Suppose a recent poll of American Househoolds about pet ownership found that for
ID: 3433120 • Letter: S
Question
Suppose a recent poll of American Househoolds about pet ownership found that for households with one pet, 39% owned a dog, 33% owned a cat, and 7% owned a bird. suppose that three households with one pet are selected randomly and with replacement.
18. [Objective: Apply the probability rules] What is the probability that at least one of the three randomly selected households own a dog? (Round to the nearest hundredth) a. 0.27 b. 0.77 c. 0.61 d. 0.70 Suppose a recent poll of American Househoolds about pet ownership found that for households with one pet, 39% owned a dog, 33% owned a cat, and 7% owned a bird. suppose that three households with one pet are selected randomly and with replacement.Explanation / Answer
Atleast one of the three owns a dog
i.e
probability ( 1 person own a dog + 2 persons own a dog + 3 persons own a dog)
= 1 - probability of zero persons owning a dog
hence , select three people from the remaining (100- 39 ) = 61 .... which is in 61 C 3 ways
Total number of ways of selecting 3 people = 100 C 3 ways
= 1 - (61 C3 / 100 C3)
=1 - 0.222
=0.777
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