Question 1150347
<br>
First solution method: using the probabilities that each card, drawn one at a time, can produce the desired result.<br>
The first card can be either an ace or a nine -- probability 8/52 = 2/13.<br>
The second card can only be a nine if the first card was an ace, or vice versa -- probability 4/51.<br>
P(an ace and a nine) = (2/13)*(4/51) = 8/663.<br>
That first solution method is easily used to find the answer for this particular problem.<br>
For more complicated probability problems, you might need a more sophisticated method.  For this problem, it could look something like this:<br>
You are choosing 2 of the 52 cards; the desired outcome is you get 1 of the 4 aces, 1 of the 4 nines, and 0 of the other 44 cards.  The probability is<br><br><pre>
   C(4,1)*C(4,1)*C(44,0)    (4*4*1)       16       8       8
   --------------------- = --------- = ------- = ----- = -----
         C(52,2)            (52*51)     26*51    13*51    663
                            -------
                             (2*1)</pre>