Respuesta :
Answer:
0.0308
Step-by-step explanation:
Let as consider
T : Student lives in state of Tennessee
O : Student lives in state of other states
I : Student is an international students
D: Students live in the dormitories.
From the given information it is clear that
[tex]P(T)=0.6, P(O)=0.3,P(I)=0.1[/tex]
[tex]P(D|T)=0.2, P(D|O)=0.5,P(D|I)=0.8[/tex]
Using conditional probability:
[tex]P(D|I)=\dfrac{P(D\cap I)}{P(I)}[/tex]
[tex]0.8=\dfrac{P(D\cap I)}{0.1}[/tex]
[tex]0.08=P(D\cap I)[/tex]
If a student does not live in the dormitory, then we need to find the probability that he/she is an international student.
We know that
[tex]P(I)=P(D\cap I)+P(D'\cap I)[/tex]
[tex]0.1=0.08+P(D'\cap I)[/tex]
[tex]0.1-0.08=P(D'\cap I)[/tex]
[tex]0.02=P(D'\cap I)[/tex]
The probability that student live in the dormitories is
[tex]P(D)=P(D\cap T)+P(D\cap O)+P(D\cap I)[/tex]
[tex]P(D)=P(D|T)P(T)+P(D|O)P(O)+P(D|I)P(I)[/tex]
[tex]P(D)=(0.2)(0.6)+(0.5)(0.3)+(0.8)(0.1)[/tex]
[tex]P(D)=0.35[/tex]
The probability that student does not live in the dormitories is
[tex]P(D')=1-P(D)\Rightarrow 1-0.35=0.65[/tex]
If a student does not live in the dormitory, then the probability that he/she is an international student is
[tex]P(I|D)=\dfrac{P(D'\cap I)}{P(D')}[/tex]
[tex]P(I|D)=\dfrac{0.02}{0.65}[/tex]
[tex]P(I|D)=0.0308[/tex]
Therefore, the required probability is 0.0308.
