If you draw 9 cards from a 52-card deck, what is the probability of one Ace and one King?
I realize you can calculate this by using the hypergeometric distribution as:
$$ p(k=2, M=52, n=8, N=9) = \frac{\binom{8}{2} \binom{44}{7}}{\binom{52}{9}} = 0.2916 $$
However, my question is if there is an easier way to calculate this, that you would for example be able to approximate in your head without pen and paper?
The only alternative method of computation for this problem that I know of is to (partially) forgo a Combinatorical approach in favor of a probability of events approach. However, I don't recommend the alternative approach, described below.
Let $E_1$ denote the event that a King is drawn on the first card.
Let $E_2$ denote the event that an Ace is drawn on the second card, given that a King was drawn on the first card.
Let $E_3$ denote the event that cards other than an Ace or King were drawn on the next $7$ cards, given that an Ace and a King were drawn on the first two cards.
Then, the probability of first drawing a King, then drawing an Ace, and then drawing $7$ other cards, besides an Ace or a King is :
$$T = p(E_1) \times p(E_2) \times p(E_3) = \frac{4}{52} \times \frac{4}{51} \times \frac{44 \times 43 \times \cdots \times 38}{50 \times 49 \times \cdots \times 44}. \tag1 $$
At this point in the analysis, I see no alternative to completing the computation other than (partially) resorting to Combinatorics.
In (1) above, $T$ specifically expressed the probability that the King was the very first card drawn, and the Ace was the second card drawn. In fact, from the problem's constraints, there are $9$ choices for which card drawn is the King, and then $8$ choices for which of the remaining cards drawn was the Ace.
So, the final computation would be
$$9 \times 8 \times T.$$
Personally, beside this approach being somewhat convoluted, I don't see this approach being easier to manually convert into the form $~\displaystyle \frac{m}{n}, ~: ~m,n \in \Bbb{Z^+},~$ than the original Combinatorics approach.
Edit
For what it's worth, I will take a crack at the mental conversion.
I would convert :
$\displaystyle (9 \times 8) \times \frac{4}{52} \times \frac{4}{51} \approx \frac{72}{169} \approx \frac{1}{2}.$
$\displaystyle \frac{44 \times \cdots \times 38}{50 \times \cdots \times 44} \approx \left[\frac{41}{47}\right]^7.$
Then, I would reason that when $a > b > 0$, that $~\displaystyle \left[1 - \frac{b}{a}\right]^2 \approx 1 - \frac{2b}{a}.$
So, I would very roughly approximate $\displaystyle ~\left[\frac{41}{47}\right]^7~$ as $\displaystyle \left[1 - \frac{5 \times 6}{47}\right] = \frac{17}{47}.$
Then, I would approximate
$$\frac{1}{2} \times \frac{17}{47} \approx \frac{8.5}{47} \approx \frac{2}{11}.$$
Frankly, I have no idea how close the computation of $~\displaystyle \frac{2}{11}~$ actually is to the right answer. $~\displaystyle \frac{2}{11}~$ would simply be my blind guess.