Statistics Quiz
Practice Applying The Addition Rule For Probability in Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A factory checks products for two features. Event A is that a randomly chosen product is waterproof, and event B is that it is scratch-resistant. Given P(A)=0.52, P(B)=0.41, and P(A∩B)=0.18, what is the probability that A or B occurs (inclusive or: waterproof, scratch-resistant, or both)?
This quiz focuses on Applying The Addition Rule For Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.
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.
A factory checks products for two features. Event A is that a randomly chosen product is waterproof, and event B is that it is scratch-resistant. Given P(A)=0.52, P(B)=0.41, and P(A∩B)=0.18, what is the probability that A or B occurs (inclusive or: waterproof, scratch-resistant, or both)?
Explanation: This problem requires the Addition Rule to find the probability that a product has at least one of two features. Adding P(A) = 0.52 and P(B) = 0.41 yields 0.93, but this counts products with both features twice. We subtract P(A∩B) = 0.18 to eliminate this double-counting. The formula gives us P(A∪B) = P(A) + P(B) - P(A∩B) = 0.52 + 0.41 - 0.18 = 0.75. This represents the probability that a product is waterproof, scratch-resistant, or both. The incorrect answer 0.93 results from forgetting to subtract the overlap. When solving these problems, sketch a mental Venn diagram to remember that the intersection must be subtracted once.
In a student club poll, event A is “a student is in the robotics club” and event B is “a student is in the debate team.” The poll shows P(A)=0.25, P(B)=0.18, and P(A∩B)=0.05. What is the probability that a student is in robotics or debate (inclusive or)?
Explanation: This question tests the Addition Rule for finding the probability of the union of two events, P(A ∪ B). Simply adding P(A) and P(B) would double-count the students in both robotics and debate, overestimating the total. Subtracting P(A ∩ B) corrects for this overlap, ensuring each student is counted only once. Using the given values, P(A ∪ B) = 0.25 + 0.18 - 0.05 = 0.38. This correct answer of 0.38 represents the inclusive probability of being in robotics or debate or both. A common mistake is forgetting to subtract the intersection, which would give 0.43, but that's too high because of the double-counting. To avoid this, mentally sketch a Venn diagram to visualize the overlap and remember to subtract it.
In a survey about commuting, event A is “a person uses public transit at least once a week” and event B is “a person bikes at least once a week.” The survey reports P(A)=0.40, P(B)=0.27, and P(A∩B)=0.10. What is P(A∪B), where “or” is inclusive (transit or biking or both)?
Explanation: This question tests the Addition Rule for finding the probability of the union of two events, P(A ∪ B). Simply adding P(A) and P(B) would double-count the people who use both transit and bike weekly, overestimating the total. Subtracting P(A ∩ B) corrects for this overlap, ensuring each person is counted only once. Using the given values, P(A ∪ B) = 0.40 + 0.27 - 0.10 = 0.57. This correct answer of 0.57 represents the inclusive probability of using transit or biking or both. A common mistake is forgetting to subtract the intersection, which would give 0.67, but that's too high because of the double-counting. To avoid this, mentally sketch a Venn diagram to visualize the overlap and remember to subtract it.
At a university, event A is “a randomly selected student is majoring in biology” and event B is “a randomly selected student is taking calculus this term.” The registrar reports P(A)=0.30, P(B)=0.42, and P(A∩B)=0.12. What is the probability that a student is majoring in biology or taking calculus (inclusive or)?
Explanation: This question tests the Addition Rule for finding the probability of the union of two events, P(A ∪ B). Simply adding P(A) and P(B) would double-count the students who are both majoring in biology and taking calculus, overestimating the total. Subtracting P(A ∩ B) corrects for this overlap, ensuring each student is counted only once. Using the given values, P(A ∪ B) = 0.30 + 0.42 - 0.12 = 0.60. This correct answer of 0.60 represents the inclusive probability of majoring in biology or taking calculus or both. A common mistake is forgetting to subtract the intersection, which would give 0.72, but that's too high because of the double-counting. To avoid this, mentally sketch a Venn diagram to visualize the overlap and remember to subtract it.
In a school survey, event A is “a randomly selected student plays a sport” and event B is “a randomly selected student is in the band.” The survey reports P(A)=0.55, P(B)=0.40, and P(A∩B)=0.18. What is P(A∪B), where “or” is inclusive (sport or band or both)?
Explanation: This question tests the Addition Rule for finding the probability of the union of two events, P(A ∪ B). Simply adding P(A) and P(B) would double-count the students who both play a sport and are in the band, overestimating the total. Subtracting P(A ∩ B) corrects for this overlap, ensuring each student is counted only once. Using the given values, P(A ∪ B) = 0.55 + 0.40 - 0.18 = 0.77. This correct answer of 0.77 represents the inclusive probability of a student playing a sport or being in the band or both. A common mistake is forgetting to subtract the intersection, which would give 0.95, but that's too high because of the double-counting. To avoid this, mentally sketch a Venn diagram to visualize the overlap and remember to subtract it.
A factory tracks two types of minor defects. Event A is “a randomly selected item has a scratch,” and event B is “a randomly selected item has a dent.” The rates are P(A)=0.18, P(B)=0.14, and P(A∩B)=0.06. What is the probability that an item has a scratch or a dent (inclusive)?
Explanation: This question tests the Addition Rule for finding the probability of a scratch or dent (inclusive). Simply adding P(A) and P(B) would double-count items with both defects. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 0.18 + 0.14 - 0.06 = 0.26. This works because it includes all scratched items, all dented items, but counts those with both only once. A common mistake is forgetting to subtract, which would give 0.32, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap.
A streaming service surveys users. Event A is “a user watches comedy,” and event B is “a user watches documentaries.” The survey finds P(A)=0.46, P(B)=0.38, and P(A∩B)=0.16. What is the probability that a user watches comedy or documentaries (inclusive)?
Explanation: This question tests the Addition Rule for finding the probability of watching comedy or documentaries (inclusive). Simply adding P(A) and P(B) would double-count users who watch both. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 0.46 + 0.38 - 0.16 = 0.68. This works because it includes all comedy watchers, all documentary watchers, but counts those who watch both only once. A common mistake is forgetting to subtract, which would give 0.84, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap.
In a student survey, event A is “a randomly selected student plays a school sport” and event B is “a randomly selected student is in the school band.” The survey results show P(A)=0.40, P(B)=0.35, and P(A∩B)=0.15. What is P(A∪B), where or is inclusive (A, B, or both)?
Explanation: This question tests the Addition Rule for finding the probability of A or B (inclusive). Simply adding P(A) and P(B) would double-count the students who both play a sport and are in the band. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 0.40 + 0.35 - 0.15 = 0.60. This works because it includes everyone in A, everyone in B, but counts the intersection only once. A common mistake is forgetting to subtract, which would give 0.75, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap.
A company inspects products. Event A is “a randomly selected product has feature X,” and event B is “a randomly selected product has feature Y.” The data show P(A)=0.55, P(B)=0.30, and P(A∩B)=0.20. What is the probability that the product has feature X or feature Y (inclusive)?
Explanation: This question tests the Addition Rule for finding the probability of feature X or Y (inclusive). Simply adding P(A) and P(B) would double-count the products with both features. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 0.55 + 0.30 - 0.20 = 0.65. This works because it includes all products with X, all with Y, but counts those with both only once. A common mistake is forgetting to subtract, which would give 0.85, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap.
From a standard 52-card deck, event A is “the card is red,” and event B is “the card is a king.” You are given P(A)=5226, P(B)=524, and P(A∩B)=522. What is P(A∪B), where or is inclusive?
Explanation: This question tests the Addition Rule for finding the probability of a red card or king (inclusive). Simply adding P(A) and P(B) would double-count the red kings. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 26/52 + 4/52 - 2/52 = 28/52. This works because it includes all red cards, all kings, but counts the red kings only once. A common mistake is forgetting to subtract, which would give 30/52, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap.
A market research poll measures opinions. Event A is “a respondent supports Proposal X,” event B is “a respondent supports Proposal Y.” The poll finds P(A)=0.41, P(B)=0.37, and P(A∩B)=0.19. What is the probability that A or B occurs (inclusive “or”)?
Explanation: This problem tests the Addition Rule. When we want P(A or B), adding P(A) = 0.41 and P(B) = 0.37 gives 0.78, but this counts respondents who support both proposals twice. We must subtract P(A∩B) = 0.19 to eliminate this double-counting. The correct calculation is P(A∪B) = 0.41 + 0.37 - 0.19 = 0.59. This gives the probability that a respondent supports at least one proposal. The common mistake of forgetting to subtract occurs because it's easy to overlook that some respondents support both proposals—visualizing a Venn diagram helps remember that the intersection appears in both P(A) and P(B).
A student survey records club participation. Event A is “a student is in the debate club,” event B is “a student is in the science club.” The survey shows P(A)=0.22, P(B)=0.28, and P(A∩B)=0.08. What is the probability that A or B occurs (inclusive “or”)?
Explanation: This problem applies the Addition Rule for probability. To find P(A or B), we start with P(A) = 0.22 and P(B) = 0.28, which sum to 0.50, but this double-counts students in both debate AND science clubs. We subtract P(A∩B) = 0.08 to remove this overlap. The correct calculation is P(A∪B) = 0.22 + 0.28 - 0.08 = 0.42. This gives the probability that a student participates in at least one of these clubs. The key is recognizing that students in both clubs are counted twice when we add P(A) and P(B)—visualizing a Venn diagram helps avoid the common error of forgetting to subtract the intersection.
In a class, event A is “a randomly selected student is taking Algebra,” and event B is “a randomly selected student is taking Biology.” Suppose P(A)=0.60, P(B)=0.45, and P(A∩B)=0.30. What is the probability that A or B occurs (inclusive “or”)?
Explanation: This problem tests the Addition Rule for probability. Simply adding P(A) = 0.60 and P(B) = 0.45 gives 1.05, which exceeds 1 and signals we've double-counted students taking both Algebra and Biology. We correct this by subtracting P(A∩B) = 0.30, representing the overlap. The proper calculation is P(A∪B) = 0.60 + 0.45 - 0.30 = 0.75. This gives the probability that a student takes at least one of these subjects. The key insight is that students in both classes appear in both P(A) and P(B), so we subtract once to count them correctly—visualizing a Venn diagram makes this clear.
A company reviews device capabilities. Event A is “a device supports 5G,” event B is “a device has at least 256GB of storage.” The review shows P(A)=0.58, P(B)=0.36, and P(A∩B)=0.22. What is the probability that A or B occurs (inclusive “or”)?
Explanation: This problem requires the Addition Rule. When finding P(A or B), we add P(A) = 0.58 and P(B) = 0.36 to get 0.94, but this counts devices with both 5G support AND 256GB+ storage twice. We must subtract P(A∩B) = 0.22 to eliminate this double-counting. The correct calculation is P(A∪B) = 0.58 + 0.36 - 0.22 = 0.72. This gives the probability that a device has at least one of these features. To avoid forgetting the subtraction, visualize a Venn diagram where the overlap region (devices with both features) is included in both P(A) and P(B), requiring us to subtract it once.
At a university, event A is “a randomly selected student lives on campus,” event B is “a randomly selected student has a part-time job.” Suppose P(A)=0.46, P(B)=0.38, and P(A∩B)=0.16. What is the probability that A or B occurs (inclusive “or”)?
Explanation: This question applies the Addition Rule for probability. To find P(A or B), we start with P(A) = 0.46 and P(B) = 0.38, which sum to 0.84, but this double-counts students who both live on campus AND have part-time jobs. We subtract P(A∩B) = 0.16 to remove this overlap. The calculation gives P(A∪B) = 0.46 + 0.38 - 0.16 = 0.68. This represents the probability that a student has at least one of these characteristics. The key insight is that students with both attributes appear in both P(A) and P(B), so we subtract once—mentally sketching a Venn diagram helps remember this crucial step.
A company tracks product features. Event A is “a product has wireless charging,” event B is “a product is water-resistant.” Data show P(A)=0.30, P(B)=0.50, and P(A∩B)=0.18. What is the probability that A or B occurs (inclusive “or”)?
Explanation: This problem requires the Addition Rule for probability. Adding P(A) = 0.30 and P(B) = 0.50 would give 0.80, but this counts products with both features twice. We must subtract P(A∩B) = 0.18 to eliminate the double-counting of products that have both wireless charging and water resistance. The correct calculation is P(A∪B) = 0.30 + 0.50 - 0.18 = 0.62. This represents the probability that a product has at least one of these features. To avoid the common error of forgetting the subtraction, sketch a mental Venn diagram showing how the overlap region is included in both P(A) and P(B).
In a school survey, event A is “a randomly selected student plays a sport,” event B is “a randomly selected student is in the music program.” The survey results show P(A)=0.55, P(B)=0.40, and P(A∩B)=0.20. What is the probability that A or B occurs (inclusive “or,” meaning A, B, or both)?
Explanation: This question tests the Addition Rule for probability. When we want P(A or B), simply adding P(A) = 0.55 and P(B) = 0.40 gives 0.95, but this double-counts students who both play sports AND are in music. To correct this overlap, we subtract P(A∩B) = 0.20, which represents students counted twice. The calculation is P(A∪B) = 0.55 + 0.40 - 0.20 = 0.75. This gives us the true probability that a student plays sports, is in music, or does both. A common mistake is forgetting to subtract the intersection—visualizing a Venn diagram helps remember that the overlapping region gets counted twice when we add P(A) and P(B).
From a standard 52-card deck, event A is “drawing a heart” and event B is “drawing a face card (J, Q, or K).” You are told P(A)=5213, P(B)=5212, and P(A∩B)=523. What is P(A∪B) (inclusive or)?
Explanation: This question tests the Addition Rule for probability with a card deck scenario. When finding P(A or B) where A is drawing a heart and B is drawing a face card, we start by adding P(A)=5213 and P(B)=5212 to get 5225. However, this double-counts the cards that are both hearts AND face cards (Jack, Queen, King of hearts). We subtract P(A∩B)=523 to remove this overlap. The calculation is P(A∪B)=5213+5212−523=5222. This correctly accounts for all hearts plus all face cards, counting the three heart face cards only once. The key insight is recognizing that simply adding probabilities overcounts the intersection; always remember to subtract P(A∩B) when using the Addition Rule.
A survey asks about exercise. Event A is “a respondent runs at least once per week” and event B is “a respondent swims at least once per week.” The probabilities are P(A)=0.36, P(B)=0.29, and P(A∩B)=0.11. What is the probability that A or B occurs (inclusive or)?
Explanation: This question tests the Addition Rule for probability with exercise habits. To find P(A or B) where A is running weekly and B is swimming weekly, we add P(A) = 0.36 and P(B) = 0.29 to get 0.65. This double-counts people who do both activities, so we must subtract the overlap. We subtract P(A∩B) = 0.11 to remove this double-counting. The calculation is P(A∪B) = 0.36 + 0.29 - 0.11 = 0.54. This gives us the correct probability that a respondent does at least one of these exercises weekly. Remember that the Addition Rule requires subtracting the intersection to avoid counting the overlapping region twice; visualizing a Venn diagram helps reinforce this concept.
At a university, event A is “a student is majoring in Biology” and event B is “a student is taking Calculus this term.” The probabilities are P(A)=0.28, P(B)=0.50, and P(A∩B)=0.12. What is the probability that A or B occurs (inclusive or)?
Explanation: This question tests the Addition Rule for probability in an academic context. To find P(A or B) where A is majoring in Biology and B is taking Calculus, we start with P(A) = 0.28 and P(B) = 0.50, which sum to 0.78. However, this double-counts Biology majors who are also taking Calculus. We subtract P(A∩B) = 0.12 to correct for this overlap. The calculation is P(A∪B) = 0.28 + 0.50 - 0.12 = 0.66. This gives us the correct probability that a student is either a Biology major, taking Calculus, or both. The common mistake is forgetting to subtract the intersection; mentally sketching a Venn diagram helps visualize why the overlapping region must be removed from the sum.