The probability of independent events. 

*Suppose that there is a 1 in 50 chance of injury on a single skydiving attempt.
(a) (3 points) If we assume that the outcomes of different jumps are independent, what is the probability
that a skydiver is injured if she jumps twice?


The solution of this question was that 1-(the probability that she is not injured).
But I think it is not reasonable. 
Let P(A)=the probability that she is not injured during two jumps.
1-P(A) contains the probability of (injured, safe), (injured, injured), (safe, injured). However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible. 
 A: The complement event of “the skydiver is injured” is “the skydiver is safe in both jumps.” The skydiver has a chance of $1/50$ to be injured in an individual jump, so the chance that the skydiver is safe in an individual jump is
$$1 -\frac{1}{50} = \frac{49}{50}. $$
Since the two jumps are independent, the probability that she is safe in both jumps is: 
$$\biggl(\frac{49}{50}\biggr)^2 = \frac{2401}{2500}. $$
Thus, the probability that the skydiver is injured in either jump is
$$1 - \frac{2401}{2500} = \frac{99}{2500} = 3.96\,\%. $$
A: To answer your question more directly, if she's injured the first jump then the desired condition is already satisfied, and it doesn't matter whether she jumps again or not. (And if she jumps again, it doesn't matter if she's injured or safe the second time.) 
So if you wish, you can consider there to be $3$ events instead of $4$:


*

*With prob $p = 1/50$ she is injured the first jump.  (The desired condition is satisfied, and it doesn't matter whether she jumps again or not.  This combines your (injured, injured) and (injured, safe) scenarios into one.  Note that I am making a mathematical point.  I am not talking about typical human behavior at all.)

*With prob $(1-p)p$ she is safe the first time and injured the second time.

*With orb $(1-p)^2$ she is safe both times.
You can find your desired prob either as $1 - (1-p)^2$ (as in the answer by @L.F.) or as $p + (1-p)p$.  They both equal $2p - p^2$.
A: 
However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible. 

If that were the case, then the events would not be independent.
You are asked to find the probability while assuming that the events are independent.
Therefore you should not consider injury to prevent (or otherwise influence) the second jump.   If you like, think of it as having sufficient time between jumps for recovery or whatever.
A: The event the skydiver is injured consists of the realized outcomes either that she is injured on her first attempt, or her second attempt, or both her first and second attempt. You should have already figured this part out by thinking about the structure of the sample space.
The union of two events $A_1$ and $A_2$ is logically equivalent to this phrase: "$A_1$ or $A_2$ or both $A_1$ and $A_2$ overall." The event of injury $A$ is therefore equivalent to the union of the event of injury on the first try, $A_1$, and on the second try, $A_2$.
Knowing now that $P(A)=P(A_1 \cup A_2)$, by the Additive Law of Probability we have it that $P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2)$. Because $A_1$ and $A_2$ are said to be independent, we can calculate $P(A_1 \cap A_2)=P(A_1)P(A_2)$. We arrive at the final answer making the necessary substitutions: $P(A)=2p-p^2$.
