College Statistics Quiz: Conditional Probability And Independence
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Conditional Probability And IndependenceQuestion 1 of 20

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?

4/51
3/51
3/52
4/52
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College Statistics Quiz

College Statistics Quiz: Conditional Probability And Independence

Practice Conditional Probability And Independence in College Statistics 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 And Independence, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

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.

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Question 1

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?

  1. 4/51 (correct answer)
  2. 3/51
  3. 3/52
  4. 4/52
Explanation: 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 4 kings remaining51 cards remaining=451\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?"

Question 2

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?

  1. 4/51 (correct answer)
  2. 3/51
  3. 3/52
  4. 4/52
Explanation: 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 4 kings remaining51 cards remaining=451\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?"

Question 3

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?

  1. 0.46 (correct answer)
  2. 0.50
  3. 0.54
  4. 0.70
Explanation: 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)=0.30P(G) = 0.30, so P(Gc)=1−0.30=0.70P(G^c) = 1 - 0.30 = 0.70 (the probability of being an undergraduate). We are also given the conditional probabilities P(F∣G)=0.60P(F|G) = 0.60 and P(F∣Gc)=0.40P(F|G^c) = 0.40. We can find the overall probability of receiving financial aid, P(F)P(F), using the Law of Total Probability: P(F)=P(F∣G)P(G)+P(F∣Gc)P(Gc)=(0.60)(0.30)+(0.40)(0.70)=0.18+0.28=0.46P(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.

Question 4

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?

  1. 0.700
  2. 0.775
  3. 0.903 (correct answer)
  4. 0.925
Explanation: 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)P(K|C). We are given P(K)=0.7P(K) = 0.7, so P(Kc)=0.3P(K^c) = 0.3 (probability of guessing). If the student knows the answer, the probability of being correct is 1, so P(C∣K)=1P(C|K) = 1. If they guess, the probability is 1/4, so P(C∣Kc)=0.25P(C|K^c) = 0.25. Using Bayes' theorem, P(K∣C)=P(C∣K)P(K)P(C)P(K|C) = \frac{P(C|K)P(K)}{P(C)}. The denominator is P(C)=P(C∣K)P(K)+P(C∣Kc)P(Kc)=(1)(0.7)+(0.25)(0.3)=0.7+0.075=0.775P(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)=0.70.775=700775=2831≈0.903P(K|C) = \frac{0.7}{0.775} = \frac{700}{775} = \frac{28}{31} \approx 0.903.

Question 5

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?

  1. 0.035
  2. 0.050
  3. 0.593 (correct answer)
  4. 0.700
Explanation: 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)=0.7P(X) = 0.7, P(Y)=0.3P(Y) = 0.3, P(D∣X)=0.05P(D|X) = 0.05, and P(D∣Y)=0.08P(D|Y) = 0.08. We want to find P(X∣D)P(X|D). Using Bayes' theorem, P(X∣D)=P(D∣X)P(X)P(D)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=0.059P(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)=0.0350.059≈0.593P(X|D) = \frac{0.035}{0.059} \approx 0.593.

Question 6

For two events, A and B, it is known that P(A)=0.6P(A) = 0.6, P(B)=0.5P(B) = 0.5, and P(A∪B)=0.9P(A \cup B) = 0.9. What is the value of P(Ac∣Bc)P(A^c | B^c), the probability of A not occurring given that B has not occurred?

  1. 0.10
  2. 0.20 (correct answer)
  3. 0.25
  4. 0.40
Explanation: First, find P(A∩B)P(A \cap B) using the addition rule: P(A∩B)=P(A)+P(B)−P(A∪B)=0.6+0.5−0.9=0.2P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.9 = 0.2. The event Ac∩BcA^c \cap B^c is the complement of A∪BA \cup B. By De Morgan's laws, P(Ac∩Bc)=P((A∪B)c)=1−P(A∪B)=1−0.9=0.1P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.9 = 0.1. We also need P(Bc)=1−P(B)=1−0.5=0.5P(B^c) = 1 - P(B) = 1 - 0.5 = 0.5. The conditional probability is then P(Ac∣Bc)=P(Ac∩Bc)P(Bc)=0.10.5=0.20P(A^c | B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{0.1}{0.5} = 0.20.

Question 7

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?

  1. 0.010
  2. 0.047 (correct answer)
  3. 0.980
  4. 0.990
Explanation: Let D be the event of having the disease and T+ be the event of testing positive. We are given P(D)=0.001P(D) = 0.001, P(T+∣D)=0.99P(T+|D) = 0.99 (sensitivity), and P(T−∣Dc)=0.98P(T-|D^c) = 0.98 (specificity). From specificity, the false positive rate is P(T+∣Dc)=1−0.98=0.02P(T+|D^c) = 1 - 0.98 = 0.02. We want to find P(D∣T+)P(D|T+). Using Bayes' Theorem, P(D∣T+)=P(T+∣D)P(D)P(T+)P(D|T+) = \frac{P(T+|D)P(D)}{P(T+)}. First, find the total probability of a positive test, P(T+)P(T+), using the Law of Total Probability: P(T+)=P(T+∣D)P(D)+P(T+∣Dc)P(Dc)=(0.99)(0.001)+(0.02)(0.999)=0.00099+0.01998=0.02097P(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+)=0.000990.02097≈0.0472P(D|T+) = \frac{0.00099}{0.02097} \approx 0.0472.

Question 8

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?

  1. 0.855
  2. 0.925
  3. 0.950
  4. 0.995 (correct answer)
Explanation: 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)=0.95P(A) = 0.95 and P(B)=0.90P(B) = 0.90. The probability that at least one system detects the fire is P(A∪B)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(Ac)=1−0.95=0.05P(A^c) = 1 - 0.95 = 0.05. The probability that the second system fails is P(Bc)=1−0.90=0.10P(B^c) = 1 - 0.90 = 0.10. Since the systems are independent, the probability that both fail is P(Ac∩Bc)=P(Ac)P(Bc)=(0.05)(0.10)=0.005P(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=0.9951 - 0.005 = 0.995.

Question 9

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?

  1. 2/9
  2. 2/7 (correct answer)
  3. 4/9
  4. 1/2
Explanation: 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)=P(A∩B)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)}. The event B is a subset of A, so A∩B=BA \cap B = B. Thus, we need to calculate P(B)P(A)\frac{P(B)}{P(A)}. The probability of drawing two red marbles is P(B)=P(R1∩R2)=P(R1)P(R2∣R1)=(5/10)(4/9)=20/90=2/9P(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(B1∩B2)=(5/10)(4/9)=20/90=2/9P(B_1 \cap B_2) = (5/10)(4/9) = 20/90 = 2/9. So, P(A)=1−P(B1∩B2)=1−2/9=7/9P(A) = 1 - P(B_1 \cap B_2) = 1 - 2/9 = 7/9. Finally, P(B∣A)=2/97/9=2/7P(B|A) = \frac{2/9}{7/9} = 2/7.

Question 10

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?

  1. 0.08
  2. 0.44 (correct answer)
  3. 0.48
  4. 0.92
Explanation: Let N be the event of subscribing to the newspaper and T be the event of owning a television. We are given P(N)=0.6P(N) = 0.6 and P(T)=0.8P(T) = 0.8. The probability of 'N or T, but not both' is P(N∩Tc)+P(Nc∩T)P(N \cap T^c) + P(N^c \cap T). Since N and T are independent, their complements are also independent. So, P(N∩Tc)=P(N)P(Tc)=(0.6)(1−0.8)=(0.6)(0.2)=0.12P(N \cap T^c) = P(N)P(T^c) = (0.6)(1 - 0.8) = (0.6)(0.2) = 0.12. And P(Nc∩T)=P(Nc)P(T)=(1−0.6)(0.8)=(0.4)(0.8)=0.32P(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=0.440.12 + 0.32 = 0.44.

Question 11

For events A and B, it is known that P(A∣B)=0.6P(A|B) = 0.6, P(A∣Bc)=0.3P(A|B^c) = 0.3, and P(B)=0.4P(B) = 0.4. What is the value of P(A)P(A)?

  1. 0.36
  2. 0.42 (correct answer)
  3. 0.45
  4. 0.48
Explanation: This problem requires the Law of Total Probability, which states P(A)=P(A∣B)P(B)+P(A∣Bc)P(Bc)P(A) = P(A|B)P(B) + P(A|B^c)P(B^c). We are given P(A∣B)=0.6P(A|B) = 0.6, P(A∣Bc)=0.3P(A|B^c) = 0.3, and P(B)=0.4P(B) = 0.4. From P(B)P(B), we can find P(Bc)=1−P(B)=1−0.4=0.6P(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=0.42P(A) = (0.6)(0.4) + (0.3)(0.6) = 0.24 + 0.18 = 0.42.

Question 12

Events A and B are such that P(A∪B)=0.7P(A \cup B) = 0.7, P(A)=0.5P(A) = 0.5, and P(B)=0.4P(B) = 0.4. Which of the following statements is true?

  1. A and B are independent and mutually exclusive.
  2. A and B are independent but not mutually exclusive. (correct answer)
  3. A and B are mutually exclusive but not independent.
  4. A and B are neither independent nor mutually exclusive.
Explanation: First, calculate the probability of the intersection of A and B using the general addition rule: P(A∩B)=P(A)+P(B)−P(A∪B)=0.5+0.4−0.7=0.2P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.7 = 0.2. Since P(A∩B)≠0P(A \cap B) \neq 0, the events are not mutually exclusive. To check for independence, we compare P(A∩B)P(A \cap B) with P(A)P(B)P(A)P(B). P(A)P(B)=(0.5)(0.4)=0.2P(A)P(B) = (0.5)(0.4) = 0.2. Since P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B), the events are independent. Therefore, A and B are independent but not mutually exclusive.

Question 13

If events E and F are independent and P(E)=0.5P(E) = 0.5, what must P(F)P(F) be for P(E∪F)=0.8P(E \cup F) = 0.8?

  1. 0.30
  2. 0.375
  3. 0.50
  4. 0.60 (correct answer)
Explanation: The formula for the union of two independent events is P(E∪F)=P(E)+P(F)−P(E)P(F)P(E \cup F) = P(E) + P(F) - P(E)P(F). We are given P(E∪F)=0.8P(E \cup F) = 0.8 and P(E)=0.5P(E) = 0.5. Let x=P(F)x = P(F). Substituting the known values gives 0.8=0.5+x−(0.5)x0.8 = 0.5 + x - (0.5)x. Simplifying the equation: 0.8=0.5+0.5x0.8 = 0.5 + 0.5x. Subtract 0.5 from both sides: 0.3=0.5x0.3 = 0.5x. Finally, solve for x: x=0.30.5=0.6x = \frac{0.3}{0.5} = 0.6. Thus, P(F)=0.60P(F) = 0.60.

Question 14

For two events, A and B, it is known that P(A)=0.6P(A) = 0.6, P(B)=0.5P(B) = 0.5, and P(A∪B)=0.9P(A \cup B) = 0.9. What is the value of P(Ac∣Bc)P(A^c | B^c), the probability of A not occurring given that B has not occurred?

  1. 0.10
  2. 0.20 (correct answer)
  3. 0.25
  4. 0.40
Explanation: First, find P(A∩B)P(A \cap B) using the addition rule: P(A∩B)=P(A)+P(B)−P(A∪B)=0.6+0.5−0.9=0.2P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.9 = 0.2. The event Ac∩BcA^c \cap B^c is the complement of A∪BA \cup B. By De Morgan's laws, P(Ac∩Bc)=P((A∪B)c)=1−P(A∪B)=1−0.9=0.1P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.9 = 0.1. We also need P(Bc)=1−P(B)=1−0.5=0.5P(B^c) = 1 - P(B) = 1 - 0.5 = 0.5. The conditional probability is then P(Ac∣Bc)=P(Ac∩Bc)P(Bc)=0.10.5=0.20P(A^c | B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{0.1}{0.5} = 0.20.

Question 15

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?

  1. 0.010
  2. 0.047 (correct answer)
  3. 0.980
  4. 0.990
Explanation: Let D be the event of having the disease and T+ be the event of testing positive. We are given P(D)=0.001P(D) = 0.001, P(T+∣D)=0.99P(T+|D) = 0.99 (sensitivity), and P(T−∣Dc)=0.98P(T-|D^c) = 0.98 (specificity). From specificity, the false positive rate is P(T+∣Dc)=1−0.98=0.02P(T+|D^c) = 1 - 0.98 = 0.02. We want to find P(D∣T+)P(D|T+). Using Bayes' Theorem, P(D∣T+)=P(T+∣D)P(D)P(T+)P(D|T+) = \frac{P(T+|D)P(D)}{P(T+)}. First, find the total probability of a positive test, P(T+)P(T+), using the Law of Total Probability: P(T+)=P(T+∣D)P(D)+P(T+∣Dc)P(Dc)=(0.99)(0.001)+(0.02)(0.999)=0.00099+0.01998=0.02097P(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+)=0.000990.02097≈0.0472P(D|T+) = \frac{0.00099}{0.02097} \approx 0.0472.

Question 16

If events E and F are independent and P(E)=0.5P(E) = 0.5, what must P(F)P(F) be for P(E∪F)=0.8P(E \cup F) = 0.8?

  1. 0.30
  2. 0.375
  3. 0.50
  4. 0.60 (correct answer)
Explanation: The formula for the union of two independent events is P(E∪F)=P(E)+P(F)−P(E)P(F)P(E \cup F) = P(E) + P(F) - P(E)P(F). We are given P(E∪F)=0.8P(E \cup F) = 0.8 and P(E)=0.5P(E) = 0.5. Let x=P(F)x = P(F). Substituting the known values gives 0.8=0.5+x−(0.5)x0.8 = 0.5 + x - (0.5)x. Simplifying the equation: 0.8=0.5+0.5x0.8 = 0.5 + 0.5x. Subtract 0.5 from both sides: 0.3=0.5x0.3 = 0.5x. Finally, solve for x: x=0.30.5=0.6x = \frac{0.3}{0.5} = 0.6. Thus, P(F)=0.60P(F) = 0.60.

Question 17

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?

  1. 0.46 (correct answer)
  2. 0.50
  3. 0.54
  4. 0.70
Explanation: 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)=0.30P(G) = 0.30, so P(Gc)=1−0.30=0.70P(G^c) = 1 - 0.30 = 0.70 (the probability of being an undergraduate). We are also given the conditional probabilities P(F∣G)=0.60P(F|G) = 0.60 and P(F∣Gc)=0.40P(F|G^c) = 0.40. We can find the overall probability of receiving financial aid, P(F)P(F), using the Law of Total Probability: P(F)=P(F∣G)P(G)+P(F∣Gc)P(Gc)=(0.60)(0.30)+(0.40)(0.70)=0.18+0.28=0.46P(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.

Question 18

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?

  1. 0.035
  2. 0.050
  3. 0.593 (correct answer)
  4. 0.700
Explanation: 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)=0.7P(X) = 0.7, P(Y)=0.3P(Y) = 0.3, P(D∣X)=0.05P(D|X) = 0.05, and P(D∣Y)=0.08P(D|Y) = 0.08. We want to find P(X∣D)P(X|D). Using Bayes' theorem, P(X∣D)=P(D∣X)P(X)P(D)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=0.059P(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)=0.0350.059≈0.593P(X|D) = \frac{0.035}{0.059} \approx 0.593.

Question 19

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?

  1. 2/9
  2. 2/7 (correct answer)
  3. 4/9
  4. 1/2
Explanation: 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)=P(A∩B)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)}. The event B is a subset of A, so A∩B=BA \cap B = B. Thus, we need to calculate P(B)P(A)\frac{P(B)}{P(A)}. The probability of drawing two red marbles is P(B)=P(R1∩R2)=P(R1)P(R2∣R1)=(5/10)(4/9)=20/90=2/9P(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(B1∩B2)=(5/10)(4/9)=20/90=2/9P(B_1 \cap B_2) = (5/10)(4/9) = 20/90 = 2/9. So, P(A)=1−P(B1∩B2)=1−2/9=7/9P(A) = 1 - P(B_1 \cap B_2) = 1 - 2/9 = 7/9. Finally, P(B∣A)=2/97/9=2/7P(B|A) = \frac{2/9}{7/9} = 2/7.

Question 20

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?

  1. 0.855
  2. 0.925
  3. 0.950
  4. 0.995 (correct answer)
Explanation: 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)=0.95P(A) = 0.95 and P(B)=0.90P(B) = 0.90. The probability that at least one system detects the fire is P(A∪B)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(Ac)=1−0.95=0.05P(A^c) = 1 - 0.95 = 0.05. The probability that the second system fails is P(Bc)=1−0.90=0.10P(B^c) = 1 - 0.90 = 0.10. Since the systems are independent, the probability that both fail is P(Ac∩Bc)=P(Ac)P(Bc)=(0.05)(0.10)=0.005P(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=0.9951 - 0.005 = 0.995.