Combination - Probability within a deck of cards The situation is that you have a 52 deck of cards, you pull 10 cards out to make a hand. What is the probability that this hand has exactly 5 red cards. I'm not sure whether this question is one that is independent or dependent. When I solve independently (5/10), I get 1/2 which makes sense. When I solve dependently, I get .15 ( I got this from 5*(1/2^5) )
Which makes no sense. Of course solving dependently makes a lot more sense here.
 A: Consider the multiplication principle and assume all "different hands" are equally likely.
What five red cards do you use?  What five nonred cards do you use?

 $\binom{26}{5}$ ways to choose the five red cards.  Same number of ways to choose the five black cards, so there are $\binom{26}{5}\cdot\binom{26}{5}$ different hands with this property total.

How many total different 10card hands are there with no restriction?

 There are $\binom{52}{10}$ number of different 10card poker hands.

So the answer is:

 $$\frac{\binom{26}{5}\cdot \binom{26}{5}}{\binom{52}{10}}$$

A: First off, here is the binomial (number of combinations) formula for your reference.
$${n \choose r} = \frac{n!}{r!(n-r)!}$$
My understanding is that this would be a hypergeometric distribution:
$$h(x,n_{hand}-x;n_{red},N_{total}-n_{red};N_{total},n_{hand})=\frac{{n_{red} \choose {x}}{ N_{total}-n_{red} \choose {n_{hand}-x}}}{N_{total} \choose n_{hand}}$$
Where in your cases, the variables take the following values:
$$N_{total}=52$$
$$n_{hand}=10$$
$$n_{red}=26$$
$$x=5$$
Gives the following probability:
$$\frac{{26 \choose 5}{26 \choose 5}}{52 \choose 10}=0.2735$$
"How does this magic work?" you might ask. The trick is to break the numerator and denominator apart. Let's look at the denominator first.
How many distinct ways of pulling 10 cards out of 52 cards without replacement are there? Since every card is unique, but we don't care about pulling all the reds, then all the blacks, we need the number of combinations: $52 \choose 10$.
How many distinct ways of pulling 5 (red) cards out of 26 (red) cards without replacement are there? Since every card is unique, but we don't care about the order, we use the number of combinations formula: $26 \choose 5$.
We apply the same argument to the black cards as we did to the red cards.
Since the number of combinations of red cards is independent of the number of combinations of black cards (because the two sets have nothing in common), we multiply the two numbers of combinations to obtain the total number of combinations of 5 red and 5 black cards.
Probability of an event is defined for discrete probability (which this is) as being the number of desired outcomes divided by the number of possible outcomes:
$$P(x)=\frac{n_{desired}}{N_{total}}$$.
Since we have the number of desired outcomes, as well as the total number of outcomes, we can directly compute the probability as a ratio.
