based on a poll, among adults who regret getting tattoos, 15% say that they were too young when they got…

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.\na. find the probability that none of \n0.2316 (round to four decimal pla\nat most 1\nthat they were too young to get tattoos\nb. find the probability that exactly 0\nless than 1\nsays that he or she was too young to get tattoos\n0.3679 (round to four decimal pla\nat least 1\nc. find the probability that the numb\n0.5995 (round to four decimal pla\nexactly 1\nming they were too young is 0 or 1\nd. if we randomly select nine adults,\nmore than 1\nnumber who say that they were too young to get tattoos?\nno, because the probability that \nof the selected adults say that they were too young is 0.05
Answer
Explanation:
Step1: Identify the probability of success and number of trials
The probability $p$ that an adult who regrets getting a tattoo says they were too young is $p = 0.15$, and the number of trials $n=9$.
Step2: Use the binomial - probability formula $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$
a. Probability that none of them say they were too young ($k = 0$)
$P(X = 0)=C(9,0)\times(0.15)^{0}\times(1 - 0.15)^{9-0}$ $C(9,0)=\frac{9!}{0!(9 - 0)!}=1$, $(0.15)^{0}=1$, $(1 - 0.15)=0.85$ $P(X = 0)=1\times1\times(0.85)^{9}\approx0.2316$
b. Probability that exactly one says they were too young ($k = 1$)
$P(X = 1)=C(9,1)\times(0.15)^{1}\times(1 - 0.15)^{9 - 1}$ $C(9,1)=\frac{9!}{1!(9 - 1)!}=\frac{9!}{1!8!}=9$ $P(X = 1)=9\times0.15\times(0.85)^{8}\approx0.3679$
c. Probability that the number is 0 or 1
$P(X=0\ or\ X = 1)=P(X = 0)+P(X = 1)$ $P(X=0\ or\ X = 1)\approx0.2316+0.3679 = 0.5995\approx0.5996$
d. To determine if it is unusual
The probability that more than 1 says they were too young is $P(X>1)=1 - P(X\leq1)=1 - 0.5996 = 0.4004>0.05$ So, it is not unusual. The probability that more than 1 of the selected adults say that they were too young is greater than 0.05.
Answer:
a. $0.2316$ b. $0.3679$ c. $0.5996$ d. more than 1; greater than