Problem
A population consists of people who are either blue or not blue. A sample of $5$ people are chosen one at a time, with replacement. Suppose that $\%20$ of the population is blue. Let $X$ be a random variable which gives the number of blue people in the sample. Write the density function, $f_X$, of $X$.
Attempted solution...
Clearly, the support set of $X$ is given by $S = \{0,1,2,3,4,5\}$. For $n\in S$, it seems to me that the probability of $n$ is given by $(0.8^{5 - n})(0.2^n)$, giving the density function:
\begin{cases} 0.8^5 & x = 0\\ (0.8^4)(0.2) & x = 1\\ (0.8^3)(0.2^2) & x = 2\\ (0.8^2)(0.2^3) & x = 3\\ (0.8^1)(0.2^4) & x = 4\\ (0.2^5) & x = 5\\ \end{cases}
This seems, at first, fairly intuitive. However, summing the probabilities yields $0.4368$, where I'd expect the sum to be $1$. I believe, then, that I've erred somewhere. I'm not sure where, however? Or perhaps, for some reason, I shouldn't expect the sum to be $1$?
What's going on here?
$(0.8)^4(0.2)$ is the probability for selecting four non-blue people, followed by one blue person, in that order.
There are $5$ ways to rearrange that ordering. That is the count for ways to select which position the blue person is placed in a selection.
The probability that there is exactly one blue person (somewhere) among the five selected is $5(0.8)^4(0.2)$
In general, there are $\binom 5k$ ways to select $k$ positions among $5$.
So your table needs to be modified using this Binomial Coefficient.
PS: The Binomial Theorem states $\displaystyle(0.8+0.2)^5=\sum_{k=0}^5 \binom 5k (0.8)^{5-k}(0.2)^k$