Selecting a number at random from a binomial distribution? 
Someone selects a number at random from a binomial distribution with $n=5$ attempts and probability of success $p=.90$. Compute the probability that the number that person selected is greater than $4.5$. 

Let $k$ be the number of successes. Then I know that to find the probability of binomial random variable is just 
$$ \binom{n}{k} p^k (1-p)^{n-k}.$$
However, after reading the question carefully "selects a number at random" I don't know what the number of successes should be. At first I thought it was just $1$ because we only needed the $4.5$ once to have a success but now I don't know.

If a random variable $X$ has a binomial distribution, then its PMF is defined as
$$P(X=k) = \binom{n}{k}p^k(1-p)^k \qquad k=0,1,\ldots,n$$
where $n$ is the number of independent trials or attempts and $p$ is the probability of success at each trial. Note that $X$ can take values $0,1,\ldots,n$, meaning that we could have $0$ successes, $1$ success, $\dots$, $n$ successes.
When we are told that someone selects at random a number from that distribution, that means that the selected number would be one of the possible numbers that $X$ can take. In that way, the question can be formulated as
$$P(X>4.5)$$
Can you follow from here?
