Conditional Probability - IB Mathematics Analysis and Approaches

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 $P(B|A)$ ?
Answer: 1 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 $P(A|B) = P(A \cap B)/P(B) = P(A)/P(B) = 0.2/0.6 = 1/3$ .
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?
Answer: $17/21$ 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 $P(A|B) = \frac{17}{21}$ .
What is the probability that a student who has a smartphone also has a laptop?
Answer: $4/7$ 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 $P(L \cup S)$ .
Given $P(A|B) = 2/3$ , $P(B|A) = 1/2$ , and $P(A \cup B) = 3/4$ . Find $P(A \cap B)$ .
Answer: $3/10$ When you encounter conditional probability problems with multiple given relationships, you need to use the fundamental formulas strategically to find connections between the known and unknown quantities. Start with the conditional probability formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ and $P(B|A) = \frac{P(A \cap B)}{P(A)}$ . From these, you can express $P(A \cap B) = P(A|B) \cdot P(B) = \frac{2}{3} \cdot P(B)$ and $P(A \cap B) = P(B|A) \cdot P(A) = \frac{1}{2} \cdot P(A)$ . Since both expressions equal $P(A \cap B)$ , you have $\frac{2}{3}P(B) = \frac{1}{2}P(A)$ , which gives you $P(A) = \frac{4}{3}P(B)$ . Now use the union formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{4}$ . Substituting your relationships: $\frac{4}{3}P(B) + P(B) - \frac{2}{3}P(B) = \frac{3}{4}$ . This simplifies to $\frac{5}{3}P(B) = \frac{3}{4}$ , so $P(B) = \frac{9}{20}$ . Therefore, $P(A \cap B) = \frac{2}{3} \cdot \frac{9}{20} = \frac{3}{10}$ , which is answer A. Answer B (3/14) likely comes from arithmetic errors in the fraction operations. Answer C (9/20) is actually P(B), not the intersection. Answer D (3/5) might result from incorrectly applying the union formula. Remember: in conditional probability problems, set up equations using the definition, then use algebraic substitution to solve the system. Always double-check that your final answer makes sense given the constraints.
A person is selected at random and tests positive. What is the probability, to two significant figures, that this person actually has the disease?
Answer: 0.17 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 $P(T \cap D)$ .
Let A and B be events such that $P(A) = 0.6$ and $P(B) = 0.3$ . If $P(A|B) = 0.4$ , find $P(A|B')$ .
Answer: $24/35$ We want to find $P(A|B') = \frac{P(A \cap B')}{P(B')}$ . First, find $P(B') = 1 - P(B) = 1 - 0.3 = 0.7$ . Next, find $P(A \cap B)$ from the given information: $P(A \cap B) = P(A|B) \times P(B) = 0.4 \times 0.3 = 0.12$ . Now find $P(A \cap B')$ . Using a Venn diagram or formula, $P(A \cap B') = P(A) - P(A \cap B) = 0.6 - 0.12 = 0.48$ . Finally, calculate the conditional probability: $P(A|B') = \frac{0.48}{0.7} = \frac{48}{70} = \frac{24}{35}$ Distractor B is $1 - P(A|B) = 1 - 0.4 = 0.6$ , which incorrectly assumes $P(A|B) + P(A|B') = 1$ . Distractor D is the value of $P(A \cap B')$ . Distractor C is from incorrectly calculating $P(A)/P(B')$ .
Given $P(A) = 0.5$ , $P(B) = 0.4$ and $P(A \cap B) = 0.1$ . Find $P(A'|B)$ .
Answer: 0.75 There are two ways to solve this. Method 1: Use the formula $P(A'|B) = 1 - P(A|B)$ . First find $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.4} = 0.25$ . Then $P(A'|B) = 1 - 0.25 = 0.75$ . Method 2: Use the direct formula $P(A'|B) = \frac{P(A' \cap B)}{P(B)}$ . Find $P(A' \cap B) = P(B) - P(A \cap B) = 0.4 - 0.1 = 0.3$ . Then $P(A'|B) = \frac{0.3}{0.4} = \frac{3}{4} = 0.75$ . Distractor B is $P(A') = 1-0.5=0.5$ , which ignores the condition. Distractor A is $P(A' \cap B)$ . Distractor C is $P(B|A') = P(A' \cap B)/P(A') = 0.3/0.5 = 0.6$ .
Events A and B are mutually exclusive with $P(A) = 0.3$ and $P(B) = 0.5$ . What is $P(A|B)$ ?
Answer: 0 If events A and B are mutually exclusive, it means they cannot occur at the same time. Therefore, the probability of their intersection is zero: $P(A \cap B) = 0$ . The formula for conditional probability is $P(A|B) = \frac{P(A \cap B)}{P(B)}$ . Substituting the values, we get $P(A|B) = \frac{0}{0.5} = 0$ . Conceptually, if we know event B has occurred, it is impossible for event A to have occurred. Distractor B is $P(A)$ . Students might confuse mutual exclusivity with independence, for which $P(A|B) = P(A)$ .
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?
Answer: 0.5 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 $P(A \cup B) = P(A) + P(B) \implies 0.7 = 0.4 + p \implies p=0.3$ .
Events E and F are independent, with $P(E) = 1/3$ and $P(F) = 3/4$ . Find $P(E | E \cup F)$ .
Answer: $2/5$ 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 $P(E|E \cup F)$ .
Given that $P(A|B) = 0.8$ , $P(B) = 0.5$ , and $P(A) = 0.6$ . Find $P(B|A)$ .
Answer: $2/3$ 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 $P(B)/P(A)$ .
Given that $P(A|B) = 0.5$ , $P(B|A) = 0.25$ , and $P(A \cap B) = 0.1$ . Find $P(A \cup B)$ .
Answer: 0.5 First, we must find $P(A)$ and $P(B)$ . From $P(A|B) = \frac{P(A \cap B)}{P(B)}$ , we have $0.5 = \frac{0.1}{P(B)}$ , which gives $P(B) = \frac{0.1}{0.5} = 0.2$ . From $P(B|A) = \frac{P(A \cap B)}{P(A)}$ , we have $0.25 = \frac{0.1}{P(A)}$ , which gives $P(A) = \frac{0.1}{0.25} = 0.4$ . Now we use the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $P(A \cup B) = 0.4 + 0.2 - 0.1 = 0.5$ . Distractor C is $P(A) + P(B) = 0.6$ , which omits the subtraction of the intersection. Distractor A is $P(A)$ .
A student is selected at random. Given that the student plays a sport, what is the probability they are also in the band?
Answer: $3/7$ 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 $P(S|B) = n(B \cap S) / n(B) = 30/60 = 1/2$ or $P(B)=60/120=1/2$ . Distractor D is $P(B)/P(S) = (60/120)/(70/120) = 6/7$ .
A fair coin is tossed three times. What is the probability of getting exactly two heads, given that at least one head was obtained?
Answer: $3/7$ 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: $P(A | \text{at least 2 H}) = (3/8)/(4/8) = 3/4$ .
Two fair six-sided dice are rolled. What is the probability that the sum of the numbers is 7, given that the sum is odd or the sum is greater than 8?
Answer: $3/11$ Let A be the event 'sum is 7'. Let E be the event 'sum is odd'. Let F be the event 'sum is greater than 8'. We want to find $P(A | E \cup F)$ . Total outcomes = 36. Event A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, so $n(A)=6$ . Event E (odd sum): sum 3, 5, 7, 9, 11. $n(E) = 2+4+6+4+2 = 18$ . Event F (sum > 8): sum 9, 10, 11, 12. $n(F) = 4+3+2+1 = 10$ . The conditioning event is $E \cup F$ . $n(E \cup F) = n(E) + n(F) - n(E \cap F)$ . $E \cap F$ is 'sum is odd and > 8', which means sum is 9 or 11. $n(E \cap F) = 4+2=6$ . So, $n(E \cup F) = 18 + 10 - 6 = 22$ . We need $P(A | E \cup F) = \frac{n(A \cap (E \cup F))}{n(E \cup F)}$ . The event A ('sum is 7') is a subset of E ('sum is odd'), so A is also a subset of $E \cup F$ . Therefore $A \cap (E \cup F) = A$ , and $n(A \cap (E \cup F)) = n(A) = 6$ . The probability is $\frac{6}{22} = \frac{3}{11}$ . Distractor A is $P(A)=6/36$ . Distractor C is $P(A|E) = n(A)/n(E) = 6/18 = 1/3$ .
Let A and B be events such that $P(A) = 0.6$ , $P(B) = 0.5$ and $P(A \cup B) = 0.9$ . Find $P(A|B)$ .
Answer: 0.4 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: $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4$
Which of the following statements is always true for any two events A and B, where $P(B) > 0$ ?
Answer: $P(A|B) + P(A'|B) = 1$ . 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.
A student is chosen at random. Given that the student studies Physics, what is the probability they also study Chemistry?
Answer: $2/5$ 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.
Given that the second marble drawn was blue, what is the probability that the first marble was red?
Answer: $1/3$ 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 $P(R1|B2)$ .
For two events C and D, $P(C) = 2/3$ and $P(D') = 1/4$ . Given that $P(C \cup D) = 5/6$ , find $P(C|D)$ .
Answer: $7/9$ First, find $P(D)$ from its complement: $P(D) = 1 - P(D') = 1 - 1/4 = 3/4$ . Next, find $P(C \cap D)$ using the addition rule: $P(C \cap D) = P(C) + P(D) - P(C \cup D)$ . $P(C \cap D) = \frac{2}{3} + \frac{3}{4} - \frac{5}{6} = \frac{8}{12} + \frac{9}{12} - \frac{10}{12} = \frac{7}{12}$ Finally, calculate $P(C|D)$ : $P(C|D) = \frac{P(C \cap D)}{P(D)} = \frac{7/12}{3/4} = \frac{7}{12} \times \frac{4}{3} = \frac{28}{36} = \frac{7}{9}$ Distractor A is $P(C \cap D)$ . Distractor B is $P(C)$ , which would be correct if C and D were independent. Distractor D is $P(D|C) = (7/12)/(2/3) = 7/8$ .

Start this quiz