Understanding Independent Events
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A website tracks visits. Let event $A$ be “a visit is from a mobile device” and event $B$ be “a visit results in a purchase.” Given that $P(A)=0.60$, $P(B)=0.20$, and $P(A\cap B)=0.15$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
No, dependent because $P(A)P(B)=0.60\times0.20=0.12$, which does not equal $P(A\cap B)=0.15$.
Yes, independent because $P(A)P(B)=0.60\times0.20=0.15$, which equals $P(A\cap B)$.
Yes, independent because device type and purchasing seem unrelated, so independence holds.
No, dependent because $P(A)+P(B)=0.60+0.20=0.80$, which does not equal $P(A\cap B)$.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = 0.60 × 0.20 = 0.12. Since this does not equal the given P(A ∩ B) = 0.15, events A and B are dependent. A tempting distractor is adding P(A) and P(B) or assuming unrelated contexts mean independence, but that's not reliable. Always compute the product and compare it directly to P(A ∩ B) for the truth. Let the probabilities guide you, not intuition about devices and purchases.
A jar contains many marbles. One marble is selected at random. Let event $A$ be “the marble is red” and event $B$ be “the marble is large.” Given that $P(A)=0.30$, $P(B)=\tfrac{2}{5}$, and $P(A\cap B)=0.12$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
Yes, independent because color and size seem unrelated, so the events must be independent.
Yes, independent because $P(A)P(B)=0.30\times0.40=0.12$, which equals $P(A\cap B)$.
No, dependent because $P(A)P(B)=0.30\times0.40=0.012$, which is not $P(A\cap B)$.
No, dependent because $P(A)+P(B)=0.30+0.40=0.70$, which does not equal $P(A\cap B)$.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = 0.30 × 0.40 = 0.12. Since this matches the given P(A ∩ B) = 0.12, events A and B are independent. A common distractor is adding probabilities or miscalculating the product, leading to wrong conclusions about dependence. Always compute the product and compare it to P(A ∩ B) to verify independence. Don't assume based on color and size seeming unrelated; confirm with the math.
A two-step experiment is performed with replacement: a ball is drawn from a bag, recorded, replaced, and then a second draw is made. Let event $A$ be “the first draw is green” and event $B$ be “the second draw is green.” Given that $P(A)=\tfrac{2}{5}$, $P(B)=0.40$, and $P(A\cap B)=0.16$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
No, dependent because the two draws happen in the same experiment, so they cannot be independent.
No, dependent because $P(A)P(B)=\tfrac{2}{5}\times0.40=0.08$, which does not equal $P(A\cap B)=0.16$.
Yes, independent because $P(A)P(B)=\tfrac{2}{5}\times0.40=0.16$, which equals $P(A\cap B)$.
Yes, independent because $P(A)+P(B)=\tfrac{2}{5}+0.40=0.80$, and $P(A\cap B)=0.16$.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = (2/5) × 0.40 = 0.40 × 0.40 = 0.16. Since this matches the given P(A ∩ B) = 0.16, events A and B are independent. One misconception is miscalculating the product, like halving it to 0.08, or confusing addition with the independence test. Always compute the product accurately and compare it to P(A ∩ B). Don't assume dependence just because it's the same experiment; with replacement, independence can hold as shown.
At a school, a student is selected at random. Let event $A$ be “the student is in the art club” and event $B$ be “the student is in the chess club.” Given that $P(A)=\tfrac{1}{4}$, $P(B)=0.30$, and $P(A\cap B)=0.075$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
Yes, independent because clubs are separate activities, so membership should be unrelated.
Yes, independent because $P(A)P(B)=0.25\times0.30=0.075$, which equals $P(A\cap B)$.
No, dependent because $P(A)+P(B)=0.25+0.30=0.55$, which does not equal $P(A\cap B)$.
No, dependent because $P(A)P(B)=0.25\times0.30=0.75$, which is not $P(A\cap B)$.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = 0.25 × 0.30 = 0.075. Since this matches the given P(A ∩ B) = 0.075, events A and B are independent. One common misconception is adding P(A) and P(B) instead of multiplying, which doesn't test independence and can mislead. Always compute the product and compare it to P(A ∩ B) to determine independence reliably. Avoid assuming independence just because the clubs seem separate; let the probabilities decide.
A student is selected at random from a large group. Let event $A$ be “the student prefers tea” and event $B$ be “the student prefers working in the morning.” Given that $P(A)=0.45$, $P(B)=\tfrac{1}{3}$, and $P(A\cap B)=0.12$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
Yes, independent because $P(A)+P(B)=0.45+\tfrac{1}{3}=0.78$, and that is greater than $P(A\cap B)$.
No, dependent because $P(A)P(B)=0.45\times\tfrac{1}{3}=0.15$, which does not equal $P(A\cap B)=0.12$.
Yes, independent because $P(A)P(B)=0.45\times0.33=0.12$, which equals $P(A\cap B)$.
No, dependent because $P(A\cap B)$ is smaller than both $P(A)$ and $P(B)$, so they cannot be independent.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = 0.45 × (1/3) ≈ 0.45 × 0.333 = 0.15. Since this does not match the given P(A ∩ B) = 0.12, events A and B are dependent. One misconception is comparing sums or assuming independence if P(A ∩ B) is smaller, but that's not the test. Always compute the product and compare it directly to P(A ∩ B) for accuracy. Avoid relying on preferences seeming unrelated; the numbers show dependence.
A quality-control team inspects items from a production line. Let event $A$ be “a randomly selected item is scratched” and event $B$ be “a randomly selected item has a loose label.” Given that $P(A)=0.20$, $P(B)=\tfrac{3}{10}$, and $P(A\cap B)=0.06$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
No, dependent because $P(A)+P(B)=0.20+0.30=0.50$, which does not equal $P(A\cap B)$.
Yes, independent because the events seem unrelated in a factory setting, so $P(A\cap B)$ should match.
No, dependent because $P(A)P(B)=0.20\times0.30=0.60$, which is not $P(A\cap B)$.
Yes, independent because $P(A)P(B)=0.20\times0.30=0.06$, which equals $P(A\cap B)$.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = 0.20 × 0.30 = 0.06. Since this matches the given P(A ∩ B) = 0.06, events A and B are independent. A common misconception is confusing independence with addition, like adding P(A) and P(B) instead of multiplying, which might lead to incorrect conclusions as seen in some distractors. Always remember to compute the product and compare it directly to P(A ∩ B) for verification. Don't rely on whether the events seem unrelated in context; use the mathematical criterion to confirm independence.
In a quality-control report, event $A$ is “a randomly selected package is underweight” and event $B$ is “a randomly selected package has a torn seal.” Given that $P(A)=0.10$, $P(B)=\tfrac{1}{2}$, and $P(A\cap B)=0.04$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
Yes, independent because $P(A)P(B)=0.10\times0.50=0.04$, which equals $P(A\cap B)$.
No, dependent because $P(A)P(B)=0.10\times0.50=0.05$, which does not equal $P(A\cap B)=0.04$.
No, dependent because $P(A)+P(B)=0.10+0.50=0.60$, which does not equal $P(A\cap B)$.
Yes, independent because $P(A\cap B)$ is less than both $P(A)$ and $P(B)$, so the product rule must hold.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = 0.10 × 0.50 = 0.05. Since this does not match the given P(A ∩ B) = 0.04, events A and B are dependent. A common distractor is adding probabilities or assuming independence if the intersection is smaller, but that's not the criterion. Always compute the product and compare it directly to P(A ∩ B) for verification. Rely on the math, not on whether package issues seem connected.
A spinner is spun once. Let event $A$ be “the spinner lands on blue” and event $B$ be “the spinner lands on an even-numbered section.” Given that $P(A)=0.40$, $P(B)=\tfrac{1}{2}$, and $P(A\cap B)=0.18$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
No, dependent because $P(A)P(B)=0.40\times0.50=0.20$, which does not equal $P(A\cap B)=0.18$.
Yes, independent because $P(A)+P(B)=0.40+0.50=0.90$, and $0.18$ is less than $0.90$.
No, dependent because $P(A)$ and $P(B)$ are close in value, so they cannot be independent.
Yes, independent because $P(A)P(B)=0.40\times0.50=0.18$, which equals $P(A\cap B)$.
Explanation
The concept here is the independence of events, which means that the occurrence of one event does not affect the probability of the other. Two events A and B are independent if the probability of both happening, P(A ∩ B), equals the product of their individual probabilities, P(A) × P(B). In this problem, calculate P(A) × P(B) = 0.40 × 0.50 = 0.20. Since this does not match the given P(A ∩ B) = 0.18, events A and B are dependent. A tempting distractor might involve adding probabilities instead of multiplying, or assuming equality based on close values, but that's incorrect. Always compute the product and compare it directly to P(A ∩ B) for accuracy. Don't rely on intuition about the spinner; the numbers tell the true story of dependence.
A school survey tracks two categories for a randomly selected student. Let $A$ be “the student takes band” and $B$ be “the student plays a school sport.” Given that $P(A)=0.28$, $P(B)=\tfrac{1}{2}$, and $P(A\cap B)=0.14$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
Yes; band and sports are different activities, so they must be independent.
No; $P(A)P(B)=0.28\cdot 0.50=0.28$, which is not $0.14$.
Yes; $P(A)+P(B)=0.28+0.50=0.78$, and $P(A\cap B)=0.14$ is less than that.
Yes; $P(A)P(B)=0.28\cdot 0.50=0.14$ and this equals $P(A\cap B)=0.14$.
Explanation
Two events are independent if P(A∩B) = P(A)P(B). This is the standard way to check. Given P(A) = 0.28 and P(B) = 0.50, P(A)P(B) = 0.28 × 0.50 = 0.14. This equals P(A∩B) = 0.14, so A and B are independent. Choice A wrongly calculates the product as 0.28 instead of 0.14, a multiplication error. Compute the product step by step to avoid such mistakes. Don't assume independence from different activities—use the probability rule every time.
A spinner has outcomes that sometimes land on blue and sometimes land on a number greater than 4. Let $A$ be “the spinner lands on blue” and $B$ be “the spinner lands on a number greater than 4.” Given that $P(A)=0.40$, $P(B)=\tfrac{1}{2}$, and $P(A\cap B)=0.18$, are $A$ and $B$ independent? Justify using probabilities by comparing $P(A\cap B)$ to $P(A)P(B)$.
Yes; the color and the number are different features, so they must be independent.
No; $P(A)P(B)=0.40\cdot 0.50=0.20$ and $0.20\ne 0.18=P(A\cap B)$.
Yes; $P(A)+P(B)=0.40+0.50=0.90$ and $P(A\cap B)=0.18$ is less than 1.
Yes; $P(A)P(B)=0.40\cdot 0.50=0.18$, matching $P(A\cap B)=0.18$.
Explanation
Independence means that knowing one event occurred doesn't change the probability of the other. The key test is whether P(A∩B) = P(A)P(B). For this spinner, P(A) = 0.40 and P(B) = 0.50, so P(A)P(B) = 0.40 × 0.50 = 0.20. However, this does not equal the given P(A∩B) = 0.18, so A and B are not independent. One tempting distractor is choice C, which miscalculates the product as 0.18 instead of 0.20, confusing the comparison. To avoid errors, always calculate the product yourself and compare it to P(A∩B). Remember, contextual intuition like 'different features' doesn't guarantee independence—use the numbers.