How Do You Calculate P(A∩B) in Probability Theory?

[karush]

View attachment 1135
this is an image of the problem as given, mostly to avoid typo's, but (b) seems to contain a notation that I don't recognize.

well for \(\displaystyle (a)\) find \(\displaystyle p(A\cap B)\)

from the counting Formula:
\(\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B)\)

replacing $n$ for $p$ and isolating \(\displaystyle p(A\cap B)\)
\(\displaystyle p(A\cup B)-p(A)-p(B)=p(A\cap B)\)
so..
$1-0.6-0.8=-0.4=p(A\cap B)$

(b) ?

 

[Jameson]

Hi karush,

For (a) you have the right idea but mixed up your signs a bit.

$P[A \cup B]=P[A]+P-P[A \cap B]$. We want to solve for $P[A \cap B]$.

$P[A \cap B]=P[A]+P-P[A \cup B]$.

Can you solve it from there?

(b) I'm going to guess that the "C" means "compliment". Normally we write this as $P[A' \cup B']$ or $P[A^c \cup B^c]$. There are some special and useful rules, called De Morgan's laws, that allow us to manipulate expressions like this. Have you seen these before?

 

[karush]

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

\(\displaystyle 0.4=0.6+0.8-1\)

i am am looking at De Morgan's laws it new to me...

I did see this

https://www.physicsforums.com/attachments/1136

 

[Jameson]

Exactly! :)

Notice that $P[A' \cup B']=P[A \cap B]'=1-P[A \cap B]$ and you'll be done.

 

[karush]

Exactly! :)

Notice that $P[A' \cup B']=P[A \cap B]'=1-P[A \cap B]$ and you'll be done.

\(\displaystyle P[A' \cup B']=P[A \cap B]'=1-P[A \cap B]\)
\(\displaystyle

0.6=1-0.4\)

really that it!