Probability and understanding solve If $P(A) = 2/5$ and $P(B)=1/3$ and $P(A \cup B) = 1/2$, find: 


*

*$P(A \cap B)$ 
Well i reasoned that I could make $2/5 + 1/3 = 1/2$
put them all same denominator like 
$12/30 +10/30 = 1/2$ which i know would be $15/30 $
I found that $22/30$ is over half by $7/30$ so that must be my intersection that was counted twice. 
like the formula $P(A) + P(B) - P(A \cap B) $

*$P(B|A)$ (Is this read probability of B given A?) 
So I think I take the intersection of them both that I found in question 1 and divide it by the restricted to A. 
I found this formula online $P(B∣A):=P(A \cap B)/P(A)$, 
So I did this: 
$7/30$ divided by $2/5$ is the same as $7/30 \cdot 5/2 = 35/60 = 7/12$ 

*$P(A|B) $
So I think I do it the same way as above. 
7/30 divided by $P(B) $
= $7/30$ divided by $1/3$ is the same as $7/30 \cdot 3/1 = 21/30 = 7/10$

*Are A and B independent events? 
I have no idea. 
when they are separate on my Venn diagram it seems so, when you start saying given I am lost. 
I am grateful for any help. 
If you do decide to help, please provide explanations as well. 
thank you

Let me take these point by point:


*

*Yes, it is true that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$; I don't like your reasoning for it, since you claimed that two numbers are equal that aren't equal, but you've got basically the right idea.  You can use this to solve for $P(A\cap B)$.

*Yes, $P(B\mid A)$ is read as the probability of $B$ given $A$, or $B$ conditioned on $A$. Your formula is correct.

*Yes.

*Let's think about what it means for two events to be independent; there are a few equivalent ways to think about it.  The most common one is that $A$ and $B$ are independent of $P(A\cap B)=P(A)\cdot P(B)$; which, you can check in this case, is not true.  Another way, more in line with what you've already solved in the problem, is that $A$ and $B$ are independent if $P(A\mid B)=P(A)$ and vice-versa (provided your events don't have probability $0$).  In other words: they are independent if knowing whether or not one of them happens doesn't affect the likelihood of the other happening.
You can see that this should be the case as a consequence of the first definition; if $A$ and $B$ are independent, then
$$
P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)\cdot P(B)}{P(B)}=P(A).
$$
