Discrete Math Quiz: Conditional Probability And Independence

17 questions · exam conditions

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Conditional Probability And IndependenceQuestion 1 of 17

A biased coin has probability $p$ of heads. The coin is flipped repeatedly until the first tail appears. Given that exactly 3 flips were needed, what is the probability that the first two flips were both heads?

p^2

\frac{p^2}{1-p}

\frac{p^2(1-p)}{(1-p)}

1

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Discrete Math Quiz

Discrete Math Quiz: Conditional Probability And Independence

Practice Conditional Probability And Independence in Discrete Math 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 Discrete Math.

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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

$p^2$
$\frac{p^2}{1-p}$
$\frac{p^2(1-p)}{(1-p)}$
$1$ (correct answer)

Explanation: Given that exactly 3 flips were needed, we know the sequence must be HHT (heads, heads, tail). There's only one way for exactly 3 flips to occur: the first two must be heads and the third must be tail. Therefore, P(first two heads | exactly 3 flips) = 1. Choice A gives the unconditional probability of two heads. Choice B attempts to use conditional probability incorrectly. Choice C shows the calculation setup but doesn't simplify correctly.

Question 2

In a database of student records, $60\%$ are undergraduate students and $40\%$ are graduate students. Among undergraduates, $30\%$ have internship experience. Among graduates, $80\%$ have internship experience. If a randomly selected student has internship experience, what is the probability they are an undergraduate?

$\frac{9}{25}$ (correct answer)
$\frac{3}{10}$
$\frac{18}{50}$
$\frac{9}{25}$

Explanation: Using Bayes' theorem: P(undergrad | internship) = P(internship | undergrad) × P(undergrad) / P(internship). First find P(internship) = 0.6 × 0.3 + 0.4 × 0.8 = 0.18 + 0.32 = 0.50. Then P(undergrad | internship) = (0.3 × 0.6) / 0.50 = 0.18 / 0.50 = 18/50 = 9/25. Choice B gives P(internship | undergrad). Choice C gives the numerator fraction before simplification. Choice D is identical to A, representing the simplified form.

Question 3

In a computer network security system, the probability that an intrusion attempt is detected is $0.8$ . Given that an intrusion attempt is detected, the probability that it is correctly classified as malicious is $0.75$ . Given that an intrusion attempt is not detected, the probability that the system remains secure (no actual breach occurs) is $0.6$ . What is the probability that the system remains secure?

$0.37$
$0.60$
$0.72$ (correct answer)
$0.95$

Explanation: The system remains secure if: (1) intrusion is detected AND correctly classified, OR (2) intrusion is not detected AND no breach occurs. P(secure) = P(detected) × P(classified | detected) + P(not detected) × P(no breach | not detected) = 0.8 × 0.75 + 0.2 × 0.6 = 0.6 + 0.12 = 0.72. Choice A incorrectly multiplies all three probabilities. Choice B uses only the conditional probability when not detected. Choice D incorrectly adds probabilities without proper weighting.

Question 4

A machine learning algorithm classifies emails as spam or legitimate. It has a false positive rate of $0.05$ (classifies legitimate email as spam) and a false negative rate of $0.15$ (classifies spam as legitimate). If $20\%$ of emails are actually spam, and the algorithm classifies an email as legitimate, what is the probability it actually is legitimate?

$\frac{79}{85}$
$\frac{79}{82}$
$\frac{76}{82}$
$\frac{76}{79}$ (correct answer)

Explanation: This is a classic Bayes' theorem problem involving conditional probability. When you see classification accuracy questions, always identify what you're looking for: here, it's P(actually legitimate | classified as legitimate). Let's define our events and use the given information systematically. We know 20% of emails are spam, so 80% are legitimate. The false positive rate of 0.05 means 5% of legitimate emails get misclassified as spam, so 95% of legitimate emails are correctly classified as legitimate. The false negative rate of 0.15 means 15% of spam emails get misclassified as legitimate. Now we can calculate how often emails get classified as legitimate overall. This happens in two ways: when legitimate emails are correctly classified (

0.8 \times 0.95 = 0.76

) and when spam emails are incorrectly classified as legitimate (

0.2 \times 0.15 = 0.03

). So P(classified as legitimate) =

0.76 + 0.03 = 0.79

. Using Bayes' theorem: P(actually legitimate | classified as legitimate) =

\frac{0.76}{0.79} = \frac{76}{79}

. Choice A (

\frac{79}{85}

) appears to use an incorrect denominator. Choice B (

\frac{79}{82}

) flips the numerator and denominator incorrectly. Choice C (

\frac{76}{82}

) uses the wrong denominator, possibly by adding 0.76 + 0.06 instead of 0.76 + 0.03. Remember: in Bayes' problems, always calculate the total probability of the observed outcome first, then use it as your denominator. The numerator is the probability of both the condition and observation occurring together.

Question 5

A discrete random process generates sequences of length 4 from alphabet $\{0,1\}$ . Each position is independently 0 with probability $\frac{2}{3}$ and 1 with probability $\frac{1}{3}$ . What is the probability that a sequence contains exactly two consecutive 1's, given that it contains exactly two 1's total?

$\frac{1}{3}$
$\frac{2}{5}$
$\frac{1}{2}$ (correct answer)
$\frac{3}{5}$

Explanation: Sequences with exactly two 1's: positions can be chosen in C(4,2) = 6 ways. These are: 1100, 1010, 1001, 0110, 0101, 0011. Sequences with consecutive 1's among these: 1100, 0110, 0011 (3 sequences). All sequences with exactly two 1's have equal probability since each has two 1's and two 0's. Therefore P(consecutive | exactly two 1's) = 3/6 = 1/2. Choice A undercounts consecutive pairs. Choice B and D use incorrect counting of total arrangements.

Question 6

Consider three events $A$ , $B$ , and $C$ where $P(A) = \frac{1}{2}$ , $P(B) = \frac{1}{3}$ , $P(C) = \frac{1}{4}$ . If $A$ and $B$ are independent, $A$ and $C$ are independent, but $B$ and $C$ are not independent with $P(B \cap C) = \frac{1}{6}$ , what is $P(A^c \cap B^c \cap C^c)$ ?

$\frac{1}{4}$ (correct answer)
$\frac{1}{3}$
$\frac{3}{8}$
$\frac{5}{12}$

Explanation: Since A and B are independent, A and C are independent, we have P(

A^c

∩

B^c

∩

C^c

) = P(

A^c

) × P(

B^c

) × P(

C^c

) only if all three are mutually independent. However, B and C are not independent. We need: P(

A^c

∩

B^c

∩

C^c

) = P(

A^c

) × P(

B^c

∩

C^c

). Since A is independent of both B and C, A^c is independent of B^c ∩ C^c. P(

B^c

∩

C^c

) = 1 - P(B ∪ C) = 1 - [P(B) + P(C) - P(B ∩ C)] = 1 - [1/3 + 1/4 - 1/6] = 1 - 5/12 = 1/2. Therefore P(

A^c

∩

B^c

∩

C^c

) = 1/2 × 1/2 = 1/4. Other choices assume full independence or make computational errors.

Question 7

A discrete mathematics exam has two sections. Students pass Section A with probability $\frac{2}{3}$ and pass Section B with probability $\frac{3}{4}$ . Among students who pass Section A, $\frac{5}{6}$ also pass Section B. What is the probability that a randomly selected student passes exactly one section?

$\frac{1}{4}$
$\frac{1}{3}$ (correct answer)
$\frac{5}{12}$
$\frac{7}{12}$

Explanation: First find P(pass both) = P(A) × P(B|A) = (2/3) × (5/6) = 5/9. Then P(pass exactly one) = P(A only) + P(B only) = [P(A) - P(both)] + [P(B) - P(both)] = (2/3 - 5/9) + (3/4 - 5/9) = 1/9 + 7/36 = 4/36 + 7/36 = 11/36 = 1/3. Choice A assumes independence incorrectly. Choice C uses P(A) × P(B̄) + P(Ā) × P(B) assuming independence. Choice D calculates total success probability instead of exactly one.

Question 8

A diagnostic test for a rare disease has sensitivity 0.95 (probability of testing positive given disease present) and specificity 0.90 (probability of testing negative given disease absent). If the disease prevalence is 0.002, and a patient tests positive, what is the approximate probability they actually have the disease?

Approximately 0.019, showing the test has poor positive predictive value despite high sensitivity (correct answer)
Approximately 0.095, indicating moderate reliability for positive results in this population
Approximately 0.950, demonstrating that high sensitivity translates to high positive predictive value
Approximately 0.475, reflecting the balance between sensitivity and specificity in the calculation

Explanation: Using Bayes' theorem: P(Disease|Positive) = P(Positive|Disease)×P(Disease)/P(Positive). P(Positive|Disease) = 0.95, P(Disease) = 0.002. P(Positive) = P(Positive|Disease)×P(Disease) + P(Positive|No Disease)×P(No Disease) = 0.95×0.002 + 0.10×0.998 = 0.0019 + 0.0998 = 0.1017. Therefore P(Disease|Positive) = (0.95×0.002)/0.1017 ≈ 0.019. This demonstrates how low prevalence leads to many false positives. Choice B uses an incorrect calculation. Choice C confuses sensitivity with positive predictive value. Choice D appears to average sensitivity and specificity, which is not the correct approach.

Question 9

Consider three events A, B, and C where P(A) = 0.6, P(B) = 0.4, P(C) = 0.5, P(A∩B) = 0.24, P(A∩C) = 0.3, and P(B∩C) = 0.2. If A and B are independent, and A and C are independent, what can be concluded about the independence of B and C?

B and C must be independent since A is independent of both, creating a transitive independence relationship
B and C are independent because P(B∩C) = 0.2 = P(B)×P(C), satisfying the independence condition (correct answer)
B and C cannot be independent because pairwise independence doesn't guarantee mutual independence of all pairs
The independence of B and C cannot be determined without additional information about P(A∩B∩C)

Explanation: To check independence of B and C, we need P(B∩C) = P(B)×P(C). Given P(B) = 0.4, P(C) = 0.5, and P(B∩C) = 0.2, we have P(B)×P(C) = 0.4×0.5 = 0.2 = P(B∩C). Therefore B and C are independent. We can verify the given independence claims: P(A)×P(B) = 0.6×0.4 = 0.24 = P(A∩B) ✓, and P(A)×P(C) = 0.6×0.5 = 0.3 = P(A∩C) ✓. Choice A incorrectly assumes transitivity of independence. Choice C makes a general statement that doesn't apply here since we can directly verify independence. Choice D is incorrect because we have sufficient information to determine B and C's independence.

Question 10

A biased coin has probability p of heads. The coin is flipped repeatedly until the first head appears. Given that it took more than 3 flips to get the first head, what is the probability that it took exactly 5 flips?

$p(1-p)^4$ since this is simply the probability of getting the first head on flip 5
$\frac{1-p}{1+(1-p)+(1-p)^2}$ based on the geometric distribution properties
$\frac{(1-p)^4}{1-(1-p)^3}$ representing the conditional probability given the constraint
$\frac{p(1-p)^4}{(1-p)^3}$ which simplifies to $p(1-p)$ after algebraic manipulation (correct answer)

Explanation: When you encounter conditional probability problems involving sequences of trials, you need to apply the fundamental formula: P(A|B) = P(A ∩ B)/P(B). Here, you want the probability of exactly 5 flips given that it took more than 3 flips. Let's define our events: A = "exactly 5 flips" and B = "more than 3 flips." For a geometric distribution, P(exactly k flips) =

p(1-p)^{k-1}

since you need k-1 tails followed by 1 head. So P(A ∩ B) = P(exactly 5 flips) =

p(1-p)^4

, and P(B) = P(more than 3 flips) =

(1-p)^3

(three consecutive tails). Therefore: P(A|B) =

\frac{p(1-p)^4}{(1-p)^3} = p(1-p)

Answer A incorrectly gives the unconditional probability of getting the first head on flip 5, ignoring the given condition entirely. Answer B uses an incorrect formula that doesn't match geometric distribution properties—the denominator represents probabilities for flips 1, 2, and 3, but we need the probability of exceeding 3 flips. Answer C omits the probability p of actually getting a head on flip 5, only accounting for the four preceding tails. Answer D correctly applies conditional probability by dividing the probability of exactly 5 flips by the probability of exceeding 3 flips, then simplifies properly. Remember: conditional probability problems require you to restrict your sample space to the given condition. Always identify both the numerator (intersection of events) and denominator (conditioning event) before calculating.

Question 11

In a computer network, packets are routed through two possible paths. Path 1 is chosen with probability 0.7, and Path 2 with probability 0.3. The probability of successful delivery is 0.9 for Path 1 and 0.8 for Path 2. If a packet is successfully delivered, what is the probability it was sent via Path 1?

$\frac{0.9}{0.9 + 0.8} = \frac{9}{17}$ based on the relative success rates
$\frac{0.7}{0.7 + 0.3} = 0.7$ since path selection is independent of delivery
$\frac{0.7 \times 0.9}{0.7 \times 0.9 + 0.3 \times 0.8} = \frac{21}{29}$ (correct answer)
$\frac{0.7 \times 0.9}{0.9} = 0.7$ representing the conditional probability calculation

Explanation: When you encounter a problem asking "given that an event occurred, what's the probability it came from a specific source," you're dealing with Bayes' theorem. This is a conditional probability problem where you need to update your initial probability based on new information. The correct approach uses Bayes' theorem:

P(\text{Path 1}|\text{Success}) = \frac{P(\text{Path 1}) \times P(\text{Success}|\text{Path 1})}{P(\text{Success})}

. First, calculate the total probability of success:

P(\text{Success}) = 0.7 \times 0.9 + 0.3 \times 0.8 = 0.63 + 0.24 = 0.87

. Then apply the formula:

\frac{0.7 \times 0.9}{0.87} = \frac{0.63}{0.87} = \frac{21}{29}

, confirming answer C. Answer A incorrectly uses only the success rates (0.9 and 0.8) while ignoring how often each path is actually chosen. Success rates alone don't determine the probability without considering path frequency. Answer B assumes the answer is simply the original probability of choosing Path 1 (0.7), but this ignores the crucial new information that the packet was successfully delivered. Different success rates for each path change the conditional probability. Answer D attempts to use conditional probability but makes a calculation error. It divides

0.7 \times 0.9

by just 0.9 instead of the total probability of success, which incorrectly cancels terms and yields the wrong result. Remember: Bayes' problems always require you to weight each pathway by both its initial probability AND the likelihood of the observed outcome. Look for the formula structure: (initial probability × likelihood) divided by (total probability of the evidence).

Question 12

In a genetics model, allele A appears with probability p in each parent. For an offspring to have trait X, it must inherit allele A from exactly one parent (not both, not neither). If an offspring is observed to have trait X, what is the probability that it inherited allele A from the mother specifically?

$\frac{p(1-p)}{2p(1-p)} = \frac{1}{2}$ using conditional probability with the constraint of exactly one A allele
$\frac{p}{p + (1-p)} = \frac{p}{1} = p$ based on the mother's probability of having allele A
$\frac{1}{2}$ by symmetry, since either parent is equally likely to contribute the single A allele (correct answer)
$\frac{p(1-p)}{p(1-p) + (1-p)p} = \frac{1}{2}$ since both scenarios for exactly one A allele are equally likely

Explanation: When you encounter genetics probability problems involving conditional probability, focus on what information the given condition provides and how it restricts the sample space. Given that an offspring has trait X (exactly one A allele), we know one parent contributed A and the other contributed the alternative allele. There are exactly two ways this can happen: mother contributes A and father doesn't, or father contributes A and mother doesn't. Since each parent independently has probability

p

of having allele A, both scenarios have equal probability

p(1-p)

. The key insight is symmetry. Since both parents have identical probability

p

of contributing allele A, and we know exactly one of them did contribute it, each parent is equally likely to be the contributor. Therefore, the probability that the mother specifically contributed the A allele is

\frac{1}{2}

. Option A shows the correct conditional probability calculation but presents it unnecessarily complicated. The numerator

p(1-p)

represents the probability mother contributed A, while the denominator

2p(1-p)

represents the total probability of having exactly one A allele, yielding

\frac{1}{2}

. Option B incorrectly uses the mother's marginal probability

p

without considering the constraint that exactly one parent contributed A. Option D shows the same calculation as A but with more explicit notation, demonstrating that both equally likely scenarios sum to give the

\frac{1}{2}

result. Remember: when dealing with conditional genetics problems, always identify what the given condition tells you about the sample space, then apply symmetry arguments when parents have identical characteristics.

Question 13

A quality control system uses two independent tests T₁ and T₂ to classify products as defective or non-defective. T₁ has a false positive rate of 0.05 and false negative rate of 0.02. T₂ has a false positive rate of 0.03 and false negative rate of 0.04. If a non-defective product is tested, what is the probability that at least one test incorrectly classifies it as defective?

$0.05 + 0.03 = 0.08$ since false positive rates are additive for independent tests
$\frac{0.05 + 0.03}{2} = 0.04$ taking the average false positive rate of the two tests
$0.05 \times 0.03 = 0.0015$ representing the probability both tests give false positives
$1 - (1-0.05)(1-0.03) = 0.0785$ using the complement of both tests being correct (correct answer)

Explanation: When you encounter probability questions involving "at least one" event occurring, think about using the complement rule. This approach is often simpler than calculating multiple scenarios directly. Since we want the probability that at least one test incorrectly classifies a non-defective product, we need to find the probability of getting one or more false positives. The complement approach calculates the probability that both tests are correct, then subtracts from 1. For a non-defective product:

T₁ correctly classifies it with probability $1 - 0.05 = 0.95$
T₂ correctly classifies it with probability $1 - 0.03 = 0.97$

Since the tests are independent, the probability both are correct is

(0.95)(0.97) = 0.9215

. Therefore, the probability at least one is wrong is

1 - 0.9215 = 0.0785

, making D correct. Option A incorrectly adds the false positive rates, but this would only work if the events were mutually exclusive (which they're not). Option B averages the rates, which has no basis in probability theory for finding "at least one" probabilities. Option C calculates the probability that both tests give false positives, but the question asks for at least one false positive, not exactly two. Study tip: For "at least one" probability problems, always consider the complement rule:

P(\text{at least one}) = 1 - P(\text{none})

. This method typically involves fewer calculations and reduces errors, especially when dealing with independent events.

Question 14

A software company runs two independent security scans on user uploads. Scan A detects malicious files with probability 0.85, and Scan B detects malicious files with probability 0.75. Given that at least one scan detected a malicious file, what is the probability that both scans detected it?

$\frac{0.85 \times 0.75}{1 - (1-0.85)(1-0.75)} = \frac{17}{31}$ (correct answer)
$\frac{0.85 \times 0.75}{0.85 + 0.75} = \frac{17}{32}$
$\frac{0.85 \times 0.75}{0.85 + 0.75 - 0.85 \times 0.75} = \frac{51}{95}$
$0.85 \times 0.75 = \frac{17}{20}$

Explanation: Let A be the event 'Scan A detects' and B be 'Scan B detects'. We want P(A∩B | A∪B). Using conditional probability: P(A∩B | A∪B) = P(A∩B)/P(A∪B). Since scans are independent, P(A∩B) = 0.85×0.75 = 0.6375. P(A∪B) = 1 - P(neither detects) = 1 - (1-0.85)(1-0.75) = 1 - 0.0375 = 0.9625. Therefore P(A∩B | A∪B) = 0.6375/0.9625 = 17/31. Choice B incorrectly uses P(A)+P(B) as denominator. Choice C uses the inclusion-exclusion formula P(A∪B) but applies it incorrectly as the answer rather than using it in the conditional probability calculation. Choice D gives just P(A∩B) without conditioning.

Question 15

In a card game, Alice draws 2 cards without replacement from a standard 52-card deck. Define events: E = 'first card is an Ace', F = 'second card is a King'. Bob claims that since there are equal numbers of Aces and Kings in the deck, P(F|E) = P(F). Which statement correctly evaluates Bob's reasoning?

Bob is correct because drawing an Ace doesn't change the total number of Kings in the remaining deck
Bob is incorrect because P(F|E) = $\frac{4}{51}$ while P(F) = $\frac{4}{52}$ , showing the events are not independent (correct answer)
Bob is incorrect because P(F|E) = $\frac{4}{52}$ while P(F) = $\frac{4}{51}$ , demonstrating dependence through deck composition
Bob is correct because independence requires only that the cards be of different ranks, not suits

Explanation: For independence, we need P(F|E) = P(F). When the first card is an Ace (event E), there are 51 cards left, still with 4 Kings, so P(F|E) = 4/51. However, P(F) considers all possible second cards across all possible first cards, which equals 4/52. Since 4/51 ≠ 4/52, the events are dependent. Choice A incorrectly focuses only on Kings remaining without considering the reduced deck size. Choice C reverses the probabilities. Choice D misunderstands independence—different ranks don't guarantee independence in sampling without replacement.

Question 16

Two fair dice are rolled. Event $A$ is "the sum is even" and event $B$ is "at least one die shows 6". Are events $A$ and $B$ independent?

Yes, because $P(A \cap B) = P(A) \times P(B)$
No, because $P(A \cap B) \neq P(A) \times P(B)$ (correct answer)
Yes, because both events have probability $\frac{11}{36}$
No, because the sum being even affects whether a 6 appears

Explanation: P(A) = 18/36 = 1/2 (half of all outcomes have even sum). P(B) = 11/36 (36 - 25 outcomes where neither die shows 6). A ∩ B occurs when sum is even AND at least one 6 appears: (1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) - but only even sums: (2,6), (4,6), (6,2), (6,4), (6,6) = 5 outcomes. P(A ∩ B) = 5/36 ≠ (1/2)(11/36) = 11/72, so not independent. Choice A incorrectly assumes independence. Choice C gives wrong probabilities. Choice D gives wrong reasoning about causation.

Question 17

Events $X$ and $Y$ satisfy $P(X) = 0.4$ , $P(Y) = 0.5$ , and $P(X \cup Y) = 0.7$ . If event $Z$ is independent of both $X$ and $Y$ with $P(Z) = 0.3$ , what is $P(X \cap Y \cap Z)$ ?

$0.06$ (correct answer)
$0.09$
$0.12$
$0.21$

Explanation: First find P(X ∩ Y) using P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y), so P(X ∩ Y) = 0.4 + 0.5 - 0.7 = 0.2. Since Z is independent of both X and Y, Z is also independent of X ∩ Y, so P(X ∩ Y ∩ Z) = P(X ∩ Y) × P(Z) = 0.2 × 0.3 = 0.06. Choice B incorrectly uses P(Z) × P(Y). Choice C incorrectly uses P(Z) × P(X). Choice D incorrectly uses P(Z) × P(X ∪ Y).