Respuesta :
Answer:
(a) The probability that all five have used a computer while under the influence of alcohol is 0.0021.
(b) The probability that at least one has not used a computer while under the influence of alcohol is 0.9979.
(c) The probability that none of the five have used a computer while under the influence of alcohol is 0.1804.
(d) The probability that at least one has used a computer while under the influence of alcohol is 0.8196.
Step-by-step explanation:
We are given that among 25- to 30-year-olds, 29% say they have used a computer while under the influence of alcohol.
Suppose five 25- to 30-year-olds are selected at random.
The above situation can be represented through the binomial distribution;
[tex]P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........[/tex]
where, n = number of trials (samples) taken = Five 25- to 30-year-olds
r = number of success
p = probability of success which in our question is probability that
people used a computer while under the influence of alcohol,
i.e. p = 29%.
Let X = Number of people who used computer while under the influence of alcohol.
So, X ~ Binom(n = 5, p = 0.29)
(a) The probability that all five have used a computer while under the influence of alcohol is given by = P(X = 5)
P(X = 5) = [tex]\binom{5}{5}\times 0.29^{5} \times (1-0.29)^{5-5}[/tex]
= [tex]1 \times 0.29^{5} \times 0.71^{0}[/tex]
= 0.0021
(b) The probability that at least one has not used a computer while under the influence of alcohol is given by = P(X [tex]\geq[/tex] 1)
Here, the probability of success (p) will change because now the success for us is that people have not used a computer while under the influence of alcohol = 1 - 0.29 = 0.71
SO, now X ~ Binom(n = 5, p = 0.71)
P(X [tex]\geq[/tex] 1) = 1 - P(X = 0)
= [tex]1-\binom{5}{0}\times 0.71^{0} \times (1-0.71)^{5-0}[/tex]
= [tex]1 -(1 \times 1 \times 0.29^{5})[/tex]
= 1 - 0.0021 = 0.9979.
(c) The probability that none of the five have used a computer while under the influence of alcohol is given by = P(X = 0)
P(X = 0) = [tex]\binom{5}{0}\times 0.29^{0} \times (1-0.29)^{5-0}[/tex]
= [tex]1 \times 1 \times 0.71^{5}[/tex]
= 0.1804
(d) The probability that at least one has used a computer while under the influence of alcohol is given by = P(X [tex]\geq[/tex] 1)
P(X [tex]\geq[/tex] 1) = 1 - P(X = 0)
= [tex]1-\binom{5}{0}\times 0.29^{0} \times (1-0.29)^{5-0}[/tex]
= [tex]1 -(1 \times 1 \times 0.71^{5})[/tex]
= 1 - 0.1804 = 0.8196
