Question

based on a poll, among adults who regret getting tattoos, 14% say that they were too young when they got their tattoos. assume that ten 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 them say that they were too young to get tattoos. 0.2213 (round to four decimal places)
b. find the probability that exactly one says that he or she was too young to get tattoos. 0.3603 (round to four decimal places)
c. find the probability that the number saying they were too young is 0 or 1. 0.5816 (round to four decimal places)
d. if we randomly select ten adults, what is the maximum usual number who say that they were too young to get tattoos?
no, because the probability that of the selected adults say that they were too young is greater than 0.05

Explanation:

Step1: Identify probability of success and failure

The probability that an adult who regrets getting a tattoo says they were too young ($p$) is $0.14$, and the probability that they don't say so ($q$) is $q = 1 - p=1 - 0.14 = 0.86$. The number of trials $n = 10$.

Step2: Use binomial probability formula for part a

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$. For $k = 0$, $C(10,0)=\frac{10!}{0!(10 - 0)!}=1$. Then $P(X = 0)=C(10,0)\times(0.14)^{0}\times(0.86)^{10}=1\times1\times(0.86)^{10}\approx0.2213$.

Step3: Use binomial probability formula for part b

For $k = 1$, $C(10,1)=\frac{10!}{1!(10 - 1)!}=\frac{10!}{1!9!}=10$. Then $P(X = 1)=C(10,1)\times(0.14)^{1}\times(0.86)^{9}=10\times0.14\times(0.86)^{9}\approx0.3603$.

Step4: Use addition - rule for part c

The probability that $X = 0$ or $X = 1$ is $P(X=0\ or\ X = 1)=P(X = 0)+P(X = 1)$. We know $P(X = 0)\approx0.2213$ and $P(X = 1)\approx0.3603$, so $P(X=0\ or\ X = 1)\approx0.2213 + 0.3603=0.5816$.

Step5: Analyze part d

The probability that $X\gt1$ is $P(X\gt1)=1 - P(X\leq1)=1 - 0.5816 = 0.4184$. Since $0.4184>0.05$, the answer is No.

Answer:

a. $0.2213$
b. $0.3603$
c. $0.5816$
d. No, because the probability that more than 1 of the selected adults say that they were too young is greater than $0.05$.