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1. ## Marcel Q11.15

Hello guys:
Problem 11.15
Among undergraduate students living on a college campus, 20% have an
automobile. Among undergraduate students living off campus, 60% have an
automobile. Among undergraduate students, 30% live on campus. Give the
probabilities of the following events when a student is selected at random:
(a) Student lives off campus
(b) Student lives on campus and has an automobile
(c) Student lives on campus and does not have an automobile
(d) Student lives on campus and/or has an automobile
(e) Student lives on campus given that he/she does not have an automobile.

My attempt:

I started by drawing the table below. Where C=campus C`=off-campus A=automobile A`= No automobile.

C C`

A 0.20 0.60

A` 0.10 0.10

I don't see any conditional statement in the question except for (e).

2. Are you asking on how to do these or just commenting?

4. Originally Posted by Dabsy
I don't see a solution so I can't answer why "we" are using conditional statements. Show your solution and if you are not getting the correct answer, we can show you where your logic is potentially flawed.

5. Ok, below is a solution not mine,but in the question to me there is nothing suggesting conditional statement,that's why I'm confused. I started solving the problem;which I gave in the table above, but got wrong answers,except the first one.
We are given: P(A|C) = 0.2 P(A|C') = 0.6 P(C) = 0.3 and P(C') = 0.7

So, P(A)=P(A|C) P(C) + P(A|C') P(C') = 0.48

Then, P(AC) = P(A|C) P(C) = 0.06

Now I am confused by the wording.

The probability that a student lives on campus AND has an automobile would be P(AC) = 0.06.

The probability that a student lives on campus OR has an automobile would be P(AUC) = P(A) + P(C) - P(AC) = 0.72

So I guess the question is referring to P(AUC) = 0.72