Mathematics, 24.06.2020 14:01 esilvar4378

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. Complete parts (a) through (d) below. (a) What is the probability that all five have used a computer while under the influence of alcohol? (Round to four decimal places as needed.) (b) What is the probability that at least one has not used a computer while under the influence of alcohol? (Round to four decimal places as needed.) (c) What is the probability that none of the five have used a computer while under the influence of alcohol? (Round to four decimal places as needed.) (d) What is the probability that at least one has used a computer while under the influence of alcohol? (Round to four decimal places as needed.)

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Answer from: isabelacarraler

(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.

(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;

$P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........$

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) = $\binom{5}{5}\times 0.29^{5} \times (1-0.29)^{5-5}$

= $1 \times 0.29^{5} \times 0.71^{0}$

= 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 $\geq$ 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 $\geq$ 1) = 1 - P(X = 0)

= $1-\binom{5}{0}\times 0.71^{0} \times (1-0.71)^{5-0}$

= $1 -(1 \times 1 \times 0.29^{5})$

= 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) = $\binom{5}{0}\times 0.29^{0} \times (1-0.29)^{5-0}$

= $1 \times 1 \times 0.71^{5}$

= 0.1804

(d) The probability that at least one has used a computer while under the influence of alcohol is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1-\binom{5}{0}\times 0.29^{0} \times (1-0.29)^{5-0}$

= $1 -(1 \times 1 \times 0.71^{5})$

= 1 - 0.1804 = 0.8196

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