# Intersections and Unions

• Oct 31st 2010, 09:57 PM
shannu82
Intersections and Unions
For events A and B we have
P(A) =0.3 P(B) = 0.8 P(A∪B) = 0.9
a) Find P(A|B) P(A'∩B) P(B'∪A').
b) Are A and B independent? Why?

Any help would be appreciated.

Here is my response that i tried... Am not sure if its correct
part a)
P(A|B) = (P(A)P(B))/P(B)
=0.3
P(A') 1-P(A) =0.7
P(B') 1-P(B) = 0.2
P(A'∩B) = P(A')P(B)=0.56
P(B'∪A') = P(A')+P(B')=0.9

part b)
Case 1
P(A∩B) = P(A)P(B)=.24
P(A|B) = P(A∩B)/P(B) =rearrange and find P(A∩B) = 0.24
Case 2
P(B|A) = P(B)=0.8
P(B|A) = P(B∩A)/P(A) = (P(B)P(A))/P(A) =.8

Since Case 1&2 are both true A&B are independent.
• Oct 31st 2010, 10:18 PM
harish21
You really neeed to check your class notes or text book.

As a matter of fact, you have been told in one of your previous posts that $P(A|B) = \dfrac{P(A \cap B}{P(B)}$
How did you find $A \cap B$??

You should have definitely covered topics like unions, intersections, complements, and independent events in your class. Without knowing them, you will never be able to do this correctly!
• Oct 31st 2010, 10:21 PM
shannu82
Re Arrange that ... In part a) i had to find P(A|B) and i was given P(B) so using that i solved for P(A∩B).. and we do know that P(A∩B) = P(A)P(B) if your asking about part a) and we are given P(A) and P(B)...
• Oct 31st 2010, 10:25 PM
harish21

to find that, you first need to figure out P(A∩B)

My question is: How did you find P(A∩B)?? For that you need to look at your book/notes.

FYI, P(A|B)=0.3 is not the correct answer.
• Oct 31st 2010, 10:37 PM
shannu82
Thank you for noticing that! I don't know what i was thinking.... Here is the revised version.... Does this make sense?
P(A|B) = P(A∩B)/P(B)
P(A∩B)= P(A∪B)-P(A)-P(B)
=-0.66

Which for part 2 makes it dependent.
• Oct 31st 2010, 10:44 PM
harish21
Quote:

Originally Posted by shannu82
Thank you for noticing that! I don't know what i was thinking.... Here is the revised version.... Does this make sense?
P(A|B) = P(A∩B)/P(B)
P(A∩B)= P(A∪B)-P(A)-P(B)
=-0.66

Which for part 2 makes it dependent

NO.

Although these facts are surely in your textbook, I will spend few precious minutes of my life writing these with a hope that you will be able to learn.

$P(A \cup B) = P(A)+P(B)-P(A \cap B)$

and

Two events are independent if $P(A \cap B) = P(A) \times P(B)$