Number and Algebra
Functions
Geometry and Trigonometry
Statistics and Probability
Calculus
In a survey of 100 people, 70 like tea, 50 like coffee, and 30 like juice. If 20 like both tea and coffee, 15 like both tea and juice, 10 like both coffee and juice, and 5 like all three, how many people like at least one of the beverages?
In a survey, respondents may choose any of three features A,B,CA,B,CA,B,C. The Venn diagram records: AAA only =25=25=25, BBB only =20=20=20, CCC only =15=15=15, A∩BA\cap BA∩B only =10=10=10, A∩CA\cap CA∩C only =8=8=8, B∩CB\cap CB∩C only =5=5=5, all three =2=2=2, and none =40=40=40. Find the total number of respondents.
In a universal set of 200 elements, subsets A,B,CA,B,CA,B,C satisfy ∣A∣=100|A|=100∣A∣=100, ∣B∣=120|B|=120∣B∣=120, ∣C∣=80|C|=80∣C∣=80, ∣A∩B∣=60|A\cap B|=60∣A∩B∣=60, ∣A∩C∣=50|A\cap C|=50∣A∩C∣=50, ∣B∩C∣=40|B\cap C|=40∣B∩C∣=40, and ∣A∩B∩C∣=30|A\cap B\cap C|=30∣A∩B∩C∣=30. How many elements lie in none of the subsets?
Given three events AAA, BBB, and CCC with probabilities P(A)=0.6P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5P(B)=0.5, P(C)=0.4P(C)=0.4P(C)=0.4, P(A∩B)=0.3P(A\cap B)=0.3P(A∩B)=0.3, P(A∩C)=0.2P(A\cap C)=0.2P(A∩C)=0.2, P(B∩C)=0.1P(B\cap C)=0.1P(B∩C)=0.1, and P(A∩B∩C)=0.05P(A\cap B\cap C)=0.05P(A∩B∩C)=0.05, calculate P(A∪B∪C)P(A\cup B\cup C)P(A∪B∪C).
In a sample of 500 outcomes, a Venn diagram for events A,B,CA,B,CA,B,C has counts: only A=150A=150A=150, only B=120B=120B=120, only C=100C=100C=100, A∩BA\cap BA∩B only =50=50=50, A∩CA\cap CA∩C only =30=30=30, B∩CB\cap CB∩C only =20=20=20, and all three =10=10=10. Verify consistency of these counts and find the probability that at least two events occur.
In a universal set of 150 elements, a Venn diagram for events AAA, BBB, and CCC has regions: only A=40A=40A=40, only B=30B=30B=30, only C=20C=20C=20, A∩BA\cap BA∩B only =10=10=10, B∩CB\cap CB∩C only =5=5=5, all three =5=5=5, and A∩CA\cap CA∩C only =x=x=x. If every element lies in at least one of AAA, BBB, or CCC, find xxx.
Given P(A)=0.6P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5P(B)=0.5, P(C)=0.4P(C)=0.4P(C)=0.4, P(A∩B)=0.3P(A\cap B)=0.3P(A∩B)=0.3, P(A∩C)=0.2P(A\cap C)=0.2P(A∩C)=0.2, P(B∩C)=0.15P(B\cap C)=0.15P(B∩C)=0.15, and P(A∩B∩C)=0.1P(A\cap B\cap C)=0.1P(A∩B∩C)=0.1, find the probability that none of the events occurs.
Events A,B,CA,B,CA,B,C satisfy P(A∪B)=0.7P(A\cup B)=0.7P(A∪B)=0.7, P(B∪C)=0.8P(B\cup C)=0.8P(B∪C)=0.8, P(A∪C)=0.75P(A\cup C)=0.75P(A∪C)=0.75, P(A)=0.4P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5P(B)=0.5, P(C)=0.6P(C)=0.6P(C)=0.6, and P(A∪B∪C)=0.9P(A\cup B\cup C)=0.9P(A∪B∪C)=0.9. Find P(A∩B∩C)P(A\cap B\cap C)P(A∩B∩C).
In a class of 120 students, 80 study math, 70 study physics, and 60 study chemistry. There are 50 who study both math and physics, 40 both math and chemistry, 30 both physics and chemistry, and 20 study all three. How many students study exactly two subjects?
Three events A,B,CA,B,CA,B,C are independent with P(A)=0.4P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5P(B)=0.5, P(C)=0.6P(C)=0.6P(C)=0.6. Calculate the probability that exactly one event occurs.
Given P(A)=0.6P(A)=0.6P(A)=0.6, P(B)=0.7P(B)=0.7P(B)=0.7, P(C)=0.5P(C)=0.5P(C)=0.5, P(A∩B)=0.45P(A\cap B)=0.45P(A∩B)=0.45, P(A∩C)=0.35P(A\cap C)=0.35P(A∩C)=0.35, P(B∩C)=0.4P(B\cap C)=0.4P(B∩C)=0.4, determine the range of possible values for P(A∩B∩C)P(A\cap B\cap C)P(A∩B∩C).
Four events A,B,C,DA,B,C,DA,B,C,D satisfy P(A)=0.3P(A)=0.3P(A)=0.3, P(B)=0.4P(B)=0.4P(B)=0.4, P(C)=0.5P(C)=0.5P(C)=0.5, P(D)=0.2P(D)=0.2P(D)=0.2, every pairwise intersection has probability 0.10.10.1, every triple intersection has probability 0.050.050.05, and P(A∩B∩C∩D)=xP(A\cap B\cap C\cap D)=xP(A∩B∩C∩D)=x. If P(at least one of the events)=0.8P(\text{at least one of the events})=0.8P(at least one of the events)=0.8, find xxx.
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Question Type 1: Constructing Venn diagrams to calculate probabilities for 2 events
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Question Type 3: Given a worded problem, converting to a tree diagram to calculate probability of an outcome