Bayes rule and base rate

Question

My homework question:

An inspector suspects that the food in the factory she is inspecting has been contaminated with a harmful chemical c. Such chemical contamination occurs in 5% of factories producing this food. The inspector has a test A for the chemical which registers positive with 100% certainty when the chemical is present, but the test also registers positive in 10% of cases where the chemical is not present. She decides to use this test to help her decide whether there is contamination.

1. Assume that the prior probability of contamination is equal to the base rate, and that the inspector’s test shows a positive result. Compute the posterior probability of contamination.
2. The inspector has another test B for chemical c which only registers positive 50% of the time when c is present, but has the advantage of never giving a false positive (i.e., if c is not present, the test will never say it is). The results of the two tests, A and B, are independent given the presence or absence of c. It turns out that when the inspector uses test B, the results are negative. In addition, the inspector knows that the factory is poorly maintained. The rate of contamination in factories with poor maintenance is twice as high as the rate in factories overall. Compute the posterior probability of contamination.

I know these should be rather basic, but I'm getting stuck. For #1 I've reached an answer, but I'm not sure it's correct:

Using Bayes rule the probability should be

$P(c|positive) = \frac{P(positive|c)P(c)}{P(positive)}$

Now I think that $P(positive)$ should be:

$P(positive) = P(positive,c)+P(positive,\neg c) = P(positive|c)P(c) + P(positive|\neg c)P(\neg c)$

Thus:

$P(c|positive) = \frac{P(positive|c)P(c)}{P(positive|c) P(c) + P(positive|\neg c) P(\neg c)}$

$P(c|positive) = \frac{1 * 0.05}{1 * 0.05 + 0.1 * 0.95} = 0.34$

Is this correct?

In part #2 I thought since they are supposed to be independent this must hold:

$ P(A_{positive} \cap B_{negative}|c) = P(A_{positive}|c)*P(B_{negative}|c) $

First I adjusted the base rate and recalculated #1:

$P(c|A_{positive}) = \frac{1 * 0.1}{1 * 0.1 + 0.1 * 0.9} = 0.52$

The questions says: "the results are negative". So B should be:

$P(c|B_{negative}) = \frac{P(B_{negative}|c)P(c)}{P(B_{negative}|c) P(c) + P(B_{negative}|\neg c) P(\neg c)} = \frac{0.5 * 0.1}{0.5 * 0.1 + 1 * 0.9} = 0.05$

$ P(c|A_{positive} \cap B_{negative}) = \frac{P(A_{positive} \cap B_{negative}|c) * P(c)}{P(A_{positive} \cap B_{negative})} = \frac{P(A_{positive}|c) * P(B_{negative}|c) * P(c)}{P(A_{positive} \cap B_{negative})} $

What should I do next?

Answer 1

It is simpler to start from scratch in your part 2, rather than proceed sequentially. I will show the answer this way, and then show where the step was "wrong" in your approach.

The independence means:

$$P(A_+B_-|c)=P(A_+|c)P(B_-|c)=P(B_-|c)=50\text{%}$$ $$P(A_+B_-|\neg c)=P(A_+|\neg c)P(B_-|\neg c)=P(A_+|\neg c)=10\text{%}$$

Where we have inserted the information given in the question (A never has false negative but 10% false positive, B never has false positive, but 50% false negative). You can use this to calculate the marginal probabilities

$$P(A_+B_-)=P(A_+B_-|c)P(c)+P(A_+B_-|\neg c)P(\neg c)=50\text{%}P(c)+10\text{%}P(\neg c)=14\text{%}$$

Where we have used $P(c)=10\text{%}$, we then have:

$$P(c|A_+B_-)=P(c)\frac{P(A_+B_-|c)}{P(A_+B_-)}=10\text{%}\frac{50\text{%}}{14\text{%}}=36\text{%}$$

Note that this is not derivable from $P(c|A_+)$ and $P(c|B_-)$ alone. The reason is that while $A_+$ and $B_-$ are conditionally independent, they are not unconditionally independent.

Now if we start from $P(c|A_+)=52\text{%}$, we need to multiply by the ratio $$\frac{P(B_-|A_+c)}{P(B_-|A_+)}=\frac{P(B_-|c)}{P(B_-|A_+)}$$

The likelihood when $c$ is true is unchanged because we have the logical relation $c\implies A_+$ which is the same thing as $c\land A_+=c$.

However, to calculate $P(B_-|A_+)$ we have:

$$P(B_-|A_+)=P(B_-|A_+c)P(c|A_+)+P(B_-|A_+\neg c)P(\neg c|A_+)$$ $$=P(B_-|c)P(c|A_+)+\frac{P(B_-A_+|\neg c)}{P(A_+|\neg c)}P(\neg c|A_+)$$ $$=50\text{%}\times 0.52+\frac{10\text{%}}{10\text{%}}\times 0.48=74\text{%}$$

And this is not the same as $P(B_-)=95\text{%}$. Plugging this in we get:

$$p(c|A_+B_-)=52\text{%}\times\frac{50\text{%}}{74\text{%}}=36\text{%}$$

Which is the same the earlier calculation I gave before.

UPDATE

@dmk has asked for some additional explanation about why this answer is intuitively the correct one not the answer given by the more straight-forward:

$$\frac{P(c|A_+B_-)}{P(\neg c|A_+B_-)}=\frac{P(c)}{P(\neg c)}\frac{P(A_+|c)}{P(A_+|\neg c)}\frac{P(B_-|c)}{P(B_-|\neg c)}$$

The answer is: they give the same answer!

$$\frac{P(c|A_+B_-)}{P(\neg c|A_+B_-)}=\frac{0.1}{0.9}\frac{1}{0.1}\frac{0.5}{1}=0.555\dots$$

Converting this to probabilities and we get $\frac{0.555}{1+0.555}=36\text{%}$

So I must retract my comments to @dmk's answer, noting that his procedure was correct, but his calculation was not. $B_-$ is twice as likely to occur if $c$ is not present compared to if it is present. So in his calculation, we modify from $10:9$ odds (after observing $A_+$) to $5:9$ odds (after observing $B_-$)

Answer 2

part 1:

1:19 (prior odds = 5%) x 10 (likelihood ratio for positive test results = for every 100 true positives there are 10 false positives, so true positive is 10x more likely) = 10:19 (posterior odds) = 34% chance that contaminating agent is present.

part 2:

test a: 1:9 (prior odds = 10%, baserate for presence of agent in poorly maintained plants) x 10 (likelihood ratio for positive test results) = 10:9 (posterior odds) = 53% chance that contaminating agent is present.

test b: 10:9 (prior odds based on test a result & baserate for poorly maintained plants) x 1 (likelihood ratio for negative result-- just as likely to be false negative as true negative) = 10:9 = 53% chance that agent is present.

Answer 3

From your independence statement,

\begin{equation} P( A_{+} \bigcap B_{+} ) = P(A_{+} ) P(B_{+}) \nonumber \end{equation} and the definitions

\begin{equation} P(A_{+} \bigcap B_{+} | c) \equiv \frac{P(A_{+} \bigcap B_{+} \bigcap c)}{P(c)} \nonumber \end{equation}

\begin{equation} P(A_{+} | c) \equiv \frac{P(A_{+} \bigcap c)}{P(c)} \nonumber \end{equation} and

\begin{equation} P(B_{+} | c) \equiv \frac{P(B_{+} \bigcap c)}{P(c)} \nonumber \end{equation} can you algebraically show that \begin{equation} P(A_{+} \bigcap B_{+} | c) \equiv \frac{P(A_{+} \bigcap B_{+} \bigcap c)}{P(c)} = \frac{P(A_{+} \bigcap c)}{P(c)} \frac{P(B_{+} \bigcap c)}{P(c)} \equiv P(A_{+} | c) P(B_{+} | c) \nonumber \end{equation}