Question

The one-time fling! Have you ever purchased an article of clothing (dress, sports jacket, etc.), worn the item once to a party, and then returned the purchase? This is called a one-time fling. About 15% of all adults deliberately do a one-time fling and feel no guilt about it! In a group of seven adult friends, find the following probabilities.. (a) No one has done a one-time fling. (Use 3 decimal places.). (b) At least one person has done a one-time fling. (Use 3 decimal places.). . ( c) No more than two people have done a one-time fling. (Use 3 decimal places.)

Answer 1

Using the binomial probability formula

b(x; n, p) = p^x * nCx * (1-P)^(n-x)

Where: 
x = flings 
n = sample size 

A) b(0; 8, .15)
= (0.15)^0 * (8C0) * (.85^8) 
= 1 * 1 *.272 
= .272 

B) = 1 - A
= 1 - 0.272
= 0.728

C) b(1; 8, .15)
= (.15^1) * (8C1) * (.85^7 )
= .385 

b(2; 8, .15)
= (.15^2) * (8C2) * (.85^6) 
= .238 

b ≤ 2 = A + b1 + b2
= .272 + .385 + .238 
= .895

Answer 2

(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: brainly.com/question/1852063

2. Learn more about problem on function 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).