How to find $P(A|B)$ when we only know $P(A)$ and $P(B)$?

Question

In case you’d want to know: I’m a 6th grade student and I am self-learning probability (that’s one of the things).

I know Bayes’ theorem:

$$ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} $$

Here’s an example:

$A$: The chance you get sick, which could be $P(A) = 0.000219$.

$B$: The chance you test negative, which could be $P(B) = 0.9993$.

Our goal is to find $P(A | B)$ which is the chance you are sick given you test negative. This is basically the false negative rate.

There are three things we need:

• The chance you get sick, which we have: it’s $0.021$.
• The chance you test negative, which we also have: it’s $0.9993$.
• The chance you test negative given you are sick, which we don’t know. This is the specificity.

To find the false negative rate, what would we do? We need $P(B | A)$ to find $P(A | B)$, but how do we find $ P(B | A)$ when we only know $P(A)$ and $P(B)$ (assuming we don’t have things like data sets at hand)?

Answer 1

You cannot. $P(B|A) = P(B \cap A)/P(A)$, so in order to know $P(B|A)$ or $P(A|B)$ you at least need knowledge of $P(A \cap B)$, which you can't in general determine using only $P(A)$ and $P(B)$.

Answer 2

Here’s another way to see why this can’t be possible.

A perfect test would have $P(B)=1-P(A)$ and $P(A\mid B)=P(B\mid A)=0$. If you could determine $P(A\mid B)$ from $P(A)$ and $P(B)$, then $P(B)=1-P(A)$ would have to yield $P(A\mid B)=0$ (since that’s what it would have to yield for a perfect test). But surely not every test that has $P(B)=1-P(A)$ (for instance uniformly drawing a random number in $[0,1]$ and giving a negative result if it’s $\ge P(A)$) has $P(A\mid B)=0$.

Answer 3

The exact value cannot be found. However, using the inequality $$\max\{P(A)+P(B)-1,0\} \le P(A\cap B) \le \min\{P(A),P(B)\},$$

and considering that

$$P(A | B) = \frac{P(A \cap B)}{P(B)},$$

you can find lower and upper bounds for $P(A | B)$ as follows:

$$\color{blue}{\max\left\{1-\frac{1-P(A)}{P(B)},0\right\} \le P(A | B) \le \min \left \{\frac{P(A)}{P(B)},1 \right\}}.$$

In your example, the false negative rate $P(A | B)$ can be bounded as follows:

$$ 0=\max\left\{1-\frac{1-0.000219}{0.9993},0\right\}\le P(A | B) \le \frac{0.000219}{0.9993}=0.0002191534,$$

which seems to be good enough for practical usage (telling the rate does not exceed 0.0002191534.)

As another example, if $P(A)=0.05$, then

$$0.0493345342 \le P(A | B) \le 0.0500350245.$$