Basic Multiplication Rule vs Conditional Probability I'm looking at old probability problems and trying to interpret them in different ways to test my understanding.  Given information in this problem: 


*

*probability of being windy = .2

*Probability rain on windy day = .3

*Probability rain on non-windy day = .8


Question: What is probability given day is both windy and rainy? 
I understand that $P(Wind \wedge Rain) = P(W)P(R|W)$ by interpreting the conditional probability as shrinking the outcome space.  Since I'm given the probability of it being windy, I thought I could condition on it like so:
$ P(WR) = P(WR | W) = P(W|W) P(R|W) = $ above answer 
Is this interpretation useful or even accurate?   It seems to return the same answer but this part of the statement bothers me $P(WR) = P(WR | W)$.  
Thank you! 
 A: "$P(WR) = P(WR | W) = P(W|W) P(R|W) =$ above answer" is not accurate


*

*The first equality claims to say that the probability of wind and rain is the probability of wind and rain given wind.  This is not correct as it ignores the possibility of no wind.  A better statement might be $P(WR) = P(WR \mid W) P(W)$

*The second equality claims to say that the probability of wind and rain given wind is the probability of wind given wind multiplied by the probability of rain given wind.  This is correct but is not particularly informative 

*The third equality claims to say that the probability of wind given wind multiplied by the probability of rain given wind is the probability of wind  multiplied by the probability of rain given wind.  This is not correct as $P(W \mid W)=1 \not = 0.2 = P(W)$
$P(WR) = P(W)P(R\mid W)$ is a basic statement of conditional probability and gives you the answer since you know $P(W)=0.2$ and  $P(R\mid W)=0.3$
A: Your instincts serve you well. Usually the notation $P(WR) = P(W \wedge R)$.
In general, $P(WR) = P(W | R) \cdot P (R) = P(R | W) \cdot P (W)$. Moreover, $P(AB|B) = P(A|B)$ since $B$ is already given.
A: We have that
$$0.3=P(R|W)=\dfrac{P(R\cap W)}{P(W)}=\dfrac{P(R\cap W)}{0.2}\implies P(R\cap W)=0.06.$$
