Difference between $P(A \cap B)$ and $P(A\mid B)$ We are in a production process where products go through a production line from two different lines, in the process of production.
$A$ : event of going through line 1 
$B$ : event of going through line 2 
It's given that $P(A)=0.2$ and $P(B)=0.8$
Also these production lines might fail and produce defective goods.
$D$ : event of producing defective goods
Now look at the sentence below :
"Production line 1 suffered mechanical difficulty and produced 5% defectives."
Is this $P(D|A)$ or $P(D \cap B)$ I'm not looking for a  mathematical answer here. I'm looking for a intuition. $P(D|A)$ is the probability of being defective given $A$. We see here that $5$% of products produced from line 1 is defective. So Isn't this $P(D|A)$. But looking the other way around : $P(D \cap A)$ means probability of $D$ and $A$ happening together. If $5$% of products are defective in line 1, wouldn't this be $P(D \cap A)?$
 A: $P(D|A)$ is the probability that D happens when we know for sure that A happens. Here we know that if a product goes through line 1 (ie if A), then we have 5% chance of it being deffective, thus $P(D|A) = 5\%$
$P(D\cap A)$  is the probability that given a product, it has gone through line 1 and is deffective.
Mathematically we have $P(D|A)P(A) = P(D\cap A)$, which one can see like this :
$P(D|A)$ is the probability of being deffective only considering products that have gone through line 1, whereas $P(D\cap A)$ is the probability of going through line 1 AND being deffective considering ALL products.
A: Just read them as


*

*$\mathsf P(D\mid A)$ as "the probability for D when A happens."

*$\mathsf P(D\cap A)$ as "the probability for D and A happening."




"Production line 1 suffered mechanical difficulty and produced 5% defectives."


Can we rephrase this into one of the two forms above?   Yes, we can:

The probability/rate for defectives when on production line 1 is $5\%$.

The statement did not tell us the probability for being defective and on line 1.  However, we can obviously find that out.
$$\mathsf P(A\cap D)~=~\mathsf P(A)\mathsf P(D\mid A) ~=~0.2\times 0.05~=~0.01$$
