Probability: drawing three cards at a time Question:
3 cards are drawn one by one without replacement. Find the probability to get 2 kings and 1 ace?
My attempt:
By fixing a postion for ace. Let fix it as first card then total favourable cases will be 4×3×4 out of 52 ×51 ×50 cases..we have three such postions so the probability should be 144/132600.
But according to the solution the answer should be 288/132600.
I am not able to understand which cases am i missing.
 A: your answer is right, I overlooked the fact that you have used a slightly unusual way of solving it
There are two standard ways of computing a general problem of this kind,

*

*combinatorial, viz $\dfrac{\binom42\binom41}{\binom{52}3}$, and


*direct probabilities, viz $\dfrac{4}{52}\dfrac{3}{51}\dfrac{4}{50} *\dfrac{3!}{2!}$
It may be better if you follow one of the above approaches. In more complex problems, there is a possibility of getting confused if you fix the first card
A: The possible combinations of an ace and two kings:
$${4\choose1}\times{4\choose2}=24$$
Arranging the combination: $3!=6$ cases.
So the probability is $144/132600$; you are right and the solution is wrong.
Double-checking, I counted all the possible permutations:
(a,b,c,d are aces and 1,2,3,4 are kings)
a12 a13 a14 a23 a24 a34
1a2 1a3 1a4 2a3 2a4 3a4
12a 13a 14a 23a 24a 34a
b12 b13 b14 b23 b24 b34
1b2 1b3 1b4 2b3 2b4 3b4
12b 13b 14b 23b 24b 34b
c12 c13 c14 c23 c24 c34
1c2 1c3 1c4 2c3 2c4 3c4
12c 13c 14c 23c 24c 34c
d12 d13 d14 d23 d24 d34
1d2 1d3 1d4 2d3 2d4 3d4
12d 13d 14d 23d 24d 34d
$72$ cases.
Including reversing the order of the kings, $144$ cases.
