Conditional Probability - IB Mathematics Applications and Interpretation

1. What is the probability that a randomly chosen tourist speaks French, given that they do not speak English?
   Answer: 0.10 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: $P(E) = 0.60$, $P(F|E) = 0.50$, and $P(F|E') = 0.10$. The question asks for the probability that a tourist speaks French, given they do not speak English, which is exactly $P(F|E')$. The value is given directly in the text as 0.10.
2. Given that a randomly selected participant supports the new policy, what is the probability that the participant is male?
   Answer: 0.368 Let M be the event the participant is male, F be female, and S be supports the policy. We have $P(M)=0.4$, $P(F)=0.6$, $P(S|M)=0.7$, and $P(S|F)=0.8$. We want to find $P(M|S)$. First, find the overall probability of supporting the policy, $P(S)$: $P(S) = P(S|M)P(M) + P(S|F)P(F) = (0.7)(0.4) + (0.8)(0.6) = 0.28 + 0.48 = 0.76$. Now, use the Bayes' theorem formula for $P(M|S)$: $P(M|S) = \frac{P(S|M)P(M)}{P(S)} = \frac{0.28}{0.76} \approx 0.368$.
3. What is the probability that a car is an SUV, given that it is electric?
   Answer: 0.30 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 $P(G) = 0.6$, which means $P(E) = 1 - 0.6 = 0.4$. We are also given $P(S|E) = 0.3$ and $P(S|G) = 0.5$. 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, $P(S|E) = 0.30$.
4. What is the probability that the first pen is green, given that the second pen is red?
   Answer: 0.429 Let G1 be the first is green, R1 be the first is red, G2 be the second is green, and R2 be the second is red. We want to find $P(G1|R2)$. We can use the formula $P(G1|R2) = \frac{P(R2|G1)P(G1)}{P(R2)}$. First, calculate the components. $P(G1) = 3/8$. $P(R2|G1) = 5/7$ (5 red pens left out of 7 total). The total probability of the second being red, $P(R2)$, is found using the law of total probability: $P(R2) = P(R2|G1)P(G1) + P(R2|R1)P(R1) = (5/7)(3/8) + (4/7)(5/8) = 15/56 + 20/56 = 35/56 = 5/8$. Now we can find $P(G1|R2)$: $P(G1|R2) = \frac{(5/7)(3/8)}{5/8} = \frac{15/56}{35/56} = \frac{15}{35} = \frac{3}{7} \approx 0.429$.
5. Find $P(D'|C)$, the probability that event D does not occur, given that event C has occurred.
   Answer: 0.6 We can first find $P(D|C)$ using the conditional probability formula: $P(D|C) = \frac{P(C \cap D)}{P(C)} = \frac{0.2}{0.5} = 0.4$. The event $D'|C$ is the complement of the event $D|C$. Therefore, $P(D'|C) = 1 - P(D|C) = 1 - 0.4 = 0.6$.
6. Are the events 'is a first-year student' and 'lives on campus' independent? Why?
   Answer: No, because $0.30 \times 0.25 = 0.075$ which is not 0.10. 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 $P(F) = 0.30$, $P(C) = 0.25$, and $P(F \cap C) = 0.10$. For the events to be independent, the condition $P(F \cap C) = P(F) \times P(C)$ must be met. Calculating the product: $P(F) \times P(C) = 0.30 \times 0.25 = 0.075$. Since $0.075 \neq 0.10$, the events are not independent (they are dependent).
7. What is the probability that the student passes exactly one of the two tests?
   Answer: 0.18 Let T1 be passing Test 1 and T2 be passing Test 2. We want to find $P(\text{Pass T1 and Fail T2}) + P(\text{Fail T1 and Pass T2})$. We have $P(T1) = 0.8$ so $P(T1') = 0.2$. We are given $P(T2|T1) = 0.9$, so $P(T2'|T1) = 1 - 0.9 = 0.1$. We are also given $P(T2|T1') = 0.5$. $P(T1 \cap T2') = P(T2'|T1)P(T1) = (0.1)(0.8) = 0.08$. $P(T1' \cap T2) = P(T2|T1')P(T1') = (0.5)(0.2) = 0.10$. The total probability of passing exactly one test is the sum of these two mutually exclusive events: $0.08 + 0.10 = 0.18$.
8. What is the probability that a student passes their math exam, given that they got at least 8 hours of sleep?
   Answer: 0.70 Let S be the event of getting at least 8 hours of sleep, and M be the event of passing the math exam. We are given $P(S) = 0.6$, $P(M) = 0.7$, and $P(S \cap M) = 0.42$. We need to find $P(M|S)$. Using the formula for conditional probability: $P(M|S) = \frac{P(M \cap S)}{P(S)} = \frac{0.42}{0.6} = 0.7$. Note that since $P(M|S) = P(M)$, the two events are independent.
9. What is the probability that a customer buys milk given that they bought bread?
   Answer: 0.67 Let B be the event of buying bread and M be the event of buying milk. We are given $P(B) = 0.6$, $P(M) = 0.5$, and $P(B|M) = 0.8$. We want to find $P(M|B)$. First, find the probability of buying both, $P(B \cap M)$, using the given conditional probability: $P(B \cap M) = P(B|M) \times P(M) = 0.8 \times 0.5 = 0.4$. Now, we can find $P(M|B)$ using its definition: $P(M|B) = \frac{P(B \cap M)}{P(B)} = \frac{0.4}{0.6} = \frac{4}{6} = \frac{2}{3} \approx 0.67$.
10. 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?
    Answer: 0.516 Let D be the event that a widget is defective. We want to find $P(M2|D)$. We use Bayes' theorem: $P(M2|D) = \frac{P(D|M2)P(M2)}{P(D)}$. We are given $P(M1)=0.6$, $P(M2)=0.4$, $P(D|M1)=0.05$, and $P(D|M2)=0.08$. The total probability of a defect is $P(D) = P(D|M1)P(M1) + P(D|M2)P(M2) = (0.05)(0.6) + (0.08)(0.4) = 0.030 + 0.032 = 0.062$. The numerator is $P(D \cap M2) = P(D|M2)P(M2) = (0.08)(0.4) = 0.032$. Therefore, $P(M2|D) = \frac{0.032}{0.062} \approx 0.516$.
11. What is the probability of A occurring given that B has occurred, $P(A|B)$?
    Answer: 0.80 By definition, two events A and B are independent if the occurrence of one does not affect the probability of the other. This means $P(A|B) = P(A)$ and $P(B|A) = P(B)$. Since events A and B are stated to be independent and $P(A) = 0.8$, it follows directly that $P(A|B) = 0.8$.
12. What is the probability that a randomly selected household has a streaming TV service?
    Answer: 0.31 Let B be the event of having broadband and S be the event of having a streaming service. We are given $P(B) = 0.70$, $P(S|B) = 0.40$, and $P(S|B') = 0.10$. The probability of not having broadband is $P(B') = 1 - 0.70 = 0.30$. We can find $P(S)$ using the law of total probability: $P(S) = P(S|B)P(B) + P(S|B')P(B')$. So, $P(S) = (0.40)(0.70) + (0.10)(0.30) = 0.28 + 0.03 = 0.31$.
13. What is the probability that the second marble drawn is red, given that the first marble drawn was also red?
    Answer: 0.556 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 $P(R2|R1)$. 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 $\frac{5}{9} \approx 0.556$.
14. Find the value of $P(Y|X)$.
    Answer: 0.24 First, find $P(X \cap Y)$ using the given conditional probability: $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$. Substituting values, $0.4 = \frac{P(X \cap Y)}{0.3}$, so $P(X \cap Y) = 0.4 \times 0.3 = 0.12$. Now, we can find $P(Y|X)$ using its formula: $P(Y|X) = \frac{P(X \cap Y)}{P(X)} = \frac{0.12}{0.5} = 0.24$.
15. Which of the following statements about events A and B is true?
    Answer: A and B are independent. 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 $P(A \cap B)$ using the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Substituting the given values: $0.76 = 0.4 + 0.6 - P(A \cap B)$, which gives us $P(A \cap B) = 0.24$. Now we can test for independence. Two events are independent if $P(A \cap B) = P(A) \times P(B)$. Let's check: $P(A) \times P(B) = 0.4 \times 0.6 = 0.24$. Since this equals our calculated $P(A \cap B) = 0.24$, events A and B are independent, making D correct. Option A is wrong because mutually exclusive events cannot occur simultaneously, meaning $P(A \cap B) = 0$. Since we found $P(A \cap B) = 0.24 \neq 0$, they're not mutually exclusive. Option B presents an incorrect formula. The correct conditional probability formula is $P(A|B) = \frac{P(A \cap B)}{P(B)}$, not $\frac{P(A)}{P(B)}$. 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 $P(A \cap B)$ first using the addition rule, then test whether $P(A \cap B) = P(A) \times P(B)$ to determine independence. This systematic approach will help you avoid common traps on probability questions.
16. What is the probability that a person has a garden, given that they do not have a pet?
    Answer: 0.50 Let P be the event of having a pet and G be the event of having a garden. We need to find $P(G|P')$. The number of people who do not have a pet is $300 - 180 = 120$. The number of people who have a garden but not a pet is the total number with a garden minus those with both: $150 - 90 = 60$. Therefore, $P(G|P') = \frac{\text{Number with Garden and no Pet}}{\text{Total Number with no Pet}} = \frac{60}{120} = 0.50$.
17. What is the probability that a randomly selected item has both Defect A and Defect B?
    Answer: 0.03 Let A be the event of having Defect A and B be the event of having Defect B. We are given $P(A) = 0.12$ and $P(B|A) = 0.25$. We need to find the probability of having both defects, which is $P(A \cap B)$. We can rearrange the conditional probability formula: $P(B|A) = \frac{P(A \cap B)}{P(A)}$. Therefore, $P(A \cap B) = P(B|A) \times P(A) = 0.25 \times 0.12 = 0.03$.
18. What is the probability that a student has brown hair, given that they do not have blue eyes?
    Answer: 0.571 Let H be the event of having brown hair and B be the event of having blue eyes. We are given $P(H) = 0.5$, $P(B) = 0.3$, and $P(H \cap B) = 0.1$. We want to find $P(H|B')$. First, we need the probability of not having blue eyes, $P(B') = 1 - P(B) = 1 - 0.3 = 0.7$. Next, we need the probability of having brown hair and not having blue eyes, $P(H \cap B')$. This can be found by $P(H) - P(H \cap B) = 0.5 - 0.1 = 0.4$. Now, we can calculate the conditional probability: $P(H|B') = \frac{P(H \cap B')}{P(B')} = \frac{0.4}{0.7} \approx 0.571$.
19. Given that the football team won their match, what is the probability that it rained?
    Answer: 0.125 Let R be the event that it rains, and W be the event the team wins. We are given $P(R) = 0.2$, $P(W|R) = 0.4$, and $P(W|R') = 0.7$. We want to find $P(R|W)$. First, find the total probability of winning, $P(W)$: $P(W) = P(W|R)P(R) + P(W|R')P(R') = (0.4)(0.2) + (0.7)(1-0.2) = 0.08 + (0.7)(0.8) = 0.08 + 0.56 = 0.64$. Now, use the formula for $P(R|W)$: $P(R|W) = \frac{P(W|R)P(R)}{P(W)} = \frac{0.08}{0.64} = 0.125$.
20. Given that a randomly selected employee uses a laptop, what is the probability they also use a desktop computer?
    Answer: 0.563 Let L be the event that an employee uses a laptop and D be the event they use a desktop. We are given $P(L) = 0.80$ and $P(D) = 0.65$. Since every employee uses at least one device, $P(L \cup D) = 1$. We can find the intersection using the addition rule: $P(L \cup D) = P(L) + P(D) - P(L \cap D)$. So, $1 = 0.80 + 0.65 - P(L \cap D)$, which gives $P(L \cap D) = 1.45 - 1 = 0.45$. We need to find $P(D|L)$, the probability they use a desktop given they use a laptop. Using the conditional probability formula: $P(D|L) = \frac{P(D \cap L)}{P(L)} = \frac{0.45}{0.80} = 0.5625$. Rounded to three significant figures, this is 0.563.

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