A classic card problem. Two cards are missing from a pack of cards, and one is drawn. The probability that this card is a king is?
My solution so far-
If no king is missing, the probability of drawing a king is $4/50$.
If one king is missing, the probability is $3/50$, and if two are missing, it becomes $2/50$.
Since all these cases are possible, we add them up, to get $9/50$. Am I missing anything here?
 A: Another way to look at this is to think of drawing three cards in a row. The first two will be designated as "missing", and you want to find the probability that the third card is a king. The probability is the same as the probability of any card being a king, which is 1/13.
A: Hint:
You cannot just add them up. 
You have to assign a weight to the result of each case (i.e. $4/50,3/50,\dots$), where the weight is the probability for that case. For instance, is the case that two kings are missing just as likely as the case that only one king is missing? Probably not. Try to work out the probabilities for each scenario.
If it is still causing you trouble, let me know. 
A: If the two cards removed contain no king then the probability of then probability of getting a king from the 50 cards is 
$$ {P_1} = \left( {\frac{{48}}{{52}}} \right)\left( {\frac{{47}}{{51}}} \right)\left( {\frac{4}{{50}}} \right) $$
If the two cards removed contain one king then the probability of then probability of getting a king from the 50 cards is 
$$ {P_2} = 2\left( {\frac{4}{{52}}} \right)\left( {\frac{{48}}{{51}}} \right)\left( {\frac{3}{{50}}} \right) $$
If the two cards removed are both kings then the probability of then probability of getting a king from the 50 cards is 
$$ {P_3} = \left( {\frac{4}{{52}}} \right)\left( {\frac{3}{{51}}} \right)\left( {\frac{2}{{50}}} \right) $$
So the probability that you will get a king from the 50 remaining cards is:
$$ {P_1} + {P_2} + {P_3} = \frac{{10200}}{{132600}} = \frac{1}{{13}} $$
as others have pointed out. 
