Probability of hitting "1" in an array. We have an array of length m, where half of the entries in the array are labeled "1" and the rest "0". What is the probability of hitting 3  "1's" in 3 tries?
I believe it should be 0.5 * 0.5 * 0.5 which is 0.125, but I am unsure if I am correct. I essentially just used the probability of hitting a "1" as 0.5 as we know half of the array is filled with "1", and since we have 3 tries it should be 0.5 for each try.
Any help/hints would be appreciated.
 A: Your argument is correct. If you choose the array location independently "at random" each time your experiment is just like flipping a coin three times.
If once you have hit a $1$ you can't hit that spot again then $m$ must be at least $6$ and must be even. Then the probability for three $1$'s in three tries would be
$$
\frac{m/2}{m}\frac{m/2 -1}{m-1}\frac{m/2 -2}{m-2}.
$$
For $m=6$ that's just $1/20$.
(From the start you should be writing your probabilities as fractions: $1/2$ instead of $0.5$. The meaning is clearer and it's easier to do the work when you encounter probabilities like $1/3$.)
A: I advise you, each time it is possible, to look for a classical distribution. This will help you a lot in the future when dealing with more complex probability issues.
Here you have a random variable $S$ (number of successes in 3 trials) following a Binomial distribution $$B(n=3,p=\frac12)$$
because it is a repeated experiment you do $n=3$ times independently, each time  with a probability of success $p=\tfrac12$.
The probability of $S$ "successes" are distributed in this way (with $q=1-p$):
$$P(S=0)=q^3, \ \ P(S=1)=3pq^2, \ \ P(S=2)=3p^2q, \ \ P(S=3)=p^3$$
(please note that the sum $q^3+3pq^2+3p^2q+p^3=(p+q)^3=1$ involving the binomial expansion, explaining the name of the distribution).
You are in the fourth case (3 "successes" in 3 attempts), giving you indeed
$$P(S=3)=p^3=\tfrac18=0.125$$
