how is the probability of event will be affected by its successive events ?? There is a deck of 52 card.
1 card fall down.
Then we take 2 cards from the rest.
These 2 cards are spade.  
What is the probablity of that fallen card is spade??
I think answer should be $\frac 14$ ... ??
 A: We can exclude the two spade cards. Then we'll have $13-2 = 11$ spade cards in $52-2=50$ card deck. So the probability is:
$$\frac{11}{50}$$
A: Informally, the fact that we picked $2$ spades makes it less likely that the missing card is a spade. Since it is easy to make a mistake, we do a formal conditional probability calculation.
Let $A$ be the event we picked $2$ spades, and $B$ the event the missing card is a spade. We find $\Pr(B|A)$. We have
$$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.$$
We calculate the two probabilities on the right.
The event $A$ can happen in two disjoint ways: (i) the missing card is a spade and we drew $2$ spades or (ii) the missing card is a non-spade and we drew $2$ spades.
For (i), the probability the missing card is a spade is $\frac{1}{4}$. Given this has happened, the probability we drew $2$ spades is $\binom{12}{2}/\binom{51}{2}$. Thus the probability of (i) is 
$$\frac{1}{4}\frac{\binom{12}{2}}{\binom{51}{2}}.$$
The same sort of reasoning shows that the probability of (ii) is 
$$\frac{3}{4}\frac{\binom{13}{2}}{\binom{51}{2}}.$$
Add. We get $\Pr(A)=\frac{1}{4}(300)\frac{1}{\binom{51}{2}}$.  We have already calculated $\Pr(A\cap B)$: it is the probability of (i).
Now we have all the information needed to calculate $\Pr(B|A)$. This simplifies to $\dfrac{11}{50}$.
