Find the probability that one card is king and the other is heart. Two cards are dealt from an ordinary deck of 52 cards (the sampling is without replacement). Find the probability that one card is king and the other is heart.
I'm having trouble figure out how to deal with the case where you pick up a king of hearts. Normally I would just multiple the separate probabilities together but as they are not independent? I'm not entirely sure how to proceed.
 A: Hint:
Try adding up

*

*probability first is King (not Hearts) and second is a Heart (not King)

*probability first is Hearts (not King) and second is a King (not Hearts)

*probability first is King of Hearts and second is a Heart (not King)

*probability first is King of Hearts and second is a King (not Hearts)

*probability first is King (not Hearts) and second is King of Hearts

*probability first is Hearts (not King) and second is King of Hearts

A: Approach this one using the basic definition: $$P(E)= \frac{n(\text{favourable cases})}{n(S)}$$ Here, $$n(S)=^{52}C_2$$ And, $$n(\text{favourable cases})= ^4C_1 \times ^{13}C_1-1$$ Explanation: In the above equation we choose one card out of four Kings, then one card out of $13$ Hearts, and then we subtract one for the one case in which both the chosen cards are the King of Hearts, since that isn't possible, but it creeps into our calculation.
To answer your question, a selection like (King of Hearts, $2$ of Hearts) is valid. And so is (King of Spades, King of Hearts), but not (King of Hearts, Ace of Spades)
Thus, your answer is simply $$\frac{^4C_1 \times ^{13}C_1-1}{^{52}C_2} = \frac{1}{26}$$
A: Upayan De's answers is slick, but maybe at tis stage you should prioritize structure over elegance.
Here is the "usual" way of attack
$\fbox{King of Hearts}\quad\fbox{3 other kings}\quad\fbox{12 hearts, no king}$
If you choose $1$ each from any two boxes, you are done, thus
$\dfrac{\binom11\binom31 + \binom11\binom{12}1 + \binom31\binom{12}1}{\binom{52}2}= \dfrac1{26}$
