Probability that a hand of 10 cards dealt from a normal pack of 52 contains...

Question

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!

Answer 1

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

$$\frac{\binom{4}{1}\binom{4}{2}\binom{44}{7}}{\binom{4}{2}\binom{48}{8}}=\frac{3952}{9729}\approx 40,62\%$$

Answer 2

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.