Applying the Addition Rule for Probability - Statistics and Probability

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 $P(A\cap B)=0.18$ , what is the probability that $A$ or $B$ occurs (inclusive or: waterproof, scratch-resistant, or both)?
Answer: 0.75 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.
A streaming service tracks preferences. Event $A$ is "a randomly selected user watches comedy," and event $B$ is "a randomly selected user watches documentaries." The data show $P(A)=0.44$ , $P(B)=0.36$ , and $P(A\cap B)=0.12$ . What is the probability that $A$ or $B$ occurs, where "or" is inclusive (comedy or documentaries or both)?
Answer: 0.68 This streaming service problem tests the Addition Rule for finding users who watch comedy or documentaries or both. Simply adding P(A) = 0.44 and P(B) = 0.36 gives 0.80, but this counts users who enjoy both genres twice. The Addition Rule corrects this by subtracting P(A∩B) = 0.12, representing users who watch both comedy AND documentaries. The calculation is P(A∪B) = 0.44 + 0.36 - 0.12 = 0.68. This gives the true probability of watching at least one of these genres. Remember that viewer preferences often overlap—mentally sketching a Venn diagram helps visualize why we must subtract the intersection to avoid double-counting.
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 $P(A\cap B)=0.22$ . What is the probability that $A$ or $B$ occurs (inclusive "or")?
Answer: 0.72 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 is a commuter," and event $B$ is "a randomly selected student has a part-time job." The registrar reports $P(A)=0.48$ , $P(B)=0.52$ , and $P(A\cap B)=0.30$ . What is $P(A\cup B)$ , where "or" is inclusive (commuter or part-time job or both)?
Answer: 0.70 This problem asks for the probability a student is a commuter or has a part-time job or both, requiring the Addition Rule. Adding P(A) = 0.48 and P(B) = 0.52 gives 1.00, suggesting every student fits one category, but this double-counts students who are both commuters AND have part-time jobs. We subtract P(A∩B) = 0.30 to correct for this overlap. Computing: P(A∪B) = 0.48 + 0.52 - 0.30 = 0.70. This properly accounts for all students with at least one of these characteristics. The key insight is recognizing that many commuter students work part-time, creating substantial overlap—a Venn diagram would show a large intersection requiring subtraction.
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 $P(A\cap B)=0.15$ . What is $P(A\cup B)$ , where or is inclusive (A, B, or both)?
Answer: $0.60$ 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.
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 $P(A\cap B)=0.20$ . What is the probability that $A$ or $B$ occurs (inclusive "or," meaning $A$ , $B$ , or both)?
Answer: 0.75 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).
A survey asks adults whether they support policy X or policy Y. Event $A$ is that a randomly selected respondent supports policy X, and event $B$ is that the respondent supports policy Y. Suppose $P(A)=0.48$ , $P(B)=0.37$ , and $P(A\cap B)=0.20$ . What is the probability that $A$ or $B$ occurs (inclusive or: supports X, supports Y, or both)?
Answer: 0.65 This question applies the Addition Rule to find P(A or B) for overlapping events. Simply adding P(A) = 0.48 and P(B) = 0.37 gives 0.85, but this counts respondents supporting both policies twice. We must subtract P(A∩B) = 0.20 to correct for this double-counting. Using the formula P(A∪B) = P(A) + P(B) - P(A∩B) = 0.48 + 0.37 - 0.20 = 0.65. This gives the probability that a respondent supports policy X, policy Y, or both. The common error of omitting the subtraction would yield 0.85 instead of 0.65. To remember this rule, visualize overlapping circles where the shared region must be counted only once.
A market survey defines event $A$ as "a randomly selected customer has bought brand X in the last month," and event $B$ as "a randomly selected customer has used a coupon in the last month." Suppose $P(A)=0.33$ , $P(B)=0.41$ , and $P(A\cap B)=0.12$ . Using inclusive or, what is $P(A\cup B)$ ?
Answer: 0.62 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 overlap where both events occur. That's why we subtract P(A ∩ B) to correct for that overlap. Here, plug in the values: 0.33 + 0.41 - 0.12 = 0.62. This correct answer of 0.62 represents the probability that a customer bought brand X or used a coupon or both, accounting for those who did both without duplication. A common mistake is forgetting to subtract the intersection, which would give 0.74, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap and ensure it's only counted once.
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 $P(A\cap B)=0.10$ . What is $P(A\cup B)$ , where "or" is inclusive (transit or biking or both)?
Answer: 0.57 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.
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 $P(A)=\tfrac{26}{52}$ , $P(B)=\tfrac{4}{52}$ , and $P(A\cap B)=\tfrac{2}{52}$ . What is $P(A\cup B)$ , where or is inclusive?
Answer: $\tfrac{28}{52}$ 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 company tracks product features. Event $A$ is "a sold laptop has a touchscreen," and event $B$ is "a sold laptop has a backlit keyboard." Suppose $P(A)=0.52$ , $P(B)=0.46$ , and $P(A\cap B)=0.30$ . Using inclusive or, what is $P(A\cup B)$ ?
Answer: 0.68 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 overlap where both events occur. That's why we subtract P(A ∩ B) to correct for that overlap. Here, plug in the values: 0.52 + 0.46 - 0.30 = 0.68. This correct answer of 0.68 represents the probability that a laptop has a touchscreen or backlit keyboard or both, accounting for those with both without duplication. A common mistake is forgetting to subtract the intersection, which would give 0.98, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap and ensure it's only counted once.
A community survey records two activities. Event $A$ is that a randomly selected person volunteers monthly, and event $B$ is that the person donates to charity yearly. Suppose $P(A)=0.28$ , $P(B)=0.62$ , and $P(A\cap B)=0.19$ . What is the probability that $A$ or $B$ occurs (inclusive or: volunteers, donates, or both)?
Answer: 0.71 This problem applies the Addition Rule to find P(volunteers or donates). Adding P(A) = 0.28 and P(B) = 0.62 gives 0.90, but this double-counts people who do both activities. We must subtract P(A∩B) = 0.19 to eliminate this overlap. The calculation is P(A∪B) = P(A) + P(B) - P(A∩B) = 0.28 + 0.62 - 0.19 = 0.71. This represents the probability that a person volunteers monthly, donates yearly, or does both. Forgetting to subtract would incorrectly give 0.90, overestimating the true probability. To remember this principle, sketch a Venn diagram mentally—the overlapping region shows why we must subtract P(A∩B) to avoid counting it twice.
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." Suppose $P(A)=0.40$ , $P(B)=0.35$ , and $P(A\cap B)=0.15$ . Using inclusive or (sport or band or both), what is $P(A\cup B)$ ?
Answer: 0.60 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 overlap where both events occur. That's why we subtract P(A ∩ B) to correct for that overlap. Here, plug in the values: 0.40 + 0.35 - 0.15 = 0.60. This correct answer of 0.60 represents the probability that a student plays a sport or is in the band or both, accounting for those who do both without duplication. A common mistake is forgetting to subtract the intersection, which would give 0.75, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap and ensure it's only counted once.
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 $P(A\cap B)=0.30$ . What is the probability that $A$ or $B$ occurs (inclusive "or")?
Answer: 0.75 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 inspects products for features. Event $A$ is "a randomly selected product is waterproof," and event $B$ is "it has wireless charging." The data show $P(A)=0.55$ , $P(B)=0.40$ , and $P(A\cap B)=0.25$ . What is the probability that $A$ or $B$ occurs, where "or" is inclusive (waterproof or wireless charging or both)?
Answer: 0.70 This problem requires the Addition Rule to find the probability of waterproof or wireless charging or both. Adding P(A) = 0.55 and P(B) = 0.40 gives 0.95, but this counts products with both features twice. We must subtract P(A∩B) = 0.25 to remove the double-counting of products that are both waterproof AND have wireless charging. The calculation is P(A∪B) = 0.55 + 0.40 - 0.25 = 0.70. This correctly accounts for all products with at least one of these features. The key insight is that when events overlap, we need to subtract their intersection—imagine a Venn diagram where the middle section would be counted twice without this correction.
A product quality check defines event $A$ as "an item has a scratch" and event $B$ as "an item has a dent." The inspection data show $P(A)=0.22$ , $P(B)=0.15$ , and $P(A\cap B)=0.06$ . What is the probability that an item has a scratch or a dent (inclusive or)?
Answer: 0.31 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 items that have both a scratch and a dent, overestimating the total. Subtracting P(A ∩ B) corrects for this overlap, ensuring each defective item is counted only once. Using the given values, P(A ∪ B) = 0.22 + 0.15 - 0.06 = 0.31. This correct answer of 0.31 represents the inclusive probability of an item having a scratch or a dent or both. A common mistake is forgetting to subtract the intersection, which would give 0.37, 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 $P(A\cap B)=0.12$ . What is the probability that a student is majoring in biology or taking calculus (inclusive or)?
Answer: 0.60 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.
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 $P(A)=\tfrac{13}{52}$ , $P(B)=\tfrac{12}{52}$ , and $P(A\cap B)=\tfrac{3}{52}$ . What is $P(A\cup B)$ (inclusive or)?
Answer: $\tfrac{22}{52}$ 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 $P(A) = \tfrac{13}{52}$ and $P(B) = \tfrac{12}{52}$ to get $\tfrac{25}{52}$ . However, this double-counts the cards that are both hearts AND face cards (Jack, Queen, King of hearts). We subtract $P(A\cap B) = \tfrac{3}{52}$ to remove this overlap. The calculation is $P(A\cup B) = \tfrac{13}{52} + \tfrac{12}{52} - \tfrac{3}{52} = \tfrac{22}{52}$ . 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 $P(A\cap B)$ when using the Addition Rule.
In a survey of adults, event $A$ is "the respondent drinks coffee daily" and event $B$ is "the respondent drinks tea daily." The results give $P(A)=0.48$ , $P(B)=0.35$ , and $P(A\cap B)=0.20$ . What is $P(A\cup B)$ , where "or" is inclusive (coffee daily or tea daily or both)?
Answer: 0.63 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 respondents who drink both coffee and tea daily, overestimating the total. Subtracting P(A ∩ B) corrects for this overlap, ensuring each respondent is counted only once. Using the given values, P(A ∪ B) = 0.48 + 0.35 - 0.20 = 0.63. This correct answer of 0.63 represents the inclusive probability of drinking coffee or tea daily or both. A common mistake is forgetting to subtract the intersection, which would give 0.83, 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 $P(A\cap B)=0.06$ . What is the probability that an item has a scratch or a dent (inclusive)?
Answer: $0.26$ 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.

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