Question #335830

The probability that a person, living in a certain city, owns a dog is estimated to be 0.3. Find the probability that the




tenth person randomly interviewed in that city is the fifth one to own a dog

1
Expert's answer
2022-05-04T14:37:10-0400

We assume that persons are independent. We consider 1010 independent random variables XiX_i, i=1,,10i=1,\ldots,10 and each of these variables has Bernoulli distribution. These variables correspond to randomly selected persons. Each variable takes value ('a person does not have a dog') with a probability 0.70.7 and 11 ('a person has a dog') with a probability 0.30.3. Suppose that a1,a2,,a9a_1,a_2,\ldots,a_9 are numbers such that 44 of them are equal to 11 and 55 of them are . The probability is equal to: p=a1,a2,,a9P(X1=a1,X2=a2,,X9=a9,X10=1)p=\sum_{a_1,a_2,\ldots,a_9}P(X_1=a_1,X_2=a_2,\ldots,X_9=a_9,X_{10}=1), where the sum is taken over all numbers a1,a2,,a9a_1,a_2,\ldots,a_9 such that 44 of them are equal to 11 and 55 of them are . We point out that P(X1=a1,X2=a2,,X9=a9,X10=1)=(0.3)5(0.7)5P(X_1=a_1,X_2=a_2,\ldots,X_9=a_9,X_{10}=1)=(0.3)^5(0.7)^5. The sum contains C94=9!5!4!C_{9}^4=\frac{9!}{5!4!} terms. Thus, p=C94(0.3)5(0.7)50.051p=C_9^4(0.3)^5(0.7)^5\approx0.051 (it is rounded to 3 decimal places)

Answer: 0.0510.051(it is rounded to 3 decimal places)


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