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I'm wondering if my answer for C would be correct and if I was understanding A correctly? I heard drawing venn diagram could help but I'm not sure how to convert the numbers to a diagram.

A quality-control program at a plastic bottle production line involves inspecting finished bottles for flaws such as microscopic holes. The proportion of bottles that actually have such a flaw is only 0.0002. If a bottle has a flaw, the probability is 0.995 that it will fail the inspection. If a bottle does not have a flaw, the probability is 0.99 that it will pass the inspection.

$P(F)=$ fails inspection
$P(P)=$ Passes inspection
$P(f)=$ has flaw.

a. If a bottle fails inspection, what is the probability that it has a flaw?
$P(F\cap f) = P(F) \times P(f|F)$

c. If a bottle passes inspection, what is the probability that it does not have a flaw? $P(P \cap f^c) = \frac{(0.99)}{(0.995)} = 0.9949$

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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.

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