Question

based on a poll, among adults who regret getting tattoos, 15% say that they were too young when they got their tattoos. assume that nine adults who regret getting tattoos are randomly selected, and find the indicated probability. complete parts (a) through (d) below.
a. find the probability that none of the selected adults say that they were too young to get tattoos.
0.2310 (round to four decimal places as needed.)
b. find the probability that exactly one of the selected adults says that he or she was too young to get tattoos.
0.3679 (round to four decimal places as needed.)
c. find the probability that the number of selected adults saying they were too young is 0 or 1.
0.5985 (round to four decimal places as needed.)
d. if we randomly select nine adults, is 1 a significantly low number who say that they were too young to get tattoos?
no, because the probability that more than 1 of the selected adults say that they were too young is 0.05

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 9$, $p=0.15$, and $1 - p = 0.85$.

Step2: Calculate the probability for part a

For $k = 0$, $C(9,0)=\frac{9!}{0!(9 - 0)!}=1$. Then $P(X = 0)=C(9,0)\times(0.15)^{0}\times(0.85)^{9}=1\times1\times0.23161694\approx0.2316$.

Step3: Calculate the probability for part b

For $k = 1$, $C(9,1)=\frac{9!}{1!(9 - 1)!}=\frac{9!}{1!8!}=9$. Then $P(X = 1)=C(9,1)\times(0.15)^{1}\times(0.85)^{8}=9\times0.15\times0.27249052\approx0.3679$.

Step4: Calculate the probability for part c

$P(X = 0\ or\ X = 1)=P(X = 0)+P(X = 1)=0.2316+0.3679 = 0.5995$.

Step5: Calculate the probability for part d

The probability that more than 1 of the selected adults say they were too young is $P(X>1)=1 - P(X = 0)-P(X = 1)=1 - 0.5995 = 0.4005$. Since $0.4005>0.05$.

Answer:

a. $0.2316$
b. $0.3679$
c. $0.5995$
d. No, because the probability that more than 1 of the selected adults say that they were too young is greater than $0.05$