In a sample of 1200 U.S. adults, 211 dine out at a resaurant more than once per

Question

In a sample of 1200 U.S. adults, 211 dine out at a resaurant more than once per week. Two U.S. adults are selected at random from the population of all U.S. adults without replacement Assuming the sample is representative of all U.S. adults, complete parts (a) through (d) (a) Find the probability that both adults dine out more than once per week. The probability that both adults dine out more than once per week is (Round to three decimal places as needed.) (b) Find the probability that neither adult dines out more than once per week The probability that neither adult dines out more than once per week is (Round to three decimal places as needed.) (c) Find the probability that at least one of the two adults dines out more than once per week. The probability that at least one of the two adults dines out more than once per week is (Round to three decimal places as needed.) (d) Which of the events can be considered unusual? Explain. Select all that apply. A. The event in part (c) is unusual because its probability is less than or equal to 0.05 B. None of these events are unusual. C. The event in part (a) is unusual because its probability is less than or equal to 0.05. D. The event in part (b) is unusual because its probability is less than or equal to 0.05

Explanation / Answer

(a) Probability that the first adult chosen dine out more than once per week = 211/ 1200 = 0.17583333

Now we are working with not replacement scenario. So since one adult has been taken out who dine out more than once per week. Hence, in that sample now number of adults who dine out more than once per week = 210

and total number adults in the sample will be 1199.

Probability that the second adult chosen dine out more than once per week = 210/ 1199 = 0.175146

So, Probability that the both adults dine out more than once per week = 0.17583333 * 0.175146 = 0.031 (three decimal places)

(b)

Probability that the first adult chosen does not dine out more than once per week = (1200-211)/ 1200 = 0.8241667

Now we are working with not replacement scenario. So since one adult has been taken out who did not dine out more than once per week. Hence, in that sample now number of adults who did not dine out more than once per week = 1200 - 211 - 1 = 988

and total number adults in the sample will be 1199.

Probability that the second adult chosen does not dine out more than once per week = 988/ 1199 = 0.82402

So, Probability that the neither adults dine out more than once per week = 0.8241667 * 0.82402 = 0.679 (three decimal places)

(c)

Probability that the atleast one of the two adults dine out more than once per week = Probability that only one adult dine out more than once per week + Probability that both adults dine out more than once per week

Probability that only one adult dine out more than once per week = Probability that first adult dine out more than once per week *  Probability that second adult does not dine out more than once per week + Probability that first adult does not dine out more than once per week *  Probability that second adult dine out more than once per week

Probability that only one adult dine out more than once per week = (211/1200)*(989/1199) + (989/1200)*(211/1199)

= 0.2900736

In the (a) part we found that Probability that the both adults dine out more than once per week = 0.031

So, Probability that the atleast one of the two adults dine out more than once per week = 0.2900736 + 0.031

= 0.321

(d) None of these events are unusual
Reason - In case of probability, No event under consideration is said to be unusual.

Even an event with probability ZERO is said to be impossible event.

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