What is the probability that two hands of 13 cards dealt from a normal shuffled pack of 52 cards contain exactly two kings and one ace? What is the probability that both contain exactly one ace given that both contain exactly two kings?
Here's my attempt:
Let K1, K2 be the events that the first hand and the second hand contain exactly two kings, respectively; A1, A2 be the events that the first hand and the second hands contain exactly one ace, respectively.
$\mathbb{P}(K1 \cap A1) = {4 \choose 2}{4 \choose 1}{44 \choose 10} \Big/{52 \choose 13}$
$\mathbb{P}(K2 \cap A2) = {2 \choose 2}{3 \choose 1}{36 \choose 10} \Big/{39 \choose 13}$
Thus $\mathbb{P}((K1 \cap A1) \cap (K2 \cap A2)) = {4 \choose 2}{4 \choose 1}{44 \choose 10}{2 \choose 2}{3 \choose 1}{36 \choose 10} \Big/\left[{52 \choose 13}{39 \choose 13}\right]$.
Second part:
$\mathbb{P}(K1) = {4 \choose 2}{48 \choose 11} \Big/ {52 \choose 13}$
$\mathbb{P}(K2) = {2 \choose 2}{37 \choose 11} \Big/ {39 \choose 13}$
$\mathbb{P}(K1 \cap K2) = {4 \choose 2}{48 \choose 11} {2 \choose 2}{37 \choose 11} \Big/ \left[{52 \choose 13}{39 \choose 13}\right]$
\begin{align} \mathbb{P}((A1 \cap A2) | (K1 \cap K2)) &= \frac{\mathbb{P}((A1 \cap A2) \cap (K1 \cap K2))}{\mathbb{P}(K1 \cap K2)}\\ &=\left\{{4 \choose 2}{4 \choose 1}{44 \choose 10}{2 \choose 2}{3 \choose 1}{36 \choose 10} \Big/\left[{52 \choose 13}{39 \choose 13}\right]\right\}\Big/ \left\{{4 \choose 2}{48 \choose 11} {2 \choose 2}{37 \choose 11} \Big/ \left[{52 \choose 13}{39 \choose 13}\right] \right\}\\ &\approx 0.000198 \end{align}
Please let me know if you think it's correct or if you have an easier solution. Thanks.
The expressions for the probabilities are correct. For the conditional probability, you had basically computed the necessary ingredients in the first part, so it is natural to recycle.
One can solve the conditional probability problem without using the conditional probability machinerty. Imagine that you have given both hands two Kings. Now let us deal the rest of the cards.
The probability that the first hand gets $1$ Ace is $\dfrac{\binom{4}{1}\binom{44}{10}}{\binom{48}{11}}$.
Given that the first hand got $1$ Ace, the probability the second hand does is $\dfrac{\binom{3}{1}\binom{34}{10}}{\binom{37}{11}}$. Now multiply.