Conditional probability and independence - Discrete Math

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?
Answer: $\frac{p(1-p)^4}{(1-p)^3}$ which simplifies to $p(1-p)$ after algebraic manipulation 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.
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 $P(X \cap Y \cap Z)$ ?
Answer: $0.06$ 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).
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?
Answer: B and C are independent because P(B∩C) = 0.2 = P(B)×P(C), satisfying the independence condition 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.
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?
Answer: Bob is incorrect because P(F|E) = $\frac{4}{51}$ while P(F) = $\frac{4}{52}$ , showing the events are not independent 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.
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?
Answer: $0.72$ 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.
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?
Answer: $\frac{1}{3}$ 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.
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?
Answer: Approximately 0.019, showing the test has poor positive predictive value despite high sensitivity 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.
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?
Answer: $\frac{1}{2}$ by symmetry, since either parent is equally likely to contribute the single A allele 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.
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?
Answer: $\frac{0.85 \times 0.75}{1 - (1-0.85)(1-0.75)} = \frac{17}{31}$ 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.
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?
Answer: $\frac{0.7 \times 0.9}{0.7 \times 0.9 + 0.3 \times 0.8} = \frac{21}{29}$ 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).
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?
Answer: $\frac{9}{25}$ 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.
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?
Answer: $\frac{76}{79}$ 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.
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?
Answer: $\frac{1}{2}$ 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.
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 $P(A^c \cap B^c \cap C^c)$ ?
Answer: $\frac{1}{4}$ 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.
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?
Answer: $1 - (1-0.05)(1-0.03) = 0.0785$ using the complement of both tests being correct 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.
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?
Answer: $1$ 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.
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?
Answer: No, because $P(A \cap B) \neq P(A) \times P(B)$ 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.

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