Probability Question - Geometric Distributions 
A jet has a $5\%$ chance of crashing on any given test flight. Once it crashes the program will be halted. Find the probability that the program lasts less than three flights.

The correct answer to this question is $0.1426$, but I can't figure out how to get it.
Here's my attempt at the question:
$p=0.05$
$q=0.95$
$x=0,1,2$
$$(0.05)(0.95)^{-1} + (0.05)(0.95)^0 + (0.05)(0.95)^1 = 0.1501$$ 
I used the geometric probability method by using the formula $pq^{x-1}$ where $p$ is the probability of successes, $q$ is the probability of failures, and $x$ is the waiting time/how many events occurred before the success occurred.
 A: $pq^{x-1}$ is the probability for $x$ trials until the first success, where $x\in\{1,2,\ldots\}$.
$pq^x$ is the probability for $x$ failures before the first success, where $x\in\{0,1,2,\ldots\}$.
Unfortunately both distributions are called Geometric but it important to not confuse them.
If you use $X$ as the count of failures before the first successful crash, you need the second distribution.  So the probability that there are two or fewer failed crashes is:
$$X\sim\mathcal{Geo}_0(0.05)\implies \mathsf P(X\leq 2) ~{= p(q^0+q^1+q^2)\\= 0.05(1+0.95+0.95^2) \\= 0.142625}$$
A: Consider this.
The probability of the jet crashing on the first is simply $5\%$, or $.05$.
The probability of the jet crashing on the second run isn't as immediate. First, the jet has to not crash on the first run, which has probability $.95$. Then it actually has to crash, with probability $.05$. Thus the probability of a crash on the second run is the probability of both of these events happening, or $.95 * .05$
Similarly, the probability of a crash on the third run is $.95 * .95 * .05$.
Thus, the probability of this crash happening within the first three runs is the probability of any of these happening, which is simply the sum of their probabilities:
$$
\begin{align}
&p = .95*.95*.05+.95*.05+.05\\
&p = .0451+.0475+.05\\
&p = .1426
\end{align}
$$
A: The probability that the jet survives after three tests: $(0.95)^3$. The probability for the other cases is $1-(0.95)^3=0.1426$.
