Question

among 19 - to 44 - year - olds, 14% say they have used a computer while under the influence of alcohol. suppose five 19 - to 44 - 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.)

Explanation:

Step1: Define the probability of using a computer under alcohol influence

Let \(p = 0.14\) (probability that a 25 - 44 year - old has used a computer under alcohol influence), and \(n=5\) (number of people).

Step2: Calculate the probability for part (a)

The probability that all \(n = 5\) have used a computer under alcohol influence is given by the binomial probability formula \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\). When \(k = 5\), \(C(5,5)=\frac{5!}{5!(5 - 5)!}=1\). So \(P(X = 5)=1\times(0.14)^{5}\times(1 - 0.14)^{0}\).

$$P(X = 5)=(0.14)^{5}=0.14\times0.14\times0.14\times0.14\times0.14=0.0005$$

Step3: Calculate the probability for part (b)

The probability that at least one has not used a computer under alcohol influence is the complement of the event that all have used. Let \(A\) be the event that all have used. Then \(P(\text{at least one has not used})=1 - P(A)\). Since \(P(A)=(0.14)^{5}\approx0.0005\), \(P(\text{at least one has not used})=1-0.0005 = 0.9995\)

Step4: Calculate the probability for part (c)

When \(k = 0\), \(C(5,0)=\frac{5!}{0!(5-0)!}=1\). Using the binomial formula \(P(X = 0)=1\times(0.14)^{0}\times(1 - 0.14)^{5}\).

$$P(X = 0)=(0.86)^{5}=0.86\times0.86\times0.86\times0.86\times0.86\approx0.4704$$

Step5: Calculate the probability for part (d)

The probability that at least one has used is the complement of the event that none have used. Let \(B\) be the event that none have used. Then \(P(\text{at least one has used})=1 - P(B)\). Since \(P(B)=(0.86)^{5}\approx0.4704\), \(P(\text{at least one has used})=1 - 0.4704=0.5296\)

Answer:

(a) \(0.0005\)
(b) \(0.9995\)
(c) \(0.4704\)
(d) \(0.5296\)