Probability of luck spell? If $p$ is the probability of success and $q = 1 - p$ is the probability of failure, what is the probability of $0,1,2, \ldots, 5$ successes in $5$ independent trials of the experiment?
I thought of using binomial random variable formula, but I am not sure.
 A: Let's start with $0$ successes. Then all $5$ trials have to be failures the probability of that happening is 
$$q\cdot q \cdot q \cdot q \cdot q = q^5$$
where each $q$ represents the probability of $i$'th failure for $i=1,2,3,4,5$.
Now if we want just one success there are $5$ different ways to obtain that. We must fail $4$ times and success once, but that success can happen on either one of the $5$ trials. So all the outcomes are
\begin{array}
 pp \cdot q \cdot q \cdot q \cdot q\\
q\cdot p \cdot q \cdot q \cdot q\\
q\cdot q \cdot p \cdot q \cdot q\\
q\cdot q \cdot q \cdot p \cdot q\\
q\cdot q \cdot q \cdot q \cdot p
\end{array}
each one happening with probability $p \cdot q^4 = p(1-p)^4$. Summing the probabilities up brings us to $5p(1-p)^4$ probability of having exactly one success.
In order to proceed we need to see what happens if we have exactly two failures. A single configuration with 2 successes will always have the same probability $p^2 (1-p)^3$. The important thing here is to count in how many different ways can we obtain two successes and three failures. In other words how many ways are there to choose $2$ elements out of a set with $5$ elements. The answer is $5\choose{2}$, you should already be familiar with this. Now in order to conclude- the probability of obtaining exactly two successes and three failures in five trials is
$$
\{ \text{number of configurations} \} \cdot \{\text{probability of a single 2 success, 3 fail configuration} \} = {5\choose 2} p^2 (1-p)^3
$$
I believe you can finish the solution by yourself from here on.
