Question 21777
<pre><font size = 3>What is the probability of obtaining a poker hand(5 cards)
containing 2 aces?

We can choose the pair of aces any of <sub>4</sub>C<sub>2</sub> ways.
For each of these choices of a pair of aces we can choose the three
non-aces any of <sub>48</sub>C<sub>3</sub> ways.

So the numerator of the desired probability is (<sub>4</sub>C<sub>2</sub>)(<sub>48</sub>C<sub>3</sub>)

The denominator of the desired probability is <sub>52</sub>C<sub>5</sub>

So the desired probability is

(<sub>4</sub>C<sub>2</sub>)(<sub>48</sub>C<sub>3</sub>)
-----------
   <sub>52</sub>C<sub>5</sub>

[(4·3)/(2·1)]·[(48·47·46)/(3·2·1)]
----------------------------------
   (52·51·50·49·48)/(5·4·3·2·1)

= .0399298181

</pre>Edwin
AnlytcPhil@aol.com