Cumulative geometric probability (greater than a value) I am learning about Cumulative Geometric Probability on Khan Academy: see this video.
Here is the word problem:

Emelia registers vehicles for the Department of Transportation. Sports
utility vehicles (SUVs) make up 12% of the vehicles she registers. Let
V be the number of vehicles Emelia registers in a day until she first
registers an SUV. Assume the type of each vehicle is independent. Find
the probability that Emelia registers more than 4 vehicles before she
registers an SUV.

At 3:50 in the video, I don't understand how come
P(MORE than 4 vehicles BEFORE she registers an SUV)
is equal to
P(FIRST 4 cars not SUVs).
"More than 4 vehicles before she registers an SUV" could be an infinite number, how could it be equal to "first 4 cars not SUVs"?
I really appreciate if someone could explain it like I'm 5.
 A: Each vehicle is selected at random, with probability of selecting an SUV being 0.12, since 12% of cars are SUVs. The geometric distribution gives the number of trials of "SUV or not SUV?", as you examine each car in turn, before the first SUV is found, so this occurs with probability $$P(k \text{ trials before first SUV})=(1-p)^{k-1}p = 0.88^{k-1} \times 0.12$$
In fact, it is better to use the "CDF of the geometric distribution", which says
$$P(\text{$k$ trials or fewer before the first success}) = 1- (1-p)^k$$
So, the probability of 4 trials or fewer before the first SUV is given by
$$P(4 \text{ trials or fewer before the first SUV})=1 - (1-p)^{4} = 1 - (1-0.12)^{4} = 0.40$$
so more than 4 vehicles is $1-0.40=0.60$.
The fact you mention

P(MORE than 4 vehicles BEFORE she registers an SUV) is equal to
  P(FIRST 4 cars not SUVs)

is the idea of reversing the probability, so more than 4 is the negation of the event less than or equal to 4, so just do $$P(\text{event occurs}) = 1 - P(\text{event does not occur})$$
which is why you go from $0.4$ to $1-0.4=0.6$ after using the CDF.
