Applying the Addition Rule for Probability
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Statistics › Applying the Addition Rule for Probability
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)?
0.25
0.31
0.37
0.94
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 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.
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 $P(A\cap B)=0.05$. What is the probability that a student is in robotics or debate (inclusive or)?
0.05
0.33
0.38
0.43
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 $P(A\cap B)=0.10$. What is $P(A\cup B)$, where “or” is inclusive (transit or biking or both)?
0.47
0.57
0.67
0.90
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 $P(A\cap B)=0.12$. What is the probability that a student is majoring in biology or taking calculus (inclusive or)?
0.12
0.48
0.60
0.72
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 $P(A\cap B)=0.18$. What is $P(A\cup B)$, where “or” is inclusive (sport or band or both)?
0.18
0.59
0.77
0.95
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.
At a bookstore, event $A$ is “a customer buys fiction” and event $B$ is “a customer buys nonfiction.” Store data show $P(A)=0.52$, $P(B)=0.33$, and $P(A\cap B)=0.17$. What is the probability that a customer buys fiction or nonfiction (inclusive or)?
0.17
0.51
0.68
0.85
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 customers who buy both fiction and nonfiction, overestimating the total. Subtracting P(A ∩ B) corrects for this overlap, ensuring each customer is counted only once. Using the given values, P(A ∪ B) = 0.52 + 0.33 - 0.17 = 0.68. This correct answer of 0.68 represents the inclusive probability of buying fiction or nonfiction or both. A common mistake is forgetting to subtract the intersection, which would give 0.85, 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)?
$0.26$
$0.06$
$0.12$
$0.32$
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 survey asks respondents about two preferences. Event $A$ is “a respondent prefers tea,” and event $B$ is “a respondent prefers coffee.” The results show $P(A)=0.48$, $P(B)=0.52$, and $P(A\cap B)=0.20$ (respondents who prefer both). What is the probability that a respondent prefers tea or coffee (inclusive)?
$0.20$
$0.60$
$1.00$
$0.80$
Explanation
This question tests the Addition Rule for finding the probability of preferring tea or coffee (inclusive). Simply adding P(A) and P(B) would double-count those who prefer both. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 0.48 + 0.52 - 0.20 = 0.80. This works because it includes everyone who prefers tea, everyone who prefers coffee, but counts those who prefer both only once. A common mistake is forgetting to subtract, which would give 1.00, 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 $P(A\cap B)=0.16$. What is the probability that a user watches comedy or documentaries (inclusive)?
$0.16$
$0.84$
$0.68$
$0.62$
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.
A product review site tracks two labels. Event $A$ is “a randomly selected review mentions ‘easy to use’,” and event $B$ is “a randomly selected review mentions ‘good value’.” The site reports $P(A)=0.33$, $P(B)=0.29$, and $P(A\cap B)=0.11$. What is the probability that a review mentions ‘easy to use’ or ‘good value’ (inclusive)?
$0.51$
$0.40$
$0.11$
$0.62$
Explanation
This question tests the Addition Rule for finding the probability of mentioning ‘easy to use’ or ‘good value’ (inclusive). Simply adding P(A) and P(B) would double-count reviews mentioning both. That's why we subtract P(A∩B) to correct for the overlap. Here, P(A∪B) = 0.33 + 0.29 - 0.11 = 0.51. This works because it includes all reviews with ‘easy to use’, all with ‘good value’, but counts those with both only once. A common mistake is forgetting to subtract, which would give 0.62, overestimating the probability. To avoid this, mentally sketch a Venn diagram to visualize the overlap.