# Probability-Playing Cards

Question. So two cards are drawn without replacements from a pack of playing cards, I've been told to work out the Probability:

a) Both cards are aces b) Only one card is an ace c) Two cards are of different suits

Then I've been told Given at least one ace is drawn, Find the probability that the two cards are different suits?

Attempt: Now I have solved a) to c), for the last part I'm not getting the same answer as the book.

Let the events A = Two cards drawn of different suits & B = At least one ace drawn.

$P(A\mid B) = {P(A\cap B)}/{P(B)}$

$P(B)= a)+ b)=33/221$

$P(A\cap B)$=....this is where I'm stuck, I keep getting different values by considering the case 1 ace is picked then adding the case where 2 aces are picked.

Any help will be appreciated, Thanks.

• Could you point out what you have to calculate ? Im a little bit confused. – callculus Apr 16 '15 at 20:22

The chance that we have both at least one ace and the cards of different suits comes from:
draw two aces(which will be of different suits): $\frac {4\cdot 3}{52 \cdot 51}$
draw an ace, then a non-ace of a different suit: $\frac {4\cdot 36}{52 \cdot 51}$
draw a non-ace, then an ace of a different suit: $\frac {48\cdot 3}{52 \cdot 51}$
The total is $\frac {300}{2652}=\frac {25}{221}$

I find it surprising that knowing an ace was drawn changes the $\frac {39}{51}=\frac {13}{17}$ chance of having two different suits, but there are many surprising dependencies in combinatorics.

Let $A$ be the event of drawing different suits, and let $X$ be the count of aces drawn.   You want to use the Law of Total Probability:

$$\mathsf P((X\geq 1)\cap A) = \mathsf P(X=1)\;\mathsf P(A\mid X=1)+\mathsf P(X=2)\;\mathsf P(A\mid X=2)$$

You say you have $\mathsf P(X=1)$ and $\mathsf P(X=2)$. Then consider the cases.

• $\mathsf P(A\mid X=1)$ If only one card is an ace, what is the probability that the other card is a different suit?
• $\mathsf P(A\mid X=2)$ If both cards are aces, what is the probability that they are different suits?