Independence of Events - Math 2

1. A medical test for a rare disease has the following characteristics: 2% of the population has the disease, 95% of people with the disease test positive, and 8% of people without the disease test positive. Are the events "having the disease" and "testing positive" independent?
   Answer: No, because $P(\text{positive|disease}) = 0.95 \neq P(\text{positive}) = 0.1074$ For independence, P(positive|disease) must equal P(positive). First calculate P(positive) = P(positive|disease) × P(disease) + P(positive|no disease) × P(no disease) = 0.95 × 0.02 + 0.08 × 0.98 = 0.019 + 0.0784 = 0.1074. Since P(positive|disease) = 0.95 ≠ 0.1074 = P(positive), the events are not independent. Choice A uses wrong reasoning about independence. Choice C misunderstands what independence means in this context. Choice D compares irrelevant conditional probabilities.
2. A weather forecasting model predicts rain and wind independently. Historical data shows $P(\text{Rain}) = 0.3$, $P(\text{Wind}) = 0.4$, and $P(\text{Rain and Wind}) = 0.15$. A meteorologist questions whether the model's independence assumption is valid. Using conditional probability, what analysis supports or refutes this assumption?
   Answer: Independence is refuted because $P(\text{Rain}|\text{Wind}) = \frac{0.15}{0.4} = 0.375 \neq 0.3 = P(\text{Rain})$ For independence, we need $P(\text{Rain}|\text{Wind}) = P(\text{Rain})$. Using conditional probability: $P(\text{Rain}|\text{Wind}) = \frac{P(\text{Rain and Wind})}{P(\text{Wind})} = \frac{0.15}{0.4} = 0.375$. Since $P(\text{Rain}|\text{Wind}) = 0.375 \neq 0.3 = P(\text{Rain})$, the independence assumption is invalid. Choice A tests the wrong conditional probability. Choice B reaches the correct conclusion but analyzes $P(\text{Wind}|\text{Rain})$ instead. Choice C incorrectly claims 0.375 equals 0.3.
3. In a lottery, tickets are numbered 1 through 1000. Let $A$ = "ticket number is divisible by 4" and $B$ = "ticket number is divisible by 6." A student calculates $P(A) = 0.25$, $P(B) = 0.166$, and $P(A \cap B) = 0.083$. Based on these calculations, what conclusion about independence is most appropriate?
   Answer: Events $A$ and $B$ are not independent because $P(B|A) = \frac{0.083}{0.25} = 0.332 \neq 0.166 = P(B)$, violating independence conditions For independence, we need $P(B|A) = P(B)$. Using the conditional probability formula: $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.083}{0.25} = 0.332$. Since $P(B|A) = 0.332 \neq 0.166 = P(B)$, the events are not independent. Note that $A \cap B$ corresponds to numbers divisible by $\text{lcm}(4,6) = 12$. Choice A incorrectly considers the values approximately equal. Choice B reaches the right conclusion but with less precise reasoning. Choice C incorrectly assumes divisibility by different numbers implies independence.
4. Two events $A$ and $B$ satisfy $P(A) = 0.4$, $P(B) = 0.3$, and $P(A \cup B) = 0.58$. A student uses the inclusion-exclusion principle to find $P(A \cap B) = 0.12$ and concludes the events are independent because $0.4 \times 0.3 = 0.12$. What additional verification should be performed?
   Answer: No additional verification needed; the multiplication rule $P(A \cap B) = P(A) \times P(B)$ is sufficient to establish independence between events The student correctly applied the independence test. Since $P(A \cap B) = 0.12 = 0.4 \times 0.3 = P(A) \times P(B)$, the events are indeed independent. This single condition is both necessary and sufficient for independence. Choice A suggests an unnecessary additional check (if $A$ and $B$ are independent, then so are $A$ and $B^c$). Choice B incorrectly confuses independence with mutual exclusivity. Choice D suggests checking complements, which would be redundant since independence of $A$ and $B$ automatically implies independence of their complements.
5. In a large corporation, 30% of employees work in sales, 20% have MBA degrees, and 8% both work in sales and have MBA degrees. The HR director wants to determine if having an MBA and working in sales are independent characteristics. Based on this information, which statement is most accurate?
   Answer: The characteristics are not independent because $P(\text{sales}) \times P(\text{MBA}) = 0.06 \neq 0.08 = P(\text{sales and MBA})$ When you encounter questions about independence of events or characteristics, you need to test whether knowing one event affects the probability of the other. Two events are independent if $P(A \text{ and } B) = P(A) \times P(B)$. Let's check this condition with the given data. We have $P(\text{sales}) = 0.30$, $P(\text{MBA}) = 0.20$, and $P(\text{sales and MBA}) = 0.08$. If these characteristics were independent, we would expect $P(\text{sales}) \times P(\text{MBA}) = 0.30 \times 0.20 = 0.06$ to equal $P(\text{sales and MBA}) = 0.08$. Since $0.06 \neq 0.08$, the characteristics are not independent. Choice A correctly identifies this relationship and concludes the characteristics are not independent, making it the right answer. Choice B incorrectly suggests independence because the joint probability is smaller than individual probabilities. This reasoning is flawed—independence isn't about relative magnitude but about the multiplicative relationship. Choice C uses conditional probability $P(\text{MBA|sales}) = \frac{0.08}{0.30} \approx 0.267$ and claims this is "close to" $P(\text{MBA}) = 0.20$. However, 0.267 is notably different from 0.20, and "close" isn't sufficient for independence anyway. Choice D correctly calculates that $P(\text{MBA|sales}) > P(\text{MBA})$, indicating dependence, but the question asks which statement is most accurate, and A provides the most direct and complete analysis. Remember: For independence questions, always check if $P(A \text{ and } B) = P(A) \times P(B)$. This is the most straightforward test and avoids subjective interpretations of conditional probabilities.
6. A survey of 200 students found that 60 play basketball, 80 play soccer, and 20 play both sports. Define events $B$ = "plays basketball" and $S$ = "plays soccer." Which conclusion about independence is justified?
   Answer: Events $B$ and $S$ are not independent because $P(B \cap S) = 0.10 \neq 0.12 = P(B) \cdot P(S)$, violating the independence criterion For independence, we need $P(B \cap S) = P(B) \cdot P(S)$. We have $P(B) = \frac{60}{200} = 0.30$, $P(S) = \frac{80}{200} = 0.40$, and $P(B \cap S) = \frac{20}{200} = 0.10$. Since $P(B) \cdot P(S) = 0.30 \times 0.40 = 0.12 \neq 0.10$, the events are not independent. Choice A incorrectly considers 0.10 and 0.12 "approximately equal." Choice C incorrectly dismisses the difference as negligible. Choice D incorrectly assumes different marginal probabilities imply dependence.
7. A medical test has a 90% sensitivity (detects 90% of positive cases) and 85% specificity (correctly identifies 85% of negative cases). In a population where 3% have the condition, let $C$ = "has condition" and $T$ = "tests positive." Which statement correctly evaluates independence?
   Answer: Events $C$ and $T$ are not independent because $P(T|C) = 0.90 \neq 0.173 = P(T)$, showing test results depend on condition status For independence, we need $P(T|C) = P(T)$. We have $P(T|C) = 0.90$. Using the law of total probability: $P(T) = P(T|C) \cdot P(C) + P(T|C^c) \cdot P(C^c) = 0.90 \times 0.03 + 0.15 \times 0.97 = 0.027 + 0.146 = 0.173$. Since $P(T|C) = 0.90 \neq 0.173$, the events are dependent. Choice A incorrectly assumes high accuracy implies independence. Choice B misunderstands what the ratio indicates about independence. Choice D reaches the correct conclusion but with less specific reasoning.
8. In a computer network, the probability that Server A fails is 0.05, Server B fails is 0.08, and both fail simultaneously is 0.004. Network engineers claim the failures are independent. A systems analyst argues this claim by computing $P(A^c \cap B^c)$. What should the analyst conclude?
   Answer: Independence is confirmed because $P(A^c \cap B^c) = 0.874$ matches the expected value of $0.95 \times 0.92 = 0.874$ under independence To check independence, we can verify if $P(A^c \cap B^c) = P(A^c) \cdot P(B^c)$. We have $P(A^c) = 1 - 0.05 = 0.95$ and $P(B^c) = 1 - 0.08 = 0.92$. Under independence, $P(A^c \cap B^c) = 0.95 \times 0.92 = 0.874$. Using $P(A^c \cap B^c) = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(A \cap B)] = 1 - [0.05 + 0.08 - 0.004] = 0.874$. Since this matches the independence calculation, the servers fail independently. Choice B uses incorrect calculation. Choice C is correct about $P(A \cap B)$ but doesn't address the analyst's approach. Choice D incorrectly uses addition instead of multiplication.
9. A researcher studying voter behavior finds that $P(\text{votes Democrat}) = 0.45$, $P(\text{college graduate}) = 0.35$, and $P(\text{votes Democrat|college graduate}) = 0.60$. What can be concluded about the independence of voting preference and education level?
   Answer: The events are dependent because $P(\text{Democrat|college}) = 0.60 \neq P(\text{Democrat}) = 0.45$ When you encounter probability questions about independence, you need to test whether one event affects the likelihood of another. Two events are independent if knowing one event occurred doesn't change the probability of the other event. The formal test for independence is: events A and B are independent if and only if $P(A|B) = P(A)$. Let's apply this test here. We have $P(\text{votes Democrat}) = 0.45$ and $P(\text{votes Democrat|college graduate}) = 0.60$. Since $0.60 \neq 0.45$, knowing someone is a college graduate changes the probability they vote Democrat from 45% to 60%. This proves the events are dependent. Option A incorrectly states the events are independent, when the fact that $P(\text{Democrat|college}) > P(\text{Democrat})$ actually proves dependence. Option B makes a calculation error—it claims $P(\text{Democrat and college}) = 0.21$ is less than $P(\text{Democrat}) \times P(\text{college}) = 0.45 \times 0.35 = 0.1575$, but $0.21 > 0.1575$. Also, we can verify: $P(\text{Democrat and college}) = P(\text{Democrat|college}) \times P(\text{college}) = 0.60 \times 0.35 = 0.21$. Option C completely misunderstands independence—it has nothing to do with whether individual probabilities are above or below 0.50. The correct answer is D because it properly applies the independence test and correctly identifies that the unequal conditional and unconditional probabilities indicate dependence. Study tip: Always check independence by comparing $P(A|B)$ to $P(A)$. If they're different, the events are dependent.
10. A bag contains red and blue marbles. Event R is "drawing a red marble on the first draw" and event B is "drawing a blue marble on the second draw" (without replacement). For these events to be independent, what must be true about the composition of the bag?
    Answer: The events can never be independent when sampling without replacement from a finite bag When you encounter problems about independence in probability, remember that two events are independent only if the outcome of the first event doesn't affect the probability of the second event. This means $P(B|R) = P(B)$ - the probability of drawing blue second must be the same whether you drew red or blue first. Let's think about what happens when sampling without replacement from a finite bag. If you draw a red marble first, you've removed one red marble, changing the composition of the remaining marbles. This necessarily changes the probability of drawing blue on the second draw compared to if you had drawn blue first. No matter how you arrange the colors, removing any marble from a finite collection changes the ratios. Looking at the answer choices: A is correct because true independence requires that the first draw doesn't change the probability structure for the second draw, which is impossible with finite sampling without replacement. B suggests an infinite bag would work, but the question specifically asks about "a bag" in practical terms. C incorrectly assumes equal numbers create independence - even with 50-50 ratios, removing one marble still changes the remaining proportions. D is wrong because if there's only one color, you can't have both red and blue marbles as described in the events. The key insight for probability problems: always check whether the sample space changes between events. When sampling without replacement from finite populations, the events are inherently dependent because each draw changes the conditions for subsequent draws.
11. Two events A and B have probabilities $P(A) = 0.4$, $P(B) = 0.3$, and $P(A \cup B) = 0.58$. A student claims that events A and B are independent. Which statement best evaluates this claim?
    Answer: The claim is correct because $P(A \cap B) = 0.12$, and $P(A) \times P(B) = 0.4 \times 0.3 = 0.12$ First, find P(A ∩ B) using P(A ∪ B) = P(A) + P(B) - P(A ∩ B), so P(A ∩ B) = 0.4 + 0.3 - 0.58 = 0.12. For independence, we need P(A ∩ B) = P(A) × P(B) = 0.4 × 0.3 = 0.12. Since this equality holds, the events are independent. Choice A misunderstands independence. Choice B incorrectly assumes independent events are mutually exclusive. Choice D incorrectly calculates P(A|B) = P(A ∩ B)/P(B) = 0.12/0.3 = 0.4 = P(A), actually confirming independence.
12. A quality control inspector finds that in a batch of electronic components, 15% are defective. She also finds that 8% of all components have a manufacturing date in January, and 1.5% of all components are both defective and made in January. What conclusion can be drawn about the relationship between being defective and being made in January?
    Answer: The events are independent because $0.15 \times 0.08 = 0.012 \neq 0.015$ For independence, P(defective ∩ January) should equal P(defective) × P(January). We have P(defective) = 0.15, P(January) = 0.08, so P(defective) × P(January) = 0.012. Since P(defective ∩ January) = 0.015 ≠ 0.012, the events are not independent. Choice A misinterprets what independence means. Choice B uses faulty reasoning about intersection probability. Choice D incorrectly suggests the values are equal within rounding when they clearly differ significantly.
13. In a certain high school, 40% of students take calculus, 25% take physics, and 15% take both calculus and physics. A guidance counselor claims that taking calculus and taking physics are independent choices. To verify this claim, which calculation should be performed?
    Answer: Check if $P(\text{both}) = 0.15$ equals $P(\text{calculus}) \times P(\text{physics}) = 0.40 \times 0.25 = 0.10$ When you encounter questions about independence in probability, remember that two events are independent if the occurrence of one doesn't affect the probability of the other. There are several equivalent ways to test this, but the most fundamental is checking whether $P(A \text{ and } B) = P(A) \times P(B)$. The correct approach is answer choice D: calculate whether $P(\text{both}) = P(\text{calculus}) \times P(\text{physics})$. If calculus and physics choices are truly independent, then $0.15$ should equal $0.40 \times 0.25 = 0.10$. Since $0.15 \neq 0.10$, the events are not independent—students who take calculus are actually more likely to take physics than the general population. Choice A incorrectly calculates the conditional probability $P(\text{physics|calculus})$ but fails to recognize that this conditional probability (0.375) being different from $P(\text{physics}) = 0.25$ actually proves dependence, not independence. Choice B makes the same error in reverse, calculating $P(\text{calculus|physics})$. Both A and B actually demonstrate dependence but frame it as a test for independence. Choice C tests whether $P(\text{calculus or physics}) = P(\text{calculus}) + P(\text{physics})$, which would only be true if the events were mutually exclusive (impossible to occur together), not independent. This confuses two completely different concepts. Remember: for independence, always check if $P(A \text{ and } B) = P(A) \times P(B)$. If they're equal, the events are independent; if not, they're dependent.
14. In a quality control process, two inspectors independently examine products. Inspector 1 rejects 8% of products, Inspector 2 rejects 12% of products, and both inspectors reject the same product 1.5% of the time. A quality manager assumes the inspections are independent. Which statistical test most directly challenges this assumption?
    Answer: Compare $P(R_2|R_1) = \frac{0.015}{0.08} = 0.1875$ with $P(R_2) = 0.12$; the difference reveals dependence When you encounter questions about statistical independence, the key concept to remember is that two events are independent if and only if the probability of one event occurring doesn't change based on whether the other event occurs. Mathematically, events A and B are independent when $P(A|B) = P(A)$ or equivalently $P(A \cap B) = P(A) \times P(B)$. The most direct way to test independence is comparing conditional probability with marginal probability. Here, we calculate $P(R_2|R_1) = \frac{P(R_1 \cap R_2)}{P(R_1)} = \frac{0.015}{0.08} = 0.1875$. If inspections were truly independent, this should equal $P(R_2) = 0.12$. Since 0.1875 ≠ 0.12, the inspections are dependent, directly challenging the independence assumption. Choice A correctly identifies this difference as evidence of dependence. Choice B uses the same calculation but incorrectly concludes the difference supports independence—this represents a fundamental misunderstanding of what independence means. Choice C compares the union probability with the sum of individual probabilities, but this tests a different property and doesn't directly address independence. Choice D compares $P(R_1 \cap R_2)$ with $P(R_1) \times P(R_2)$, which is a valid independence test, but the conditional probability approach in choice A is more direct and intuitive. Remember: To test independence, always check if $P(A|B) = P(A)$. If these probabilities differ significantly, the events are dependent, which challenges any assumption of independence.
15. Two dice are rolled simultaneously. Let A be the event "the first die shows an even number" and B be the event "the sum of both dice is 8." A student calculates $P(A) = \frac{1}{2}$, $P(B) = \frac{5}{36}$, and $P(A \cap B) = \frac{2}{36}$. What should the student conclude about independence?
    Answer: Events A and B are not independent because $P(A) \times P(B) = \frac{5}{72} \neq \frac{2}{36} = \frac{4}{72}$ To check independence, compare P(A) × P(B) with P(A ∩ B). We have P(A) × P(B) = (1/2) × (5/36) = 5/72, while P(A ∩ B) = 2/36 = 4/72. Since 5/72 ≠ 4/72, the events are not independent. Choice A makes an error in decimal conversion. Choice B incorrectly claims the values are approximately equal when they're clearly different. Choice D uses the correct alternative method (conditional probability) and reaches the right conclusion, but choice C more directly addresses the standard independence test.
16. In a standard deck of 52 cards, let A be the event "drawing a red card" and B be the event "drawing a face card." If these events are independent, what must be true about the conditional probability $P(A|B)$?
    Answer: $P(A|B) = \frac{26}{52} = \frac{1}{2}$ because independence requires $P(A|B) = P(A)$ For independent events, P(A|B) = P(A). Since P(A) = 26/52 = 1/2 (probability of drawing a red card), if the events are independent, then P(A|B) must equal 1/2. We can verify: P(A|B) = P(A∩B)/P(B) = (6/52)/(12/52) = 6/12 = 1/2, confirming independence. Choice A gives the right value but wrong reasoning. Choice C confuses conditional probability calculation. Choice D reverses the conditional probability.
17. Let $S$ = "participant has high stress" and $P$ = "participant has poor sleep." Which statement correctly analyzes the independence of these events?
    Answer: Events $S$ and $P$ are not independent because $P(P|S) = 0.30 \neq 0.24 = P(P)$, demonstrating conditional dependence between stress and sleep For independence, we need $P(P|S) = P(P)$. We have $P(S) = \frac{180}{500} = 0.36$, $P(P) = \frac{120}{500} = 0.24$, and $P(S \cap P) = \frac{54}{500} = 0.108$. Therefore, $P(P|S) = \frac{P(S \cap P)}{P(S)} = \frac{0.108}{0.36} = 0.30$. Since $P(P|S) = 0.30 \neq 0.24 = P(P)$, the events are not independent. Choice A incorrectly considers 0.108 and 0.0864 "reasonably close." Choice B reaches the correct conclusion but focuses on the multiplication rule. Choice C incorrectly dismisses the difference as minimal.
18. A manufacturing company produces widgets with a defect rate of 8%. Quality control tests each widget independently with a detection accuracy of 95% (correctly identifies 95% of defective widgets as defective and 95% of non-defective widgets as non-defective). If events $D$ = "widget is defective" and $T$ = "test indicates defective," which statement about independence is correct?
    Answer: Events $D$ and $T$ are not independent because $P(T|D) = 0.95 \neq P(T) = 0.122$, violating the independence condition For independence, we need $P(T|D) = P(T)$. First, calculate $P(T)$ using the law of total probability: $P(T) = P(T|D) \cdot P(D) + P(T|D^c) \cdot P(D^c) = 0.95 \cdot 0.08 + 0.05 \cdot 0.92 = 0.076 + 0.046 = 0.122$. Since $P(T|D) = 0.95 \neq 0.122 = P(T)$, the events are not independent. Choice A incorrectly assumes different marginal probabilities indicate independence. Choice B incorrectly assumes high accuracy guarantees independence. Choice C uses the wrong value for $P(T)$.
19. In a card game, player A draws a card from a standard deck and keeps it. Then player B draws from the remaining cards. Let $R_A$ = "A draws red" and $R_B$ = "B draws red." A student claims these events are independent because "each player draws randomly." What is the flaw in this reasoning?
    Answer: Independence requires $P(R_B|R_A) = P(R_B)$, but $P(R_B|R_A) = \frac{25}{51} \neq \frac{26}{52} = P(R_B)$ due to sampling without replacement Independence requires $P(R_B|R_A) = P(R_B)$. When A draws red, 25 red cards remain out of 51 total, so $P(R_B|R_A) = \frac{25}{51}$. Initially, $P(R_B) = \frac{26}{52} = \frac{1}{2}$. Since $\frac{25}{51} \neq \frac{1}{2}$, the events are dependent. Choice A incorrectly assumes random drawing guarantees independence. Choice C confuses the definition of independence with equal probabilities. Choice D incorrectly states the multiplication rule as addition.
20. A survey of 200 students found that 80 students like pizza, 60 students like burgers, and 30 students like both pizza and burgers. If a student is randomly selected from this group, what can be concluded about the events "the student likes pizza" and "the student likes burgers"?
    Answer: The events are not independent because $P(\text{pizza and burgers}) \neq P(\text{pizza}) \times P(\text{burgers})$ To check independence, we need to verify if P(A ∩ B) = P(A) × P(B). Here, P(pizza) = 80/200 = 0.4, P(burgers) = 60/200 = 0.3, and P(pizza and burgers) = 30/200 = 0.15. Since P(pizza) × P(burgers) = 0.4 × 0.3 = 0.12 ≠ 0.15, the events are not independent. Choice A incorrectly states they are independent. Choice C shows a misunderstanding that overlap prevents independence. Choice D incorrectly states the condition for independence.

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