b.) find the probability that neither adults dines out more than once per week.
ID: 3352102 • Letter: B
Question
b.) find the probability that neither adults dines out more than once per week.
the probability that neither adults dines out more than once per week is_____%?
(round to three decimal places as needed)
c.) find the probabilty that at least one of the two adults dines out more than once poer week.
the probability that at least one of the two adults dines out more than once poer week is ____%
(round to three demial places as needed)
d.) which of the events can be considered unusual? explain. select all that apply.
a. THe even in part (c) is unusual because its probability is less than or equal to 0.05
b.the event in part (a) is unusual because its probability is less than or equal to 0.05
c. none of these events are unususal
d. the event in part (b) is unusual because its prbability is less than or equal to 0.05
c.)
In a sample of 1000 U.S. adults, 219 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.)Explanation / Answer
p=219/1000 =0.219
n=2
q=1-p=1-0.219 =0.781
a) P(both ) =P(x=2) =nCr p^r q^(n-r)
=2C2 (0.219)^2 (0.781)^(2-2)
= 1*0.047961 * 0.781^0
=0.048
b) x=0
P(x=0) =nCr p^r q^(n-r)
=2C0 (0.219)^0 (0.781)^(2-0)
=1*1*0.781^2
=0.610
c) P(atleast one ) =1 -P(none)
=1-0.610
=0.39
d) choice B,
the event in part (a) is unusual because its probability is less than or equal to 0.05
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