In SmallTown USA, 65% of the adults own a cat. A local veterinary office sent ou

Question

In SmallTown USA, 65% of the adults own a cat. A local veterinary office sent out postcards that remind people that cats need to be vaccinated. An informal poll showed that of the cat owners, 85% received a postcard while of the non-cat owners, 20% received a postcard. Let C denote the event that a randomly chosen adult in small Town USA is a cat owner. Let R note the event that a random chosen adult in smallTown USA received a postcard from the veterinary office. a. What is Prob(C)? Prob (C bar)? Prob (R|C)? Prob (R|C bar)? b. Briefly explain why Prob (R) Prob (R intersection C) Prob (R intersection C)? c. Using conditional probabilities, compute Prob (R intersection C), Prob (R intersection C bar) and then use part (b) to determine Prob (R). d. Finally, what is the probability that someone who received a postcard is a cat owner?

Explanation / Answer

a) Prob(C) = 0.65, P(C') = 0.35, P(R|C) = 0.85, P(R|C') = 0.2

bThis is using the law of probability, where Law of Total Probability:
If B1,B2,B3 is a partition of the sample space S, then for any event A we have

P(A)= P(A|Bi)P(Bi) = P(ABi). In this case, since the sample space is split into
Prob(R) = Prob(RC) + Prob(RC')

c) Prob(RC) = Prob(R|C)*Prob(C) = 0.85*0.65 = 0.5525
Prob(RC') = Prob(R|C')*Prob(C') = 0.2*0.35 = 0.07
Prob(R) = 0.5525 + 0.07 = 0.6225

d) Prob(C|R) = [P(R|C)*P(C)]/[P(R|C)*P(C) + P(R|C')*P(C')] = [0.5525/[0.5525+0.07] = 0.8876

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