Question 15532

If 10 cards are drawn without replacement from an ordinary deck, find the probability of obtaining exactly 3 aces or exactly 3 kings (or both). I tried
using (52 choose 10) for the denominator and (10 choose 3)^4 for the numerator(4 for each suite). It seems as though I'm missing something. Please help asap.
Total number of possibilities. N = ...taking 10 out of 52 cards = 52C10
 number of successes....M =...taking 3 aces and rest seven other than aces.
   taking 3 aces out of 4 aces ..=4C3
   taking 7 cards out of rest other than aces...=48C7
   M = 4C3 * 48C7 
  probability = M/N = 4C3 * 48C7 /52C10 = (4*48*47*46*45*44*43*42)*(10*9*8*7*6*5*4*3*2*1)/(7*6*5*4*3*2*1)*(52*51*50*49*48*47*46*45*44*43)
   FOr 3 kings only the answer is same 
for both 3 kings and 3 aces you can try your self on the above basis