You draw at random five cards from a standard deck of 52 cards.
What is the probability that there is an ace among the five cards and a king
or queen?
I want to do this by counting. There are 3 mutually exclusive events that need to be summed: ace and king with no queen, ace and queen with no king, ace and both.
Ace and king with no queen:
$$4 * 4 * {46 \choose 3} $$
One of 4 aces, one of 4 kings, 3 from the 46 other possible cards. Same thing for ace with queen and no kings.
Ace with both:
$$4 * 4 * 4 * {49 \choose 2} $$
One of 4 aces, one of 4 kings, one of 4 queens, 2 from the 49 cards remaining.
There are a total of $52 \choose 5$ ways to draw hands of 5 cards, so the answer is:
$$ \frac{2 * 4 * 4 * {46 \choose 3} + 4 * 4 * 4 * {49 \choose 2}}{52 \choose 5} = 0.215$$
…which is wrong. The answer is $0.1765$, which was obtained by taking $1 – P(A\cup B)$, where $A$ is no aces and $B$ is no king or queen. Where did I go wrong?
Best Answer
As users have pointed out in the comments, counting might complicate matters. You would have to first determine all the cases and then be very careful in your computations to avoid over-counting.
Here I present an alternate approach which, in my opinion, makes this question easier to tackle.
Consider the three events $A$,$B$,$C$ defined as follows:
$A-$ The $5$ cards picked contain at least one Ace
$B-$ The $5$ cards picked contain at least one King
$C-$ The $5$ cards picked contain at least one Queen
We are interested in determining $P(A \cap(B\cup C)) $.
Now, using De Morgan's Law: $$(A \cap(B\cup C))^c =A^c \cup(B\cup C)^c $$
Therefore, $$P(A \cap(B\cup C)) $$$$=1-P(A^c \cup(B\cup C)^c)$$
$$=1-\bigg(P(A^c)+P((B\cup C)^c)-P(A^c\cap(B\cup C)^c)\bigg)$$
$$=1-\bigg(\frac{{48 \choose 5}}{52\choose 5}+\frac{{44 \choose 5}}{52\choose 5}-\frac{{40 \choose 5}}{52\choose 5}\bigg)$$$$=0.1765$$