What helps here is to draw a tree to keep track of what is going on. But that is much easier with pencil and paper than on q computer screen , so we use a poor substitute, formulas.
Problem (c) asks for a conditional probability. The wording of the question makes it clear that we want $\Pr(f^c|P)$. We use the familiar formula $\Pr(f^c|P)\Pr(P)=\Pr(f^c\cap P)$. So if we can find $\Pr(P)$ and $\Pr(f^c\cap P)$ we will be finished. It will turn out that the answer is different from the one you produced.
Passing inspection can happen in two ways: (i) there is a flaw, and the bottle passes inspection or (ii) there is no flaw, and the bottle passes.
For (i), the probability of a flaw is $0.0002$. Given there is a flaw, the probability of passing is $0.005$, so the probability of (i) is $(0.0002)(0.005)$. For (ii), the probability of no flaw is $1-0.0002$. Given there is no flaw, the probability of passing is $0.99$, so the probability of (ii) is $(0.9998)(0.99)$.
Thus $\Pr(P)=(0.0002)(0.005)+(0.9998)(0.99)$.
During our calculation of $\Pr(P)$, we calculated $\Pr(f^c\cap P)$: it is just the probability of (ii). So
$$\Pr(f^c|P)=\frac{(0.9998)(0.99)}{(0.0002)(0.005)+(0.9998)(0.99)}.$$
Note that the term (i) makes a microscopic contribution. The required conditional probability is $1$ for all practical purposes. Makes sense, there are hardly any flaws to begin with.
Remark: For (a), the equation you wrote down is the relevant one. You want $\Pr(f|F)$. Now you need to do a calculation much like the one we did above. The numerical answer may surprise you.
I'm assuming a very large population of units from which to select, so
that the distribution is binomial, and that 'have issues' means 'are defective." Then we have the number of defective
units seen $X \sim Bin(n=10, p=.1)$ Thus, (1) $E(X) = np = 1$ as you say.
(2) V(X) = np(1-p) = .9, so SD(X) = \sqrt(.9) = 0.9487.$; I think you
overlooked a decimal point.
(3) $P\{Rej\} = P\{X \ge 3\} = 1 - P\{X \le 2\} = 0.0701,$ where
$P\{X \le 2\}$ is the sum of three terms, the first of which is 0.3487.
Use the formula for individual binomial probabilities (usually called
PMF or PDF) to fill in the gaps in the computation. Alternatively, see if your book has a binomial table for $n = 10, p=.01.$ or use software.
In R statistical software, you can get the answer from code '1 - pbinom(2, 10, .1)'.
Best Answer
The number of defective toys is a binomial $(250,0.04)$ random variable - each toy is treated as an independent Bernoulli trial where the "success" is voicebox defect. It suffices to use the standard formula for variance of a binomial variable.