Answer:
a) The probability that both adults dine out more than once per week = 0.0253
b) The probability that neither adult dines out more than once per week = 0.7069
c) The probability that at least one of the two adults dines out more than once per week = 0.2931
d) Of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.
Step-by-step explanation:
In a sample of 1200 U.S. adults, 191 dine out at a restaurant more than once per week.
Assuming this sample.is a random sample and is representative of the proportion of all U.S. adults, the probability of a randomly picked U.S. adult dining out at a restaurant more than once per week = (191/1200) = 0.1591666667 = 0.1592
Now, assuming this probability per person is independent of each other.
Two adults are picked at random from the entire population of U.S. adults, with no replacement, thereby making sure these two are picked at absolute random.
a) The probability that both adults dine out more than once per week.
Probability that adult A dines out more than once per week = P(A) = 0.1592
Probability that adult B dines out more than once per week = P(B) = 0.1592
Probability that adult A and adult B dine out more than once per week = P(A n B)
= P(A) × P(B) (since the probability for each person is independent of the other person)
= 0.1592 × 0.1592
= 0.02534464 = 0.0253 to 4 d.p.
b) The probability that neither adult dines out more than once per week.
Probability that adult A dines out more than once per week = P(A) = 0.1592
Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408
Probability that adult B dines out more than once per week = P(B) = 0.1592
Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408
Probability that neither adult dines out more than once per week = P(A' n B')
= P(A') × P(B')
= 0.8408 × 0.8408
= 0.70694464 = 0.7069 to 4 d.p.
c) The probability that at least one of the two adults dines out more than once per week.
Probability that adult A dines out more than once per week = P(A) = 0.1592
Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408
Probability that adult B dines out more than once per week = P(B) = 0.1592
Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408
The probability that at least one of the two adults dines out more than once per week
= P(A n B') + P(A' n B) + P(A n B)
= [P(A) × P(B')] + [P(A') × P(B)] + [P(A) × P(B)]
= (0.1592 × 0.8408) + (0.8408 × 0.1592) + (0.1592 × 0.1592)
= 0.13385536 + 0.13385536 + 0.02534464
= 0.29305536 = 0.2931 to 4 d.p.
d) Which of the events can be considered unusual? Explain.
The event that can be considered as unusual is the event that has very low probabilities of occurring, probabilities of values less than 5% (0.05).
And of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.
Hope this Helps!!!
