What’s the probability of getting a full house given a full house or 3 of a kind? p(full house| full house or 3 of a kind) I have no idea where to start for this question, I tried using conditional probabilities where p(a|b) = p(ab)/p(b) but don’t know where to go from there I found the probability of a full house by doing (13)(12)(4 choose 3)(4 choose 2)/(52 choose 5) And 3 of a kind (13)(4 choose 3)(12 choose 2)(4)(4)/(52 choose 5)

Let $A$ be the event that you get a full house and $B$ the event that you get a three of a kind.

You correctly found $Pr(A)=\dfrac{13\cdot 12\cdot \binom{4}{3}\cdot\binom{4}{2}}{\binom{52}{5}}$ and $Pr(B)=\dfrac{13\cdot\binom{4}{3}\cdot\binom{12}{2}\cdot 4\cdot 4}{\binom{52}{5}}$

Recognize that $A$ and $B$ are disjoint and so $Pr(A\cup B) = Pr(A)+Pr(B)$

> Remember that in general $Pr(A\cup B)=Pr(A)+Pr(B)-Pr(A\cap B)$ and that the $Pr(A\cap B)$ term in this should not be forgotten. It is a special situation when $Pr(A\cap B)=0$ and in particular when $A\cap B=\emptyset$. When $A\cap B=\emptyset$ we say the events are disjoint, that is to say the two events cannot happen simultaneously.

We have then $Pr(A\mid A\cup B) = \dfrac{Pr(A\cap (A\cup B))}{Pr(A\cup B)} = \dfrac{Pr(A)}{Pr(A)+Pr(B)}$. Using the two values you found earlier, plugging them in finishes the problem.