A deck of 52 cards is revealed 1 at a time till an ace appears. Let $X$ be the number of cards revealed until the ace. Find the PMF and Expectation(Can leave as a summation).
Let me know if any of my thought process is wrong
Since it's asking for a PMF it's going to be probabilities clumped up at different Xs(aka not a continuous function). The probability of drawing an Ace on the first draw is 4/52 then assuming you didn't draw it on the first draw it becomes 4/51 and this keeps going so I assume the PMF=P(X=k)=$\prod_{i=1}^k$(48-i/(52-i)) for $1\le k<48$. 0 otherwise.
With the PMF we know the expectation is going to be E[X]=P(X)*X. So that gives us $\sum^{47}_{k=1}k\prod_{i=1}^k(48-i/(52-i))$ because in worst can we'll have to draw 47 cards before we get an Ace and that's the case that the last 4 cards are aces.
Now that the answer is in a comment, I'll write it as an answer with some extra stuff.
So we want to find $P(X=k)$, for $1\le k \le 49$. Let's take $k=4$ as an example. We first need a non-ace card, which has probability $\frac{48}{52}$. Then another one, $\frac{47}{51}$, and another, $\frac{46}{50}$. Then we need an ace, $\frac{4}{49}$. So $P(X=4)=\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdot\frac{4}{49}$. This generalises to: $$ P(X=k) = \frac{4}{53-k}\cdot\prod_{j=0}^{k-2}\frac{48-j}{52-j} $$ (Note that if $k=1$, the product is the empty product, which is $1$).
You seem to have no problem writing up the expectation, so I won't go into that (it turns out to be $53/5=10.6$ by the way). Note however that the possible values for $k$ are $\{1,\ldots,49\}$, not $\{1,\ldots,47\}$.
Now for the extra bit. We can rewrite the probability as: $$ P(X=k) = \frac{4}{52}\cdot\frac{52-k}{51}\cdot\frac{51-k}{50}\cdot\frac{50-k}{49} $$ This form actually gives another interpretation. We think about the positions of the aces in the deck. The $k$'th card needs to be an ace, which has probabailty $\frac{4}{52}$. The other three aces have to be placed after position $k$, which gives the rest of the fractions (e.g. there are 51 spots left for the second ace, $52-k$ of these being after position $k$).