
Jon S. answered 08/09/21
Patient and Knowledgeable Math and English Tutor
Event A = have a dog
Event B = have a cat
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.2 + 0.32 - 0.08 = 0.44
Jerry N.
asked 08/08/21Suppose that 20% of people have a dog, 32% of people have a cat, and 8% of people own both. What is the probability that someone owns a dog or a cat?
The probability of a person having a dog or a cat is _____.
(Type an integer or a decimal.
Jon S. answered 08/09/21
Patient and Knowledgeable Math and English Tutor
Event A = have a dog
Event B = have a cat
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.2 + 0.32 - 0.08 = 0.44
Raymond B. answered 08/09/21
Math, microeconomics or criminal justice
You might draw a Venn diagram, with 2 overlapping circles
the overlapping part is .08
those who own only a dog are .2-.08 = .12
those who own only a cat are .32-.08 = .24
add up .12+.24+.08 = .44 = probability of someone owning a cat or a dog
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