Respuesta :
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
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
(a). The probability that no one has done one time fling is [tex]\boxed{\bf 0.377}[/tex].
(b). The probability that at least one person has done one time fling is [tex]\boxed{\bf 0.623}[/tex].
(c). The probability that no more than two person has done one time fling is [tex]\boxed{\bf 0.98279}[/tex].
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 [tex]15\%[/tex] adults do one time fling.
Concept used:
The probability of an event [tex]E[/tex] can be calculated as follows:
[tex]\boxed{P(E)=\dfrac{n(E)}{n(S)}}[/tex]
Here, [tex]n(E)[/tex] is the number of favorable outcomes in an event [tex]E[/tex] and [tex]n(S)[/tex] is the number of element in sample space [tex]S[/tex].
The probability of exactly [tex]r[/tex] success in [tex]n[/tex] trial can be expressed as follows:
[tex]\boxed{P(F)=^{n}C_{r}p^{r}q^{n-r}}[/tex]
Here, [tex]p[/tex] is the probability of success in an event and [tex]q[/tex] is the probability of failure.
Calculation:
Part (a):
The probability that adults do one time fling is [tex]0.15[/tex].
The probability that no adult do one time fling is calculated as follows:
[tex]\boxed{1-0.15=0.85}[/tex]
There are [tex]7[/tex] adult friends in a group.
Consider [tex]A[/tex] as an event that no one has done one time fling in a group of seven friends and [tex]P(A)[/tex] as the probability of an event [tex]A[/tex].
The probability [tex]P(A)[/tex] can be calculated as follows:
[tex]\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}[/tex]
Therefore, the probability [tex]P(A)[/tex] is [tex]\boxed{\bf 0.377}[/tex].
Part (b):
Consider [tex]A'[/tex] as a complement event of an event [tex]A[/tex].
The complement of event [tex]A[/tex] is the event that at least one person has done one time fling in a group of [tex]7[/tex] friends.
The probability of event [tex]A'[/tex] can be calculated as follows:
[tex]\boxed{P(A')=1-P(A)}[/tex] …… (1)
Substitute [tex]P(A)=0.377[/tex] in the equation (1) to obtain the probability as follows:
[tex]\begin{aligned}P(A')&=1-0.377\\&=0.623\end{aligned}[/tex]
Therefore, the probability that at least one person has done one time fling in a group of [tex]7[/tex] friends is [tex]\boxed{\bf 0.623}[/tex].
Part (c):
The probability that no one has done one time fling is [tex]0.377[/tex].
Consider [tex]B[/tex] as an event that one person has done one time fling in a group of seven friends and [tex]P(B)[/tex] as the probability of an event [tex]B[/tex].
The probability [tex]P(B)[/tex] can be calculated as follows:
[tex]\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}[/tex]
Consider [tex]C[/tex] as an event that exactly two person has done one time fling in a group of seven friends and [tex]P(C)[/tex] as the probability of an event [tex]C[/tex].
The probability [tex]P(C)[/tex] can be calculated as follows:
[tex]\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}[/tex]
Now, the probability that no more than two person has done one time fling in a group of [tex]7[/tex] friends can be calculated as follows:
[tex]\boxed{P(X)=P(A)+P(B)+P(C)}[/tex] …… (2)
Substitute [tex]P(A)=0.377[/tex], [tex]P(B)=0.396[/tex] and [tex]P(C)=0.209[/tex] in the equation (1) to obtain the probability [tex]P(X)[/tex] as follows:
[tex]\begin{aligned}P(X)&=0.377+0.396+0.20979\\&=0.98279\end{aligned}[/tex]
Therefore, the probability that no more than two person has done one time fling is [tex]\boxed{\bf 0.98279}[/tex].
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).