events A,C disjoint, A,B independent, P(A) = 0.5,P(B) = 0.3,

Angela123

Junior Member
Joined
Oct 9, 2008
Messages
54
A sample space S has three events, A, B,and C. The events A and C are disjoint, and the events A and B are independent. In addition, it is the case that
Pr[ A ] = 0.50 Pr[ B ? C ] = 0.05.
Pr[ B ] = 0.30
Pr[ C ] = 0.20
Which of the following represents Pr[ A’ ? B ? C’ ] ?

I used a venn diagram to draw out this problem, but I do not know the difference between disjoint and independent. I know the answer is .1, but I have tried and cannot see where that answer is coming from.
 
Re: Intersections and Unions

Angela123 said:
A sample space S has three events, A, B,and C. The events A and C are disjoint, and the events A and B are independent. In addition, it is the case that
Pr[ A ] = 0.50 Pr[ B ? C ] = 0.05.
Pr[ B ] = 0.30
Pr[ C ] = 0.20
Which of the following represents Pr[ A’ ? B ? C’ ] ?

I used a venn diagram to draw out this problem, but I do not know the difference between disjoint and independent. I know the answer is .1, but I have tried and cannot see where that answer is coming from.

Exclusive (ie. disjoint) events capture the intuition of non-compatible outcomes. Not compatible outcomes cannot happen at the same time. This is not the same as independent outcomes. If A, B are disjoint and you know that A occurred, then you do know a lot about B. Namely you know that B cannot occur. Thus there is an interaction between A and B. Knowing whether A occurred influences chances of B, which is not possible under independence.

Do a google search - you will many more explanation with examples.
 
Re: Intersections and Unions

So from what I read, if they are independent that means that you can't get any information about A from B and vice versa, and if they are disjoint that means that they cannot occur together at the same time. However, how does that affect how I make my venn diagram? I am still getting .25 for my answer.
 
Re: Intersections and Unions

Because A & B are independent P(A^B)=0.15.
So you had better redo the Venn diagram.
I get an answer of 0.1.
 
Re: Intersections and Unions

How did you figure that out that (A?B)=.15? I just can't see where that comes from.
 
Re: Intersections and Unions

Angela123 said:
How did you figure that out that (A?B)=.15? I just can't see where that comes from.
That is what it means to be independent.
If A & B are independent events then \(\displaystyle P\left( {A \cap B} \right) = P(A)P(B)\).

What I don’t understand is: “Why in the world were you asked to do this problem if you know nothing about independent events”?
 
Re: Intersections and Unions

Oh, I see where that .15 came from now. Well, I have a really bad teacher. Tomorrow is the midterm, and even today kids were asking what it meant to be independent, but we never get a straight answer. Thanks for explaining it to me. Is there a rule for disjoint too?
 
Re: Intersections and Unions

Angela123 said:
Oh, I see where that .15 came from now. Well, I have a really bad teacher. Tomorrow is the midterm, and even today kids were asking what it meant to be independent, but we never get a straight answer. Thanks for explaining it to me. Is there a rule for disjoint too?
No! If events A & B are disjoint then \(\displaystyle P\left( A \cap B \right) = 0\).
 
Top