Probability that a hand of 10 cards dealt from a normal pack of 52 contains... Exactly one ace given that it contains exactly two queens?
I am struggling how to solve this maths problem. I know that a hand of $10$ cards dealt from a normal pack of $52$ is $52\choose 10$ $=$ $15820024220$. And, I know that there are $4$ queens in a normal pack of $52$, hence, the problem states that there are exactly two queens, which is $4\choose2$ $= 6$.
Any help is much appreciated - thanks!
 A: You can use the definition of conditional probability.
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

*

*$P(A \cap B)$ is the probability to have 1A and 2Q in 10 cards. So you have

*

*$\binom{4}{1}$ choices to get the Aces

*$\binom{4}{2}$ choices to get two Queens

*$\binom{44}{7}$ choices to get 7 cards different from Queens and Aces



Thus the probability of $P(A \cap B)=\frac{\binom{4}{1}\binom{4}{2}\binom{44}{7}}{\binom{52}{10}}$
Do the same brainstoming to calculate the probability to have exactly 2Q and then divide the two results:
Numerator: #cases to have exactly 1A and 2Q
Denominator: #cases to have exactly 2Q
$$\frac{\binom{4}{1}\binom{4}{2}\binom{44}{7}}{\binom{4}{2}\binom{48}{8}}=\frac{3952}{9729}\approx 40,62\%$$
A: I favor the shortcut of equating the problem to the chances of
a hand of 8 cards having exactly 1 Ace from a pack of 48 cards.
I would calculate this as
$\displaystyle \frac{\binom{4}{1} \times \binom{44}{7}}{\binom{48}{8}}.$
The reason that I regard the problems as equivalent is that you can distribute
the 10 cards out of 52 in three steps:
(1) split the deck into two groups, 48 cards and 4 cards
(2) split the 10 cards that will be dealt into 2 groups, 8 cards and 2 cards.
(3) you are given that the group of 2-cards will automatically have 2 queens.
Consequently, my intuition indicates that the two problems are equivalent.
