How come $P(AB|C)=P(A|C).P(B|C)$ for events $A,B, C$ under any condition

Question

If a machine is correctly set up, it produces $90\%$ acceptable items. If it is incorrectly set up, it produces only $40\%$ acceptable items. Past experience shows that $80\%$ of the set ups are correctly done. If after a certain set up, the machine produces $2$ acceptable items, find the probability that the machine is correctly setup.

This is solved in my reference as

$A:$ machine produces acceptable item, $B:$ machine is correctly setup, $C:$ machine produces $2$ acceptable items $$ P(B)=80/100=\frac{4}{5},\;\;P(B')=\frac{1}{5}\\ P(C|B)=90\%.90\%=\frac{9}{10}.\frac{9}{10}\\ P(C|B')=40\%.40\%=\frac{2}{5}.\frac{2}{5}\\ P(C)=P(C|B)P(B)+P(C|B')P(B')=\frac{9}{10}\frac{9}{10}\frac{4}{5}+\frac{2}{5}\frac{2}{5}\frac{1}{5}=\frac{85}{125}=\frac{17}{25}\\P(B|C)=\frac{P(B)P(C|B)}{P(C)}=\frac{\frac{4}{5}.\frac{9}{10}\frac{9}{10}}{\frac{17}{25}}=\frac{4.9.9}{17.20}=\frac{81}{85}=0.95 $$

Why not $P(C)=P(AA)=P(A).P(A)$ where $P(A)=P(A|B)P(B)+P(A|B')P(B')$ ?.

How come we define $P(C|B)=P(AA|B)=P(A|B).P(A|B)$ ?. How do I prove it and where does this come from ?

Answer 1

The problem is with $P(C) = P(A).P(A)$ since two acceptable items being produced are not independent of each other, but only conditionally independent if we know $B$, which we don't here.

Two acceptable items being produced given that we know whether the machine is correctly setup or not are independent events, so their joint probability $P(C | B)$ equals the product of the individual events $P(A|B).P(A|B)$.

So $$P(C | B) = P(A | B).P(A | B)$$ and $$P(C | B') = P(A | B').P(A | B')$$ but $$\begin{align} P(C) &= P(C | B).P(B) + P(C | B').P(B') \\ &= P(A | B).P(A | B).P(B) + P(A | B').P(A|B').P(B') \\ &\ne P(A).P(A) \end{align}$$

In terms of where $P(C | B) = P(A|B).P(A|B)$ comes from, this statement assumes that producing two acceptable items on a correctly setup machine happen independently. This assumption would be incorrect, for example, if one failure to produce an item had a knock-on effect and made it more likely that the next item would also fail to be acceptable.

If we assume that this is not the case — that there is no correlation / dependence between whether or not two consecutively produced items are acceptable — we can derive $P(C | B) = P(A|B) P(A|B)$ like below. Note that for clarity, we distinguish between the two $A$s, labelling them $A_1$ (the first item is acceptable) and $A_2$ (the second item is acceptable).

$$\begin{align} P(C | B) &= P(A_1, A_2 | B)\\ &= P(A_1 | A_2, B).P(A_2 | B) \qquad\text{(chain rule)}\\ &= P(A_1 | B).P(A_2 | B) \qquad\text{(assumption: $A_1$ and $A_2$ independent)}\\ &= P(A | B).P(A | B) \end{align}$$

Answer 2

There is more going on here than what meets the eye.

Let's consider $C \mid B$. This is the probability that, given the machine is correctly set up, two acceptable items are produced.

This is not the best wording I've seen for this type of situation, but let's try to explain why it would be $(0.9)(0.9)$.

Suppose you grabbed two items that this machine produced, say items $I_1$ and $I_2$.

There are four possible situations:

• $I_1$ and $I_2$ both acceptable, with probability $(0.9)(0.9)$ assuming independence of each item's acceptability. Note that I do NOT use the same symbol for each of the two items (the symbol "$AA$" is confusing in this situation); I use $I_1$ to mean acceptability of the first item, and $I_2$ to mean acceptability of the second item.*
• $I_1$ and $I_2$ both not acceptable, with probability $(0.1)(0.1)$.
• $I_1$ is acceptable and $I_2$ is not acceptable, with probability $(0.9)(0.1)$.
• $I_2$ is acceptable and $I_1$ is not acceptable, with probability $(0.9)(0.1)$.

In case you are familiar with this terminology, this situation outlines a binomial experiment with 2 trials.

Thus, it is clear that the probability $P(C \mid B) = (0.9)(0.9)$, because it is the only situation above where two items are acceptable.

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As for your second question, the reason why you would not just compute $(0.9)(0.9)$ to get $P(C)$ is because $(0.9)(0.9)$ assumes that the machine is correctly set up, which is not necessarily true. You must also take into account the possibility that the machine is not correctly set up and calculate that probability, and then weigh them according to what is in your solution.

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*To elaborate on the edited version of the question: why do we assume independence? The main reason is because it's something that intuitively makes sense: the acceptability of one item should not influence the acceptability of another item. However, in practice, this is often not the case.