What is the probability that 4 cards drawn randomly from a deck are all hearts, given that there are at least three hearts?

Here is the question:

$4$ cards are drawn at random from a standard $52$-card deck. At least $3$ of them are hearts. What is the probability that they are all hearts?

I was able to obtain the correct solution as follows:

$\frac{ \binom{13}{4}}{\binom{13}{4} + 39 \binom{13}{3}} = \frac{5}{83}$.

since there are $\binom{13}{3} \cdot 39$ ways to choose a heart and a non-heart.

I want to solve this using the Principle of Inclusion-Exclusion, but I'm having difficulty arriving at the same answer. To compute the number of 4-card hands with at least three hearts, I tried the following:

$\binom {4}{3} \binom{13}{3}49 - 3\binom{13}{4}$

but this is giving me the wrong answer.

Reasoning:

$\binom {4}{3}$ is to choose which three cards out of four are hearts, $\binom{13}{3}$ is to count which ranks of hearts are selected, and then there are 49 choices for the last (fourth) card. Since we have over-counted four-heart hands, we subtract $3\binom{13}{4}.$

I think you made a typo in your answer it should be $\frac{ \binom{13}{4}}{\binom{13}{4} + 39 \binom{13}{3}}$. Now about your other approach of finding number of ways of choosing 4 cards with at least 3 hearts note that it will be $\binom{13}{3}*49 - 3 \binom{13}{4}$, which is same as above. The reasoning goes as follows, first note that order does not matter, now there are $\binom{13}{3}*49$ ways of choosing at least three hearts with double counting of cases which have all hearts. The cases with all hearts are counted 4 times so we subtract those to get the right answer.
You were doing great, only notice that the number of ways to obtain exactly three hearts is $\binom{13}{3}\times 38$, as there are $\binom{13}{3}$ ways to pick the three hearts and $38$ ways to pick the non-heart.
So the answer is $\frac{\binom{13}{4}}{\binom{13}{4}+38\binom{13}{3}}$