I came across this problem in an additional exercises section without any solutions and I am wondering if my approach is correct. More specifically, with regard to the 'may or may not' phrasing as below.
Question: An electrician must check lightbulbs to detect defects. The employee must check exactly 100 lightbulbs on a given day. It is assumed that lightbulbs may be defects with probability p = 0.015, and that each lightbulb may or may not be a defect independently from the others. Each lightbulb is valued at $5
The loss that occurred on a given day is denoted L. This corresponds to the cost for the process of checking n lightbulbs, which is equal to $\sqrt{n}$ dollars, plus the value of any defect lightbulbs found. Calculate the expected loss.
My solution:
n = 100 lightbulbs, p = 0.015, x = number of defect lightbulbs. (Binomial)
\begin{equation*} \begin{aligned} P_{X}\;({X = x})\; &= \binom{n}{x} \cdot p^x(1-p)^{n-x} \\[5pt] &= \binom{100}{x} \cdot 0.015^x \cdot (1-0.015)^{100-x} \\[5pt] &= \binom{100}{x} \cdot 0.015^x \cdot 0.985^{100-x} \end{aligned} \end{equation*}
Cost of checking 100 lightbulbs = $\sqrt{100}$ = $10
\begin{equation*} \begin{aligned} Cost = \sum_{x=0}^{100} x \cdot P_{X}\, ({\{X = x}\}) \cdot 5 &= 100 * 0.015 * 5 = $7.5 \end{aligned} \end{equation*}
Expected Cost = 10 + 7.5 = $17.50