Probability: 5 cards drawn at random from a well-shuffled pack of 52 cards A poker hand consists of 5 cards drawn at random from a well-shuffled pack of 52 cards. Then, the probability that a poker hand consists of a pair and a triple of equal face values (for example, 2 sevens and 3 kings or 2 aces and 3 queens, etc.) is?
I'm a bit confused how to go about this.Some hints and suggestions might help!
 A: So a full house is five cards where three are of one rank and two are of another. How many ways can we get three of a kind?
There are 13 ranks and we need to get three out of four cards from that rank: $$\binom{13}{1}\binom{4}{3}$$
Now we need a pair. How many ranks do we have to choose from?
Can you use binomial coefficients as above to find the number of possible poker hands?
Finally, once we have the total number of hands which feature a three of a kind and a pair of another rank, how do we get the probability?
Answer: 

 $$\begin{align}\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}}=\frac{6}{4165} \approx .00144\end{align}$$

A: What is the total number of combinations $5$ cards out of $52$?

 $C(52, 5)$

This will be our denominator.
Now, let's count the number of combinations for the pair and triple separately.
Consider $4$ cards of the same face value. How many combinations of $2$ cards are there?

 $C(4, 2)$

Now consider another $4$ cards of the same face value. How many combinations of $3$ cards are there?

 $C(4, 3)$

How many combinations of a pair and a triple are there, if we only consider the $2$ sets of $4$ cards from above?

 $C(4,2) \cdot C(4, 3)$

So far, we only considered two sets of $4$ cards with equal face values. However, there are $13$ such sets. How do we take this into account?

 For each combination of $2$ of the $13$ sets, we have $C(4,2) \cdot C(4, 3)$  possibilities of a pair and a triple. Therefore, with $13$ sets, we have $C(13, 2) \cdot C(4,2) \cdot C(4, 3)$ possible combinations of a pair and a triple.

Now for each pair of ranks, we can have a pair of cards from the first rank and a triple from the second rank, or a triple from the first rank, and a pair from the second rank. We need to multiply our previous result by $2$ (Thanks to joriki for this step)
Therefore, the final answer is

 $\dfrac {2 \cdot C(13, 2) \cdot C(4,2) \cdot C(4, 3)}{C(52, 5)} \approx 0.00144$

A: So for a hint:
I would use this formula - given two events A and B P(A$\bigcap$B) = P(A) + P(B) + P(A$\bigcup$B). You want the left hand side while A is the event of having a pair and B is the event of having a triple.  
A: Count all the possible hands which satisfy your premise then divide that number by all the possible hands(2598960). 
