1. ## 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.

2. 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 $\displaystyle P(A|B) = \dfrac{P(A \cap B}{P(B)}$
How did you find $\displaystyle A \cap B$??

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!

3. 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)...

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.

5. 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.

6. 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.

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

and

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