Problem in Multiplication theorem. 
An urn contains $10$ black and $5$ white balls. Two black
  balls are drawn from the urn and one after other.
  $E$ and $F$ denote respectively the events that first and second ball drawn are black.

How do I find $P(F|E)$?
I think $P(F|E) =$ the probability of occurrence of $F$ when $E$ is taken as sample space.
When $E$ took place we had a black ball which should now be sample space for $F$. So, now $P(F|E)$ should be  $1$.
But book says $P(F|E) = 9/14$.
where am I wrong.
In all questions I did in previous exercise of my book I used $P(A|B) =$ Probability of occurrence of $A$ when $B$ is taken as sample space.
So if I already have done event $E$, now I should have only a black ball in my hand. Now $P(F|E)$ should mean I take out next black black ball from my black ball obtained in $E$. This gives probability of $1$.
Another way to find $P(F|E)$ can be by $n\left\lvert F \cap E\right\lvert / n\left\lvert E \right\lvert$ .
Can you please explain using this method?
 A: The urn contained $10$ black and $5$ white balls, then two were drawn without replacement.  

In all questions I did in previous exercise of my book I used $P(A\mid B)=$ Probability of occurrence of $A$ when $B$ is taken as sample space. So if I already have done event $E$, now I should have only a black ball in my hand. Now $P(F|E)$ should mean I take out next black black ball from my black ball obtained in $E$. This gives probability of $1$.

No, the second ball must be drawn from those remaining in the urn.
There are $140$ ways to obtain event $E$:   That is to draw one of the $10$ black balls then draw any of the remaining $14$ balls.   That is your sample space.   $\mathsf P(F\mid E)$ is thus the probability measure that given we have one of these $140$ equally-probable outcomes we also obtain event $F$ - that the second ball is also black.
$$\mathsf P(F\mid E) =\frac{\mathsf P(E\cap F)}{\mathsf P(E)} = \frac{90}{140}$$
A: Close. Conditioning on $E$ means that I tell you on the first draw, you got a black ball. Thus in the next turn, what is the probability that you get a black?
Well, there are only 14 balls left, and 9 of them are black. Thus
$$P(F|E) = \frac{9}{14}$$
using the logic above. You can check by hand using Bayes' rule.
