(a). The probability that no one has done one time fling is \(\boxed{\bf 0.377}\).
(b). The probability that at least one person has done one time fling is \(\boxed{\bf 0.623}\).
(c). The probability that no more than two person has done one time fling is \(\boxed{\bf 0.98279}\).
Further explanation:
Given:
If a person purchased an article of clothing and worn the cloths once then returned the clothes this is known as one time fling.
About \(15\%\) adults do one time fling.
Concept used:
The probability of an event \(E\) can be calculated as follows:
\(\boxed{P(E)=\dfrac{n(E)}{n(S)}}\)
Here, \(n(E)\) is the number of favorable outcomes in an event \(E\) and \(n(S)\) is the number of element in sample space \(S\).
The probability of exactly \(r\) success in \(n\) trial can be expressed as follows:
\(\boxed{P(F)=^{n}C_{r}p^{r}q^{n-r}}\)
Here, \(p\) is the probability of success in an event and \(q\) is the probability of failure.
Calculation:
Part (a):
The probability that adults do one time fling is \(0.15\).
The probability that no adult do one time fling is calculated as follows:
\(\boxed{1-0.15=0.85}\)
There are \(7\) adult friends in a group.
Consider \(A\) as an event that no one has done one time fling in a group of seven friends and \(P(A)\) as the probability of an event \(A\).
The probability \(P(A)\) can be calculated as follows:
\(\begin{aligned}P(A)&=^7C_{0}\cdot (0.15)^{0} \cdot (0.15)^{7-0}\\&=\dfrac{7!}{0!\cdot 7!}\cdot 1\cdot (0.85)^{7}\\&=1\cdot 1\cdot 0.377\\&=0.377\end{aligned}\)
Therefore, the probability \(P(A)\) is \(\boxed{\bf 0.377}\).
Part (b):
Consider \(A'\) as a complement event of an event \(A\).
The complement of event \(A\) is the event that at least one person has done one time fling in a group of \(7\) friends.
The probability of event \(A'\) can be calculated as follows:
\(\boxed{P(A')=1-P(A)}\) …… (1)
Substitute \(P(A)=0.377\) in the equation (1) to obtain the probability as follows:
\(\begin{aligned}P(A')&=1-0.377\\&=0.623\end{aligned}\)
Therefore, the probability that at least one person has done one time fling in a group of \(7\) friends is \(\boxed{\bf 0.623}\).
Part (c):
The probability that no one has done one time fling is \(0.377\).
Consider \(B\) as an event that one person has done one time fling in a group of seven friends and \(P(B)\) as the probability of an event \(B\).
The probability \(P(B)\) can be calculated as follows:
\(\begin{aligned}P(B)&=^7C_{1}\cdot (0.15)^{1} \cdot (0.85)^{7-1}\\&=\dfrac{7!}{1!\cdot 6!}\cdot (0.15) \cdot (0.85)^{6}\\&=\dfrac{7!}{6!}\cdot 0.15 \cdot 0.37714\\&=7\cdot 0.0565\\&=0.396\end{aligned}\)
Consider \(C\) as an event that exactly two person has done one time fling in a group of seven friends and \(P(C)\) as the probability of an event \(C\).
The probability \(P(C)\) can be calculated as follows:
\(\begin{aligned}P(C)&=^7C_{2}\cdot (0.15)^{2} \cdot (0.85)^{7-2}\\&=\dfrac{7!}{2!\cdot 5!}\cdot (0.15)^{2}\cdot (0.85)^{5}\\&=\dfrac{7\cdot 6}{2}\cdot 0.0225\cdot 0.444\\&=21\cdot 0.00999\\&=0.20979\end{aligned}\)
Now, the probability that no more than two person has done one time fling in a group of \(7\) friends can be calculated as follows:
\(\boxed{P(X)=P(A)+P(B)+P(C)}\) …… (2)
Substitute \(P(A)=0.377\), \(P(B)=0.396\) and \(P(C)=0.209\) in the equation (1) to obtain the probability \(P(X)\) as follows:
\(\begin{aligned}P(X)&=0.377+0.396+0.20979\\&=0.98279\end{aligned}\)
Therefore, the probability that no more than two person has done one time fling is \(\boxed{\bf 0.98279}\).
Learn more:
1. Learn more about problem on numbers: https://brainly.com/question/1852063
2. Learn more about problem on function https://brainly.com/question/3225044
Answer details:
Grade: Senior school
Subject: Mathematics
Chapter: Probability
Keywords: Probability, exact event, sample space, number of element, complement event, success, failure, favorable, trial, one time fling, clothes, person, P(E)=n(E)/n(S).
