Question

60% of the student body at UTC is from the state of Tennessee (T), 30% are from other states (O), and the remainder constitutes international students (I). Twenty percent of students from Tennessee lives in the dormitories, whereas, 50% of students from other states live in the dormitories. Finally, 80% of the international students live in the dormitories. Given that a student does not live in the dormitory, what is the probability that he/she is an international student?

Answer 1

0.0308

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

`P(T)=0.6, P(O)=0.3,P(I)=0.1`

`P(D|T)=0.2, P(D|O)=0.5,P(D|I)=0.8`

Using conditional probability:

`P(D|I)=(P(D\cap I))/(P(I))`

`0.8=(P(D\cap I))/(0.1)`

`0.08=P(D\cap I)`

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

`P(I)=P(D\cap I)+P(D'\cap I)`

`0.1=0.08+P(D'\cap I)`

`0.1-0.08=P(D'\cap I)`

`0.02=P(D'\cap I)`

The probability that student live in the dormitories is

`P(D)=P(D\cap T)+P(D\cap O)+P(D\cap I)`

`P(D)=P(D|T)P(T)+P(D|O)P(O)+P(D|I)P(I)`

`P(D)=(0.2)(0.6)+(0.5)(0.3)+(0.8)(0.1)`

`P(D)=0.35`

The probability that student does not live in the dormitories is

`P(D')=1-P(D)\Rightarrow 1-0.35=0.65`

If a student does not live in the dormitory, then the probability that he/she is an international student is

`P(I|D)=(P(D'\cap I))/(P(D'))`

`P(I|D)=(0.02)/(0.65)`

`P(I|D)=0.0308`

Therefore, the required probability is 0.0308.