Find the probability that both adults and more than once per week. The probabili

Question

Find the probability that both adults and more than once per week. The probability that both adults out more than once per week is Find the probability that neither adult out more than once per week. The probability that neither adults out more than once per week. Find the probability that at least one of the two adults out more than once per week is The probability that at least one of the two adults out more than once per week is Which of the events can be considered unusual? Explain. Select all that apply. The event is part (b) is unusual because its probability is less than or equal is 0.05. None of these events are unusual The event is part (a) is unusual because its probability is less than or equal is 0.05.

Explanation / Answer

Since out of 1100 213 dine out so

P(dine out) = 213 /1100

(a)

Both selected adults are independent from each other so probability that both dine out is

P(both dine out) = P(dine out)P(dine out)= (213/1100) (213/1100) = 0.037

(b)

Both selected adults are independent from each other so probability that neither dine out is

[1-P(dine out)][1-P(dine out)]= [1-(213/1100)][ 1-(213/1100)] = 0.650

(c)

The probability that one dine out is

P(one dine out) = C(2,1)[ P(dine out)][1-P(dine out)] = 2 * (213/1100) * (887 /1100) = 0.312

So the probability that at least one dine out is

P(at least one dine out) = P(one dine out) + P(both dine out)=0.312 + 0.037 = 0.349

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