Events A and B with complements

#1
Hi all,

I will show the quest first and my work after.

Given: P(A) = 0.3, P(B) = 0.8, P(AUB) = 0.9

A) Find P(A|B), P(A'nB), P(B'UA')
B) Are A and B independant? Why?

My Work so far...

I know that P(A') = 0.7, P(B') = 0.2

To find P(A|B) I need to use the formula P(A|B) = P(AnB)/P(B).
I dont have P(AnB) so I must find it first using
P(AUB) = P(A)+P(B)-P(AnB) becomes
P(AnB) = P(A)+P(B)-P(AUB)
P(AnB) = 0.3 + 0.8 - 0.9
P(AnB) = 0.2

Now I can use P(A|B) = P(AnB)/P(B) = 0.2/0.8 = 0.25

Can I use the P(AnB) = P(A)P(B|A) to find P(A'nB)? If I just sub in the complements? Like this? P(A'nB) = P(A')P(B|A') If I can't do this, what should I do?

Finding P(B'UA') is a similar problem. Unforunatly this problem is just using numbers so I cannot think of anything in real life to help me. I also thing drawing a venn diagram may help but I'm not sure where to start.

B) I used the P(AUB) = P(A)+P(B)-P(AnB) formula to find P(AnB) = 0.2 however the events can only be independant if P(A)*P(B) = P(AnB).
Actually doing the multiplication given the starting info 0.3*0.8 = .24 means is not indepentant.

Tjvelcro
 

BGM

TS Contributor
#2
The events [math] \{B, B^c\} [/math] actually forms a partition over the
sample space [math] \Omega [/math].

So [math] P(A) = P(A\cap\Omega) = P(A\cap(B\cup B^c))
= P((A\cap B)\cup(A\cap B^c)) = P(A\cap B) + P(A\cap B^c) [/math]

And also using De Morgan's Law, you can quickly simplify:
[math] P(A^c \cup B^c) = P((A \cap B)^c) [/math]