A package of 16 identical components contains exactly one faulty part, which is too light. This should now be found by means of a balance. To do this, one of the parts is randomly removed from the package and weighed (and then put aside) until the faulty part is found. Let X be the number of measurements required.
a) Determine the expected value and variance of the random variable X.
Tip: By the 15th measurement at the latest, you know which part is the lighter one. But first, model the simpler case, which also weighs the 16th part, if the first 15 parts were not all that light.
b) Is there a method where the variance of the necessary measurements is zero?
Trial answers:
a) I'll just calculate the variance and expected value of 16 balls.
E(x) = 1/16(1+2+...+16) = 8.5
Var(x) = (E(x^2)) - (E(x))^2 = 1/16(2+4+...+256) - (8.5)^2 = 21.25
Am I right until this point?
b) Maybe using negative binomial distribution? but how?
I hope someone can help me to answer these questions, thanks a lot :)
(a) looks good.
Just an alternate method:
There are formulas for the expected value and variance of the number trials until the first success when dealing without replacement. Let $g$ be a good draw (faulty), $b$ be a bad draw, and let $N$ be the total amount of objects.
$$E(X)=\frac{N+1}{g+1}=\frac{16+1}{1+1}=8.5$$
\begin{align} Var(X) &=\frac{2b(b-1)}{(g+1)(b+2)}+\frac{b}{g+1}-\left(\frac{b}{g+1}\right)^2\\\\ &=\frac{gb(g+b+1)}{(g+1)^2(g+2)}\\\\ &=\frac{1\cdot15\cdot(1+15+1)}{({1+1})^2\cdot({1+2})}\\\\ &=21.25 \end{align}
These both agree with your results.
For (b)
I don't know if this is allowed. Maybe you could draw $2$ balls. If one is lighter than the other, then replace the one that is lighter (pretending like you don't know that it is the faulty one). If they are the same weight, then neither is faulty, so randomly replace one of them. Repeat this method until you only have $2$ balls left, the faulty one and a non-faulty one, guaranteeing $X$ to be $16$.