Question

Suppose that 32% of people have a dog, 27% of people have a cat, and 12% have both.

a) What is the probability that someone owns a dog but not a cat?

b) What is the probability that two independently selected people have cats?

c) What is the probability that two independently selected people have a cat or dog but not both?

d) What is the probability that someone owns neither a dog nor cat?

Answer 1

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Let A and B are two event A: people have a dog Bi people have a cat PCA) = 0:32 PCB)= 0.27 PLADB)- 012 a) probability that someone owns a dog but not cat is PC ANB() = P(A)- PLANB) = 0°32 - 0:12 = 0.2 b) probability that two independently selected people have cats is PCB) 2. 0.272 = 0.0729 probability that two independent selected people have cat or dog but not both is

PCA)2 + PCB)2 – (PLANB))? - 09322+ 0.242. 0.122 = 0.1609 d) probability that some one owns neigther cat nor dog is PCACNBC) MCA UB)9) = E- PLAUB) EA-PLA)..PCB) + PlanB) SL-0:32-0-27+ 0.12 = 0.53 = 053

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