Simple probability calculations Question1: Every time you fire an arrow at the target, you have a 1/5 chance of hitting it. What is the probability that you succeed in hitting the target at least once in two tries?
my attempt: chance of hitting $1/5$, probability of missing $1-(1/5)= 4/5$
$p=1-(4/5)^2= 9/25$ or $0.360$
is this correct the way I did it or I'm wrong? 
Question2: What is the probability that you hit the target the second time given that you didn't hit the target the first time?
I'm confused on how to start with this one
 A: Looks good to me! Just a check using an alternative approach:
$$\begin{align*}
P(\text{hitting at least once}) 
&=P(\text{hitting twice})+P(\text{hitting once})\\\\
&= \frac{1}{5}^2 + {2 \choose 1}\cdot\frac{1}{5} \cdot\frac{4}{5}\\\\
&=0.36
\end{align*}$$
For the second one, whether or not you hit the first shot has no effect on whether you'll hit the second shot. The probability is just $\frac{1}{5}$.
A: Your answer to number 1 looks correct!
For number 2, it is important to note that the two events are independent of each other.  Therefore the probability of hitting the second one is always 0.2.  However, if the two events were not independent, this would not be the case.  You would have to use conditional probability.  Ex: what is probability that the sum of two fair die is $s$ given that the value of the first dice rolled was $x$.  You can see that $P(s)$ depends on $x$!
A: You have correctly calculated the probability that you do not happen to miss the target in both attempts.   That is indeed exactly what you want.$$1-(1-\tfrac 15)^2$$

Now, what is the probability that you hit the target the second time given that you didn't hit the target the first time, and knowing that every time you fire an arrow at the target, you have a 1/5 chance of hitting it?   Mmm...
