}{7/10} = \frac{7}{30} \times \frac{10}{7} = \frac{10}{30} = \frac{1}{3}

. Distractor A is

P(R1)

, ignoring the condition. Distractor D is

P(B2|R1)

, confusing it with

P(R1|B2)

Question 3

In a school of 120 students, 70 students play a sport, 60 students are in the band, and 30 students do both.

A student is selected at random. Given that the student plays a sport, what is the probability they are also in the band?

$1/4$
$3/7$ (correct answer)
$1/2$
$6/7$

Explanation: Let S be the event that a student plays a sport, and B be the event that a student is in the band. We are looking for

P(B|S)

. The formula for conditional probability using counts is

P(B|S) = \frac{n(B \cap S)}{n(S)}

. We are given

n(S) = 70

and

n(B \cap S) = 30

. Therefore,

P(B|S) = \frac{30}{70} = \frac{3}{7}

. Distractor A is

P(B \cap S) = 30/120 = 1/4

. Distractor C is

P(S|B) = n(B \cap S) / n(B) = 30/60 = 1/2

P(B)=60/120=1/2

. Distractor D is

P(B)/P(S) = (60/120)/(70/120) = 6/7

Question 4

A test for a disease is 99% accurate for people who have the disease. It gives a false positive 5% of the time for people who do not have the disease. It is known that 1% of the population has the disease.

A person is selected at random and tests positive. What is the probability, to two significant figures, that this person actually has the disease?

0.0099
0.17 (correct answer)
0.83
0.99

Explanation: Let D be the event that a person has the disease, and T be the event that they test positive. We are given:

P(D) = 0.01

P(T|D) = 0.99

, and

P(T|D') = 0.05

. We want to find

P(D|T)

. Using Bayes' theorem:

P(D|T) = \frac{P(T|D)P(D)}{P(T)}

First, we find

P(T)

using the law of total probability:

P(T) = P(T|D)P(D) + P(T|D')P(D')

P(T) = (0.99)(0.01) + (0.05)(1-0.01) = 0.0099 + (0.05)(0.99) = 0.0099 + 0.0495 = 0.0594

. The numerator is

P(T \cap D) = P(T|D)P(D) = 0.0099

. So,

P(D|T) = \frac{0.0099}{0.0594} = \frac{99}{594} = \frac{1}{6} \approx 0.1666...

. To two significant figures, this is 0.17. Distractor D, 0.99, is

P(T|D)

, a common error of confusing the conditional probabilities. Distractor A is the numerator

P(T \cap D)

Question 5

In a class, 60% of students have a laptop, 70% have a smartphone, and 10% have neither.

What is the probability that a student who has a smartphone also has a laptop?

0.4
$4/7$ (correct answer)
$2/3$
0.9

Explanation: Let L be the event a student has a laptop, and S be the event a student has a smartphone. We are given

P(L)=0.6

P(S)=0.7

, and

P(L' \cap S')=0.1

. We want to find

P(L|S) = \frac{P(L \cap S)}{P(S)}

. First, find

P(L \cup S)

. By De Morgan's laws,

P(L' \cap S') = P((L \cup S)') = 0.1

. So,

P(L \cup S) = 1 - P((L \cup S)') = 1 - 0.1 = 0.9

. Now, find

P(L \cap S)

using the addition rule:

P(L \cap S) = P(L) + P(S) - P(L \cup S) = 0.6 + 0.7 - 0.9 = 0.4

. Finally,

P(L|S) = \frac{0.4}{0.7} = \frac{4}{7}

. Distractor A is

P(L \cap S)

. Distractor C is

P(S|L) = 0.4/0.6 = 2/3

. Distractor D is

P(L \cup S)

Question 6

In a school of 120 students, 70 students play a sport, 60 students are in the band, and 30 students do both.

A student is selected at random. Given that the student plays a sport, what is the probability they are also in the band?

$1/4$
$3/7$ (correct answer)
$1/2$
$6/7$

Explanation: Let S be the event that a student plays a sport, and B be the event that a student is in the band. We are looking for

P(B|S)

. The formula for conditional probability using counts is

P(B|S) = \frac{n(B \cap S)}{n(S)}

. We are given

n(S) = 70

and

n(B \cap S) = 30

. Therefore,

P(B|S) = \frac{30}{70} = \frac{3}{7}

. Distractor A is

P(B \cap S) = 30/120 = 1/4

. Distractor C is

P(S|B) = n(B \cap S) / n(B) = 30/60 = 1/2

P(B)=60/120=1/2

. Distractor D is

P(B)/P(S) = (60/120)/(70/120) = 6/7

Question 7

A person is selected at random and tests positive. What is the probability, to two significant figures, that this person actually has the disease?

0.0099
0.17 (correct answer)
0.83
0.99

Explanation: Let D be the event that a person has the disease, and T be the event that they test positive. We are given:

P(D) = 0.01

P(T|D) = 0.99

, and

P(T|D') = 0.05

. We want to find

P(D|T)

. Using Bayes' theorem:

P(D|T) = \frac{P(T|D)P(D)}{P(T)}

First, we find

P(T)

using the law of total probability:

P(T) = P(T|D)P(D) + P(T|D')P(D')

P(T) = (0.99)(0.01) + (0.05)(1-0.01) = 0.0099 + (0.05)(0.99) = 0.0099 + 0.0495 = 0.0594

. The numerator is

P(T \cap D) = P(T|D)P(D) = 0.0099

. So,

P(D|T) = \frac{0.0099}{0.0594} = \frac{99}{594} = \frac{1}{6} \approx 0.1666...

. To two significant figures, this is 0.17. Distractor D, 0.99, is

P(T|D)

, a common error of confusing the conditional probabilities. Distractor A is the numerator

P(T \cap D)

Question 8

In a class, 60% of students have a laptop, 70% have a smartphone, and 10% have neither.

What is the probability that a student who has a smartphone also has a laptop?

0.4
$4/7$ (correct answer)
$2/3$
0.9

Explanation: Let L be the event a student has a laptop, and S be the event a student has a smartphone. We are given

P(L)=0.6

P(S)=0.7

, and

P(L' \cap S')=0.1

. We want to find

P(L|S) = \frac{P(L \cap S)}{P(S)}

. First, find

P(L \cup S)

. By De Morgan's laws,

P(L' \cap S') = P((L \cup S)') = 0.1

. So,

P(L \cup S) = 1 - P((L \cup S)') = 1 - 0.1 = 0.9

. Now, find

P(L \cap S)

using the addition rule:

P(L \cap S) = P(L) + P(S) - P(L \cup S) = 0.6 + 0.7 - 0.9 = 0.4

. Finally,

P(L|S) = \frac{0.4}{0.7} = \frac{4}{7}

. Distractor A is

P(L \cap S)

. Distractor C is

P(S|L) = 0.4/0.6 = 2/3

. Distractor D is

P(L \cup S)

Question 9

In a group of 40 students, 20 study Chemistry, 25 study Physics, and 5 study neither.

A student is chosen at random. Given that the student studies Physics, what is the probability they also study Chemistry?

$2/5$ (correct answer)
$1/4$
$1/2$
$7/8$

Explanation: When you encounter a question asking for the probability that one event occurs "given that" another has occurred, you're dealing with conditional probability. This requires you to focus only on the subset where the given condition is true. First, let's organize the information using a Venn diagram approach. With 40 total students, 5 study neither subject, so 35 students study at least one subject. If 20 study Chemistry and 25 study Physics, then the total would be 45 students—but we only have 35 studying at least one subject. This means

45 - 35 = 10

students study both subjects. Now we can break down the Physics students: 25 total Physics students, with 10 studying both subjects and 15 studying only Physics. For conditional probability, we use:

P(\text{Chemistry}|\text{Physics}) = \frac{\text{Students studying both}}{\text{Total Physics students}} = \frac{10}{25} = \frac{2}{5}

Looking at the wrong answers: B)

\frac{1}{4}

incorrectly uses 10 out of 40 total students rather than focusing on Physics students only. C)

\frac{1}{2}

might come from using 10 out of 20 Chemistry students, confusing the direction of the conditional probability. D)

\frac{7}{8}

appears to use an incorrect calculation, possibly

\frac{35}{40}

(students studying at least one subject). The correct answer is A)

\frac{2}{5}

. Strategy tip: For conditional probability questions, always restrict your sample space to only those cases where the given condition is met. Draw a Venn diagram when dealing with overlapping groups—it prevents calculation errors and clarifies your thinking.

Question 10

An arithmetic sequence has first term $u_1$ and common difference $d$ . The values of $u_1$ and $d$ are determined by rolling two independent fair six-sided dice.

What is the probability that the third term of the sequence, $u_3$ , is greater than 10, given that the second term, $u_2$ , is greater than 6?

$17/36$
$1/2$
$7/12$
$17/21$ (correct answer)

Explanation: Let A be the event

u_3 > 10

and B be the event

u_2 > 6

. We want

P(A|B) = \frac{n(A \cap B)}{n(B)}

u_2 = u_1 + d > 6

and

u_3 = u_1 + 2d > 10

. The sample space for

(u_1, d)

has

6 \times 6 = 36

outcomes. First, find the number of outcomes in B (

u_1+d > 6

). By listing pairs

(u_1, d)

: (6,1), (5,2), (6,2), (4,3), (5,3), (6,3), (3,4), (4,4), (5,4), (6,4), (2,5), (3,5), (4,5), (5,5), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6). There are

1+2+3+4+5+6 = 21

outcomes. So

n(B)=21

. Next, find the outcomes in

A \cap B

, which are the pairs from the list above that also satisfy

u_1 + 2d > 10

. Checking the list: (5,3), (6,3), (3,4), (4,4), (5,4), (6,4), (2,5), (3,5), (4,5), (5,5), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6). There are

2+4+5+6 = 17

outcomes. So

n(A \cap B) = 17

. The required probability is

P(A|B) = \frac{17}{21}

Question 11

In a group of 40 students, 20 study Chemistry, 25 study Physics, and 5 study neither.

A student is chosen at random. Given that the student studies Physics, what is the probability they also study Chemistry?

$2/5$ (correct answer)
$1/4$
$1/2$
$7/8$

45 - 35 = 10

students study both subjects. Now we can break down the Physics students: 25 total Physics students, with 10 studying both subjects and 15 studying only Physics. For conditional probability, we use:

P(\text{Chemistry}|\text{Physics}) = \frac{\text{Students studying both}}{\text{Total Physics students}} = \frac{10}{25} = \frac{2}{5}

Looking at the wrong answers: B)

\frac{1}{4}

incorrectly uses 10 out of 40 total students rather than focusing on Physics students only. C)

\frac{1}{2}

might come from using 10 out of 20 Chemistry students, confusing the direction of the conditional probability. D)

\frac{7}{8}

appears to use an incorrect calculation, possibly

\frac{35}{40}

(students studying at least one subject). The correct answer is A)

\frac{2}{5}

Question 12

A bag contains 3 red and 7 blue marbles. Two marbles are drawn at random without replacement.

Given that the second marble drawn was blue, what is the probability that the first marble was red?

$3/10$
$1/3$ (correct answer)
$7/10$
$7/9$

Explanation: Let R1 be the event 'first marble is red' and B2 be the event 'second marble is blue'. We want to find

P(R1|B2)

. Using the formula for conditional probability,

P(R1|B2) = \frac{P(R1 \cap B2)}{P(B2)}

. First, find the probability of the intersection,

P(R1 \cap B2)

. This is the probability of drawing red then blue:

P(R1 \cap B2) = P(R1) \times P(B2|R1) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90} = \frac{7}{30}

. Next, find the overall probability of the second marble being blue,

P(B2)

, using the law of total probability:

P(B2) = P(B2|R1)P(R1) + P(B2|B1)P(B1)

P(B2) = (\frac{7}{9})(\frac{3}{10}) + (\frac{6}{9})(\frac{7}{10}) = \frac{21}{90} + \frac{42}{90} = \frac{63}{90} = \frac{7}{10}

. Finally,

P(R1|B2) = \frac{7/30}{7/10} = \frac{7}{30} \times \frac{10}{7} = \frac{10}{30} = \frac{1}{3}

. Distractor A is

P(R1)

, ignoring the condition. Distractor D is

P(B2|R1)

, confusing it with

P(R1|B2)

Question 13

Events E and F are independent, with $P(E) = 1/3$ and $P(F) = 3/4$ . Find $P(E | E \cup F)$ .

$1/4$
$1/3$
$2/5$ (correct answer)
$5/6$

Explanation: We need to calculate

P(E | E \cup F) = \frac{P(E \cap (E \cup F))}{P(E \cup F)}

. Since E is a subset of

E \cup F

, the intersection

E \cap (E \cup F)

is simply E. So,

P(E | E \cup F) = \frac{P(E)}{P(E \cup F)}

. We need to find

P(E \cup F)

. Since E and F are independent,

P(E \cap F) = P(E)P(F) = (1/3)(3/4) = 1/4

. Using the addition rule:

P(E \cup F) = P(E) + P(F) - P(E \cap F) = 1/3 + 3/4 - 1/4 = 1/3 + 2/4 = 1/3 + 1/2 = 5/6

. Finally,

P(E | E \cup F) = \frac{1/3}{5/6} = \frac{1}{3} \times \frac{6}{5} = \frac{6}{15} = \frac{2}{5}

. Distractor B is

P(E)

. This would be the answer for

P(E|F)

, not

P(E|E \cup F)

Question 14

Events A and B are such that $P(A) = 0.4$ and $P(A \cup B) = 0.7$ . Let $P(B) = p$ . For which value of $p$ are events A and B independent?

0.3
0.4
0.5 (correct answer)
0.7

Explanation: For events A and B to be independent, we must have

P(A \cap B) = P(A)P(B)

. First, we express

P(A \cap B)

using the addition rule:

P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + p - 0.7 = p - 0.3

. For independence, we set this equal to

P(A)P(B)

p - 0.3 = (0.4)p

0.6p = 0.3

p = \frac{0.3}{0.6} = 0.5

Distractor A assumes A and B are mutually exclusive, so

P(A \cup B) = P(A) + P(B) \implies 0.7 = 0.4 + p \implies p=0.3

Question 15

An arithmetic sequence has first term $u_1$ and common difference $d$ . The values of $u_1$ and $d$ are determined by rolling two independent fair six-sided dice.

What is the probability that the third term of the sequence, $u_3$ , is greater than 10, given that the second term, $u_2$ , is greater than 6?

$17/36$
$1/2$
$7/12$
$17/21$ (correct answer)

Explanation: Let A be the event

u_3 > 10

and B be the event

u_2 > 6

. We want

P(A|B) = \frac{n(A \cap B)}{n(B)}

u_2 = u_1 + d > 6

and

u_3 = u_1 + 2d > 10

. The sample space for

(u_1, d)

has

6 \times 6 = 36

outcomes. First, find the number of outcomes in B (

u_1+d > 6

). By listing pairs

(u_1, d)

: (6,1), (5,2), (6,2), (4,3), (5,3), (6,3), (3,4), (4,4), (5,4), (6,4), (2,5), (3,5), (4,5), (5,5), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6). There are

1+2+3+4+5+6 = 21

outcomes. So

n(B)=21

. Next, find the outcomes in

A \cap B

, which are the pairs from the list above that also satisfy

u_1 + 2d > 10

. Checking the list: (5,3), (6,3), (3,4), (4,4), (5,4), (6,4), (2,5), (3,5), (4,5), (5,5), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6). There are

2+4+5+6 = 17

outcomes. So

n(A \cap B) = 17

. The required probability is

P(A|B) = \frac{17}{21}

Question 16

Let A and B be events such that $P(A) = 0.6$ , $P(B) = 0.5$ and $P(A \cup B) = 0.9$ . Find $P(A|B)$ .

0.2
$1/3$
0.4 (correct answer)
0.5

Explanation: First, we find

P(A \cap B)

using the formula

P(A \cup B) = P(A) + P(B) - P(A \cap B)

0.9 = 0.6 + 0.5 - P(A \cap B)

P(A \cap B) = 1.1 - 0.9 = 0.2

Now, we use the conditional probability formula:

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4

Question 17

Given that $P(A|B) = 0.8$ , $P(B) = 0.5$ , and $P(A) = 0.6$ . Find $P(B|A)$ .

0.4
0.5
$2/3$ (correct answer)
$5/6$

Explanation: First, find

P(A \cap B)

using the given conditional probability:

P(A \cap B) = P(A|B) \times P(B) = 0.8 \times 0.5 = 0.4

Next, use this to find

P(B|A)

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.6} = \frac{4}{6} = \frac{2}{3}

Distractor A is the value of

P(A \cap B)

. Distractor B is

P(B)

, which would be the answer if A and B were independent. Distractor D is

P(B)/P(A)

Question 18

Let A and B be two events such that A is a subset of B ( $A \subset B$ ). Given $P(A) = 0.2$ and $P(B) = 0.6$ . What is $P(B|A)$ ?

0.2
$1/3$
0.6
1 (correct answer)

Explanation: We want to find

P(B|A) = \frac{P(B \cap A)}{P(A)}

. Since A is a subset of B (

A \subset B

), the intersection of A and B is A itself. That is,

A \cap B = A

. Therefore,

P(B \cap A) = P(A) = 0.2

. Substituting this into the conditional probability formula:

P(B|A) = \frac{P(A)}{P(A)} = \frac{0.2}{0.2} = 1

Conceptually, if event A has occurred, then event B must also have occurred, since every outcome in A is also in B. Thus, the probability is 1. Distractor B is the value of

P(A|B) = P(A \cap B)/P(B) = P(A)/P(B) = 0.2/0.6 = 1/3

Question 19

A fair coin is tossed three times. What is the probability of getting exactly two heads, given that at least one head was obtained?

$3/8$
$3/7$ (correct answer)
$1/2$
$3/4$

Explanation: Let A be the event of getting exactly two heads. The outcomes for A are {HHT, HTH, THH}. There are 3 such outcomes, so

P(A) = 3/8

. Let B be the event of getting at least one head. The only outcome not in B is {TTT}. So there are

8-1=7

outcomes in B, and

P(B) = 7/8

. We want to find

P(A|B) = \frac{P(A \cap B)}{P(B)}

. The event A (exactly two heads) is a subset of event B (at least one head). Therefore,

A \cap B = A

. So,

P(A|B) = \frac{P(A)}{P(B)} = \frac{3/8}{7/8} = \frac{3}{7}

. Distractor A is

P(A)

, ignoring the condition. Distractor D is the probability of exactly two heads given at least two heads:

P(A | \text{at least 2 H}) = (3/8)/(4/8) = 3/4

Question 20

Let A and B be events such that $P(A) = 0.6$ , $P(B) = 0.5$ and $P(A \cup B) = 0.9$ . Find $P(A|B)$ .

0.2
$1/3$
0.4 (correct answer)
0.5

Explanation: First, we find

P(A \cap B)

using the formula

P(A \cup B) = P(A) + P(B) - P(A \cap B)

0.9 = 0.6 + 0.5 - P(A \cap B)

P(A \cap B) = 1.1 - 0.9 = 0.2

Now, we use the conditional probability formula:

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4