How does bayes rule guarentee a value between 0 and 1?

Question

Probabilities are values between 0 and 1. In Bayes Rule:

$$ \Pr(A\mid B) = \Pr(B \mid A) \cdot \frac{\Pr(A)}{\Pr(B)}$$

$\Pr(A)$ is independent of $\Pr(B)$. So $\Pr(B)$ can be very small (lets say 0.1) and $\Pr(A)$ can be large (lets say 0.5) and the $\Pr(B \mid A)$ can also be large (lets say 1.0). Then the $\Pr(A\mid B)$ is $> 1$ (5 in the example). How then is this a probability?

Where am I messing this up. I must be missing something obvious.

Answer 1

All 3 "possibilities" you mention cannot occur simultaneously. Recall that $$ \Pr[B\mid A] = \frac{\Pr[A\cap B]}{\Pr A} \leq \frac{\Pr B}{\Pr A} $$ so if $\Pr A$ large and $\Pr B$ is small, $\Pr[B\mid A]$ has to be even smaller.

(the inequality comes from the fact that $A\cap B\subseteq B$).

Answer 2

So P(B) can be very small (lets say 0.1) and P(A) can be large (lets say 0.5) and the P(B|A) can also be large (lets say 1.0).

If P(B)=0.1 and P(A)=0.5 then P(B|A) is at most 0.2.

To show this note that P(B|A)=P(A∩B)/P(A) and P(A∩B)≤P(B) hence P(B|A)≤P(B)/P(A).