Conditional Probability & Independence - College Statistics

1. A student is taking a multiple-choice test where each question has 4 options. For any given question, the probability that the student knows the answer is 0.7. If the student does not know the answer, they guess randomly. Given that the student answered a question correctly, what is the probability that they knew the answer?
   Answer: 0.903 Let K be the event that the student knew the answer, and C be the event that they answered correctly. We want to find $P(K|C)$. We are given $P(K) = 0.7$, so $P(K^c) = 0.3$ (probability of guessing). If the student knows the answer, the probability of being correct is 1, so $P(C|K) = 1$. If they guess, the probability is 1/4, so $P(C|K^c) = 0.25$. Using Bayes' theorem, $P(K|C) = \frac{P(C|K)P(K)}{P(C)}$. The denominator is $P(C) = P(C|K)P(K) + P(C|K^c)P(K^c) = (1)(0.7) + (0.25)(0.3) = 0.7 + 0.075 = 0.775$. Therefore, $P(K|C) = \frac{0.7}{0.775} = \frac{700}{775} = \frac{28}{31} \approx 0.903$.
2. Events A and B are such that $P(A \cup B) = 0.7$, $P(A) = 0.5$, and $P(B) = 0.4$. Which of the following statements is true?
   Answer: A and B are independent but not mutually exclusive. First, calculate the probability of the intersection of A and B using the general addition rule: $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.7 = 0.2$. Since $P(A \cap B) \neq 0$, the events are not mutually exclusive. To check for independence, we compare $P(A \cap B)$ with $P(A)P(B)$. $P(A)P(B) = (0.5)(0.4) = 0.2$. Since $P(A \cap B) = P(A)P(B)$, the events are independent. Therefore, A and B are independent but not mutually exclusive.
3. Given two events A and B, which of the following expressions represents the probability that event A occurs but event B does not?
   Answer: $P(A) - P(A \cap B)$ This question tests your understanding of set operations and probability relationships. When you see "A occurs but B does not," you're looking at the intersection of event A with the complement of event B. The correct answer is A: $P(A) - P(A \cap B)$. Think of this using a Venn diagram approach. The probability of A includes all outcomes where A happens, whether B occurs or not. To find just the cases where A occurs but B doesn't, you subtract the overlap where both A and B happen together. This gives you exactly the region where A is true but B is false. Option B, $P(A|B^c)$, represents a conditional probability - the probability of A given that B doesn't occur. This would only equal our desired probability if A and B were independent, which isn't stated. Option C, $P(A)P(B^c)$, assumes independence between A and the complement of B. This product rule only works when events are independent, which again isn't given. Option D, $P(A) / P(B^c)$, has no standard probabilistic interpretation for this scenario. Division of probabilities doesn't represent intersection or set difference operations. Study tip: When you see "but" or "and not" in probability problems, think set operations. "A but not B" means $A \cap B^c$. You can always calculate this as $P(A) - P(A \cap B)$ since the probability of A splits into two mutually exclusive parts: A with B, and A without B.
4. A rare disease affects 1 in 1000 people. A test for the disease has a 99% sensitivity (it correctly identifies 99% of people who have the disease) and a 98% specificity (it correctly identifies 98% of people who do not have the disease). If a randomly selected person tests positive, what is the approximate probability that they actually have the disease?
   Answer: 0.047 Let D be the event of having the disease and T+ be the event of testing positive. We are given $P(D) = 0.001$, $P(T+|D) = 0.99$ (sensitivity), and $P(T-|D^c) = 0.98$ (specificity). From specificity, the false positive rate is $P(T+|D^c) = 1 - 0.98 = 0.02$. We want to find $P(D|T+)$. Using Bayes' Theorem, $P(D|T+) = \frac{P(T+|D)P(D)}{P(T+)}$. First, find the total probability of a positive test, $P(T+)$, using the Law of Total Probability: $P(T+) = P(T+|D)P(D) + P(T+|D^c)P(D^c) = (0.99)(0.001) + (0.02)(0.999) = 0.00099 + 0.01998 = 0.02097$. Then, $P(D|T+) = \frac{0.00099}{0.02097} \approx 0.0472$.
5. It is known that 30% of the students in a certain university are graduate students. Of the graduate students, 60% are receiving financial aid. Of the undergraduate students, 40% are receiving financial aid. If a student is selected at random, what is the probability that they are receiving financial aid?
   Answer: 0.46 Let G be the event that a student is a graduate student, and F be the event that a student receives financial aid. We are given $P(G) = 0.30$, so $P(G^c) = 1 - 0.30 = 0.70$ (the probability of being an undergraduate). We are also given the conditional probabilities $P(F|G) = 0.60$ and $P(F|G^c) = 0.40$. We can find the overall probability of receiving financial aid, $P(F)$, using the Law of Total Probability: $P(F) = P(F|G)P(G) + P(F|G^c)P(G^c) = (0.60)(0.30) + (0.40)(0.70) = 0.18 + 0.28 = 0.46$.
6. Let A and B be events with $P(A) = 0.5$ and $P(A \cup B) = 0.8$. If A and B are independent, what is $P(B|A^c)$?
   Answer: 0.60 First, find $P(B)$. Since A and B are independent, $P(A \cup B) = P(A) + P(B) - P(A)P(B)$. Let $x = P(B)$. Then $0.8 = 0.5 + x - 0.5x$, which simplifies to $0.3 = 0.5x$, so $x = P(B) = 0.6$. We need to find $P(B|A^c)$. If A and B are independent, then B and $A^c$ are also independent. Therefore, $P(B|A^c) = P(B)$. So, $P(B|A^c) = 0.6$.
7. Let A and B be independent events with $0 < P(A) < 1$ and $0 < P(B) < 1$. Which of the following statements is NOT necessarily true?
   Answer: $P(A|B) = P(B|A)$ If A and B are independent, $P(A|B) = P(A)$ and $P(B|A) = P(B)$. Therefore, the statement $P(A|B) = P(B|A)$ would imply that $P(A) = P(B)$. This is not necessarily true for any two independent events. The other three statements are fundamental properties of independent events: (A) if A and B are independent, so are their complements; (B) this is the general addition rule with the independence substitution $P(A \cap B) = P(A)P(B)$; (C) this is a primary definition of independence.
8. For events A and B, it is known that $P(A|B) = 0.6$, $P(A|B^c) = 0.3$, and $P(B) = 0.4$. What is the value of $P(A)$?
   Answer: 0.42 This problem requires the Law of Total Probability, which states $P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$. We are given $P(A|B) = 0.6$, $P(A|B^c) = 0.3$, and $P(B) = 0.4$. From $P(B)$, we can find $P(B^c) = 1 - P(B) = 1 - 0.4 = 0.6$. Substituting these values into the formula gives: $P(A) = (0.6)(0.4) + (0.3)(0.6) = 0.24 + 0.18 = 0.42$.
9. Refer to the table provided. Given that a randomly selected employee does not use public transit, what is the probability that this employee is in the Engineering department?
   Answer: 0.357 Let E be the event that an employee is in Engineering, and let NPT be the event that an employee does not use public transit. We want to find $P(E|NPT)$. The total number of employees who do not use public transit is the sum of those who use a Car and those who Bike: $200 + 80 = 280$. The number of employees in the Engineering department who do not use public transit is the sum of engineers who use a Car and engineers who Bike: $50 + 50 = 100$. The conditional probability is the ratio of these two numbers: $P(E|NPT) = \frac{100}{280} = \frac{10}{28} = \frac{5}{14} \approx 0.357$.
10. Let E and F be two events such that $P(E) = 0.4$, $P(F) = 0.3$, and $P(E \cup F) = 0.7$. Which statement accurately describes the relationship between events E and F?
    Answer: They are mutually exclusive but not independent. First, check for mutual exclusivity. Events are mutually exclusive if they cannot occur at the same time, meaning $P(E \cap F) = 0$. Using the addition rule, $P(E \cap F) = P(E) + P(F) - P(E \cup F) = 0.4 + 0.3 - 0.7 = 0$. So, the events are mutually exclusive. Next, check for independence. Events are independent if $P(E \cap F) = P(E)P(F)$. We have $P(E)P(F) = (0.4)(0.3) = 0.12$. Since $P(E \cap F) = 0 \neq 0.12$, the events are not independent. Therefore, they are mutually exclusive but not independent.
11. For two events, A and B, it is known that $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cup B) = 0.9$. What is the value of $P(A^c | B^c)$, the probability of A not occurring given that B has not occurred?
    Answer: 0.20 First, find $P(A \cap B)$ using the addition rule: $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.9 = 0.2$. The event $A^c \cap B^c$ is the complement of $A \cup B$. By De Morgan's laws, $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.9 = 0.1$. We also need $P(B^c) = 1 - P(B) = 1 - 0.5 = 0.5$. The conditional probability is then $P(A^c | B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{0.1}{0.5} = 0.20$.
12. A warehouse is protected by two independent fire alarm systems. The first system has a 95% probability of detecting a fire, and the second system has a 90% probability of detecting a fire. If a fire occurs, what is the probability that at least one of the systems detects it?
    Answer: 0.995 Let A be the event that the first system detects the fire, and B be the event for the second system. We are given $P(A) = 0.95$ and $P(B) = 0.90$. The probability that at least one system detects the fire is $P(A \cup B)$. A simpler approach is to find the probability of the complement event—that neither system detects the fire—and subtract it from 1. The probability that the first system fails is $P(A^c) = 1 - 0.95 = 0.05$. The probability that the second system fails is $P(B^c) = 1 - 0.90 = 0.10$. Since the systems are independent, the probability that both fail is $P(A^c \cap B^c) = P(A^c)P(B^c) = (0.05)(0.10) = 0.005$. Therefore, the probability that at least one detects the fire is $1 - 0.005 = 0.995$.
13. Two fair six-sided dice are rolled. Let A be the event that the sum of the dice is 7. Let B be the event that the first die shows a 4. Which statement is true about events A and B?
    Answer: A and B are independent. When analyzing relationships between events in probability, you need to distinguish between independence, mutual exclusivity, and conditional probabilities. These concepts are fundamentally different and often confused. To determine if events A and B are independent, check if $P(A \cap B) = P(A) \times P(B)$. Event A (sum equals 7) occurs in 6 ways out of 36 possible outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), so $P(A) = \frac{6}{36} = \frac{1}{6}$. Event B (first die shows 4) occurs in 6 ways out of 36, so $P(B) = \frac{6}{36} = \frac{1}{6}$. The intersection A ∩ B occurs only when the first die shows 4 AND the sum is 7, which happens only with (4,3), so $P(A \cap B) = \frac{1}{36}$. Since $\frac{1}{36} = \frac{1}{6} \times \frac{1}{6}$, the events are independent, making A correct. B is wrong because mutually exclusive events cannot occur simultaneously, but A and B can both happen (outcome (4,3)). C is incorrect because for independent events, $P(A|B) = P(A)$, not greater than P(A). D is false because independence also means $P(B|A) = P(B)$, not less than P(B). Remember: Independence means one event doesn't affect the probability of the other. Always verify by checking if $P(A \cap B) = P(A) \times P(B)$, and remember that independent events can occur together, unlike mutually exclusive events.
14. In a certain city, 60% of residents subscribe to the local newspaper and 80% own a television. Assuming these two events are independent, what is the probability that a randomly selected resident subscribes to the newspaper or owns a television, but not both?
    Answer: 0.44 Let N be the event of subscribing to the newspaper and T be the event of owning a television. We are given $P(N) = 0.6$ and $P(T) = 0.8$. The probability of 'N or T, but not both' is $P(N \cap T^c) + P(N^c \cap T)$. Since N and T are independent, their complements are also independent. So, $P(N \cap T^c) = P(N)P(T^c) = (0.6)(1 - 0.8) = (0.6)(0.2) = 0.12$. And $P(N^c \cap T) = P(N^c)P(T) = (1 - 0.6)(0.8) = (0.4)(0.8) = 0.32$. The total probability is $0.12 + 0.32 = 0.44$.
15. Two events A and B are independent. If $P(A) = p$ and $P(B) = 2p$, and $P(A \cup B) = 0.52$, what is the value of $p$?
    Answer: 0.20 For independent events, the addition rule is $P(A \cup B) = P(A) + P(B) - P(A)P(B)$. Substituting the given values, we get $0.52 = p + 2p - (p)(2p)$. This simplifies to the quadratic equation $0.52 = 3p - 2p^2$, or $2p^2 - 3p + 0.52 = 0$. Using the quadratic formula, $p = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(0.52)}}{2(2)} = \frac{3 \pm \sqrt{9 - 4.16}}{4} = \frac{3 \pm \sqrt{4.84}}{4} = \frac{3 \pm 2.2}{4}$. This gives two possible solutions: $p = \frac{5.2}{4} = 1.3$ and $p = \frac{0.8}{4} = 0.2$. Since p represents a probability, it cannot be greater than 1. Thus, $p=0.20$.
16. If events E and F are independent and $P(E) = 0.5$, what must $P(F)$ be for $P(E \cup F) = 0.8$?
    Answer: 0.60 The formula for the union of two independent events is $P(E \cup F) = P(E) + P(F) - P(E)P(F)$. We are given $P(E \cup F) = 0.8$ and $P(E) = 0.5$. Let $x = P(F)$. Substituting the known values gives $0.8 = 0.5 + x - (0.5)x$. Simplifying the equation: $0.8 = 0.5 + 0.5x$. Subtract 0.5 from both sides: $0.3 = 0.5x$. Finally, solve for x: $x = \frac{0.3}{0.5} = 0.6$. Thus, $P(F) = 0.60$.
17. An urn contains 5 red and 5 blue marbles. Two marbles are drawn in succession without replacement. If it is known that at least one of the marbles drawn is red, what is the probability that both marbles are red?
    Answer: 2/7 Let A be the event 'at least one red marble is drawn' and B be the event 'both marbles are red'. We want to find $P(B|A) = \frac{P(A \cap B)}{P(A)}$. The event B is a subset of A, so $A \cap B = B$. Thus, we need to calculate $\frac{P(B)}{P(A)}$. The probability of drawing two red marbles is $P(B) = P(R_1 \cap R_2) = P(R_1)P(R_2|R_1) = (5/10)(4/9) = 20/90 = 2/9$. The event A, 'at least one red', is the complement of drawing two blue marbles. The probability of drawing two blue marbles is $P(B_1 \cap B_2) = (5/10)(4/9) = 20/90 = 2/9$. So, $P(A) = 1 - P(B_1 \cap B_2) = 1 - 2/9 = 7/9$. Finally, $P(B|A) = \frac{2/9}{7/9} = 2/7$.
18. From a standard 52-card deck, two cards are drawn without replacement. What is the probability that the second card is a king, given that the first card was NOT a king?
    Answer: 4/51 When you encounter conditional probability problems involving "without replacement," you need to carefully track how the sample space changes after each draw. Here, you're finding P(second card is king | first card is NOT a king). Since the first card drawn was not a king, it was one of the 48 non-king cards in the deck. After removing this non-king card, you're left with 51 cards total, and crucially, all 4 kings are still in the deck. So the probability that the second card is a king equals $\frac{4 \text{ kings remaining}}{51 \text{ cards remaining}} = \frac{4}{51}$ Let's examine why the other answers miss the mark. Answer B (3/51) incorrectly assumes that one king was already removed, but the condition states the first card was NOT a king, so all 4 kings remain available. Answer C (3/52) makes the same error about kings remaining while also incorrectly using 52 as the denominator, ignoring that one card has already been drawn. Answer D (4/52) correctly identifies that 4 kings remain but fails to account for the reduced sample space—after drawing one card, only 51 cards are left. The key insight is that conditional probability problems require you to update both your numerator (favorable outcomes) and denominator (total possible outcomes) based on the given condition. Always ask yourself: "What information does the condition give me, and how does it change what's left in my sample space?"
19. A manufacturer receives 70% of its widgets from Supplier X and 30% from Supplier Y. It is known that 5% of widgets from Supplier X are defective, and 8% of widgets from Supplier Y are defective. If a randomly selected widget is found to be defective, what is the probability it came from Supplier X?
    Answer: 0.593 Let X be the event that a widget is from Supplier X, and Y from Supplier Y. Let D be the event a widget is defective. We are given $P(X) = 0.7$, $P(Y) = 0.3$, $P(D|X) = 0.05$, and $P(D|Y) = 0.08$. We want to find $P(X|D)$. Using Bayes' theorem, $P(X|D) = \frac{P(D|X)P(X)}{P(D)}$. First, calculate the total probability of a defective widget, $P(D) = P(D|X)P(X) + P(D|Y)P(Y) = (0.05)(0.7) + (0.08)(0.3) = 0.035 + 0.024 = 0.059$. Now, we can find the conditional probability: $P(X|D) = \frac{0.035}{0.059} \approx 0.593$.
20. For three events A, B, and C, it is known that $P(A \cap B \cap C) = 0.1$ and $P(A \cap B) = 0.2$. The probability $P(A \cap C) = 0.3$ is also known. What is the value of $P(C | A \cap B)$?
    Answer: 1/2 The conditional probability $P(X|Y)$ is defined as $\frac{P(X \cap Y)}{P(Y)}$. In this case, the event 'X' is C, and the event 'Y' is $A \cap B$. So, $P(C | A \cap B) = \frac{P(C \cap (A \cap B))}{P(A \cap B)}$. The numerator $P(C \cap A \cap B)$ is the same as $P(A \cap B \cap C)$. We are given $P(A \cap B \cap C) = 0.1$ and $P(A \cap B) = 0.2$. Therefore, $P(C | A \cap B) = \frac{0.1}{0.2} = 0.5 = 1/2$. The information $P(A \cap C) = 0.3$ is extraneous.

Start this quiz