Following Bayes example is taken from here

A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test?

The Bayes formula for this example is:

$P(S|F)$ = $\frac{P(S)*P(F|S)}{P(F)}$

Using the following formula:

$P(S \cap F)$ = $P(S)*P(F|S)$

the above formula can be rewritten to this:

$P(S|F)$ = $\frac{P(S \cap F)}{P(F)}$

after assigning proper numbers this can be computed and this is the result:

$P(S|F)$ = $\frac{0.25}{0.42} = 0.60$

My question is if is possible to compute also $P(S)$ and $P(F|S)$ (as a compounds of $P(S \cap F)$) when $P(S|F)$ is known already?

  • $\begingroup$ PS: Can I somewhere find the documentation or cheat sheet for stackexchange equation editor? $\endgroup$ – Wakan Tanka Jun 3 '15 at 11:46
  • 1
    $\begingroup$ Don't know about stats.SE but there is a very useful MathJax basic tutorial and quick reference over on meta.math.SE. $\endgroup$ – Dilip Sarwate Jun 3 '15 at 12:58

If you know only P(S|F) and P(F), you can't definetely compute P(S) and P(F|S). Lets use your example: we have P(S|F)=0.6 and P(F)=0.42. Here we have Solution1: P(S) = 0.5, P(F|S) = 0.5 fits. But Solution2: P(S) = 0.25, P(F|S) = 1 fits as well. So the only thing you can tell is that P(S∩F) = 0.25 but you can't definetely compute P(S) and P(F|S) from this inputs.

  • $\begingroup$ Thank you. Is it possible to somehow extend the assignment so the P(S) and P(F|S) will be computable? $\endgroup$ – Wakan Tanka Jun 3 '15 at 11:45
  • 1
    $\begingroup$ Yes, now you have 2 variables only one equation for it. To solve variables, you have to add another equation. For example, this could be done by determining P(S U F). Given P(S U F) = 0.67 you can solve for P(S) = P(S U F) - P(S) - P(S∩F) = 0.5. Then you have P(F|S) = P(S∩F)/P(S) = 0.5. $\endgroup$ – hvedrung Jun 3 '15 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.