Conditional probability (cards question).

Question

Given a piece of cards with $52$ total cards, $13$ of each one from the $4$ options ($\heartsuit, \spadesuit, \diamondsuit, \clubsuit$) . You and your friend getting randomly $7$ cards each,

what is the probability that your friend have one heart card given that you have three heart cards?

*note that I mean to the amount of cards from type hearts not the number 1$\heartsuit$.

My attempt

Let $A$ and $B$ be two events :

$A$- my friend have $1$ heart cards

$B$- I have $3$ heart cards

we need to find $$P(A|B)$$ I want to use the formula $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

now I am stuck how to calculate $P(B)$ and $P(A\cap B)$

In the book there is 4 options $$1. \ \frac{12!}{3!\cdot9!}\cdot\frac{32!}{5!\cdot27!}\cdot\frac{8!\cdot36!}{44!}$$ $$2.\ \frac{12!}{3!\cdot9!}\cdot\frac{33!}{4!\cdot29!}\cdot\frac{7!\cdot38!}{45!}$$ $$3.\ \frac{10!}{1!\cdot9!}\cdot\frac{34!}{7!\cdot27!}\cdot\frac{8!\cdot36!}{44!}$$ $$4.\ \frac{10!}{1!\cdot9!}\cdot\frac{35!}{6!\cdot29!}\cdot\frac{7!\cdot38!}{45!}$$

I did not understand this pattern with the factorial, how did they build it?

Answer 1

Actually, you don't want to treat this as a conditional probability question. Just count combinations. You have $45$ cards available for your friend to choose of which $10$ are hearts. So of the possible combinations, what fraction have exactly $1$ heart?

There are $\binom{45}{7}$ possible hands your friend could hold. For a "good" hand, there are $10$ different hearts he could hold and $\binom{35}{6}$ ways to choose the remaining cards. So the answer is:

$$\frac{10 \binom{35}{6}}{\binom{45}{7}}.$$

That's choice $4$.

Answer 2

I agree with Robert Shore's answer. I further agree with his comment that for this particular problem, it is most readily solved without using Bayes Theorem. However, I still think that it is important to understand how to set up the conditional probability mathematics.

Let $S$ denote the event that your 7 card hand has exactly 3 hearts.
Let $T$ denote the event that your friend's 7 card hand has exactly 1 heart.

Then $$p(T|S) ~=~ \frac{p(S ~\text{and} ~T ~\text{both occur})}{p(S)}.$$

$${p(S ~\text{and} ~T ~\text{both occur})} ~=~ \frac{\binom{13}{3}\binom{39}{4}}{\binom{52}{7}} \times \frac{\binom{10}{1}\binom{35}{6}}{\binom{45}{7}}.$$

$${p(S)} ~=~ \frac{\binom{13}{3}\binom{39}{4}}{\binom{52}{7}}.$$