A sharpshooters hit a target with a probability of $\frac{3}{4}$ Assuming independence  find the probabilities of getting
A. Two hits and a miss in any order.
What I tried
I know these are the possible trees of of probability
H=HIT
M=MISS
H H H
H H M
H M H
H T T
M H H
M H M
M M H
M M M
So there are three possibilities for two misses and one hit
But I am not sure how to calculate this part.
You have as one possibility 
$(\frac{1}{4}\frac{3}{4}\frac{3}{4})$
which is $\frac{9}{16}$ but adding this three times give me 27/16 and this cannot be correct.

Final answer is :
(.75)^2(.25)+(.75)^2(.25)+(.75^2)(.25)=$\frac{27}{64}$
 A: The probability of getting two hits and one miss in three independent shots is given by $\left[\frac{3}{4}\right]^2\times\frac{1}{4}=\frac{9}{64}$. Then as you correctly pointed out two hits and a miss can occur in $\binom{3}{2}=3$ ways. The probability you are looking for then is $3 \times \frac{9}{64}=\frac{27}{64}$.
More generally you can tackle a problem like that with the Binomial Distribution:

$$ P\left(X=x \right)=\binom{n}{x}p^x \times \left( 1-p \right)^{n-x}$$

In your case the random variable $X$ is the number of hits and its range is $\{0,1,2,3\}$
Hope this helps.
A: Let's say you have $N$ objects, of which you want to pick $k$. How many possible configurations of $k$ objects can you choose out of the selection of $N$?
Well for your first pick you have $N$ objects so there are $N$ choices. For your second pick you have $N-1$ objects so for the first two there are $N \times (N-1)$ choices. For the $k$th object you have $N - (k-1)$ choices. So overall there are $$N \times (N-1) \times \cdots \times (N - k + 1) = \frac{N!}{(N-k)!}$$ possible ways of picking $k$ objects out of $N$. BUT... don't forget that for any set of $k$ objects there are $k!$ possible ways of ordering them, and all those ways of ordering have been accounted for in the number of ways to pick a set as I described above. So the actual number of ways of choosing $k$ unique objects out of $N$, regardless of the order in which you choose them is $$\frac{N!}{k! (N-k)!} \equiv {N \choose k}$$ also known as "$N$ choose $k$". That is just the number of ways of picking them divided by the number of possible orderings for each choice.
Now let's say you want to run a certain trial $N$ times, and each trial has a probability $p$ of success. You want to know what the probability is of getting $k$ and only $k$ successes. Well the probability of that many successes is equal to the joint probability of each success. That is the product of the probabilities of each trial. For $k$ successes, that means $N-k$ failures (which has probability $1-p$), so the probability of all those events is $$p^k (1-p)^{N-k}.$$ However that is the probability for a particular set of successes and failures, given a particular order. Given that there are many possible configurations in which we have $k$ successes out of $N$ trials, it seems like we must multiply this probability by the number of possible ways we get that outcome. Well that sounds like the number of possible ways that we can choose $k$ items out of a set $N$ (where in this analogy an "item" is one of the trials in the set of $N$ trials). Well that number is $N$ choose $k$ as we've established.
Thus the overall probability of getting $k$ successes out of $N$ trials is $$P = \frac{N!}{k! (N-k)!} p^k (1-p)^{N-k},$$
where $p$ is the probability of success for an individual trial. This is called the "Binomial Distribution".
