Explanation of Geometric Distribution Graph I understand what the geometric distribution is and that it is calculating the probability of the number of trials needed up to and including the first success. Expressed as the following:
$$P(X = k) = (1-p)^{k-1}p$$
I asked a similar question years ago and for the life of me I still cannot wrap my head around the interpretation of the graph from the example:

What I mean by not wrapping my head around the interpretation is for example, let's choose $x = 10$. What I'm interpreting in my mind is "the probability of winning the lottery after purchasing 10 tickets is roughly 0.19". But this cannot be right to me. Interpreting it in this fashion alludes to purchasing more tickets should mean a higher probability of winning the lottery. This is what would be the case if I was thinking of the CDF, but I'm not. So what is the way to interpret the PMF of the geometric distribution in this example?
Question that I asked previously:   Understanding the geometric distribution
 A: Your interpretation is correct: $(1-p)^{k-1}p$ is the probability of first success on the $k$th independent attempt.
The problem lies in the statement more tickers should mean a higher probability of winning the lottery: 
You are raising the power $k-1\; (>1)$ of a fraction $(1-p)$. As the power increases (i.e. you purchase more tickets), the value $(1-p)^{k-1}$ decreases for a fixed value of $p$.
A: The probability $P(X=10)$ is the probability of needing to buy exactly 10 tickets in order for the 10th ticket to be your first winner.  Put another way, it's the probability of buying 9 losing tickets in a row before buying your first winning ticket.
If it helps, consider the similar experiment of flipping a coin until the first head.  The probability that the very first toss gives a head is $P(Y=1)=1/2$.   The probability $P(Y=2)$ represents the probability of failing to get heads on your first toss (probability 1/2) and then furthermore succeeding in getting heads on your second toss.  That's probability 1/2 to first fail followed by probability 1/2 to then success, for an overall probability of $P(Y=2)=1/2  * 1/2 = 1/4$.  Similarly, to get $P(Y=3)$, you have to fail twice (probability $1/2 * 1/2 = 1/4$) and then furthermore succeed on your third toss (probability $1/2$), for a total probability of $P(Y=3)=1/2 * 1/2 * 1/2 = 1/8$.
The decreasing probabilities in both the lottery and coin examples result from the fact that, for example, $P(Y=10)$ needs 9 failures (and it's tough to lose so many times!) before you get your first success on the tenth try.
