Is This Conditional Probability Problem asking for Pr(A|B) or Pr(A and B)? A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black? the source
It is the English of this problem that causes me problems.  

The probability of selecting a black marble and then a white marble is 0.34

Shouldn't that be P(White | Black) not P(Black and White) because P(Black and White) = P(White and Black).  $P(B \cap W) = P(W \cap B)$  The reason I think this is because

The probability of selecting a black marble and then a white marble is 0.34 

The site says that means P(Black and White).   I would agree with that if then was removed from the sentence. 
Conditional Probability word problems always confuse me because the wording is always non-standard. 
Is there a standard grammatical syntax for conditional probability problems to distinguish  between P(A and B) and P(A | B).  
I understand P(A and B) means that both events occur while P(A | B) means that given (knowing) event B occurred what is the probability that event A occurs(follows) 
 A: The events labelled "Black" and "White" are really "first marble black" and "second marble white" (I think this is confusing: it would be better to label these as something like "B1" and "W2").  If they just said "a black marble and a white marble" it would also include the case of a white marble first and a black marble second.   
A: There are four possible outcomes, $bb,bw,wb$, and $ww$. The probability of selecting a black marble and then a white marble is the probability of the outcome $bw$. Let $X$ represent the first draw and $Y$ the second; then symbolically this is $\Bbb{P}(X=b\land Y=w)$ (where $\land$ is ‘and’). The word then has to be there in order to distinguish the outcome $bw$ from the outcome $wb$: these are not the same outcome, since we’re keeping track of the order of the draws.
The probability of selecting a black marble and a white marble would be the probability of getting $bw$ or $wb$, which is larger than the probability of getting specifically $bw$, i.e., first a black and then a white marble.
Neither of these is a conditional probability: each is simply the probability of a certain outcome. On the other hand, the probability of selecting a white marble on the second draw, given that the first marble drawn was black is a conditional probability, $\Bbb{P}(Y=w\mid X=b)$. This is signalled by the words given that: the probability of A given B is $\Bbb{P}(A\mid B)$.
Once you’ve correctly interpreted the problem, the rest is straightforward, but I’ll include it for the sake of completeness. The basic formula
$$\Bbb{P}(X=b\land Y=w)=\Bbb{P}(X=b)\Bbb{P}(Y=w\mid X=b)$$
becomes $0.34=0.47\,\Bbb{P}(Y=w\mid X=b)$, so $$\Bbb{P}(Y=w\mid X=b)=\frac{0.34}{0.47}=\frac{34}{47}\approx 0.7234\;.$$
