0
$\begingroup$

Given

P(A|B)
P(not A| not B)
P(B)

I can find out

P(A | not B) = 1 -P(not A| not B)
P(A AND B) = P(A|B) * P(B)
P(A AND B) = P(B|A) * P(A)

Can I get P(A)?

$\endgroup$
  • $\begingroup$ This question is more appropriate on math.stackexchange. This site is primarily focused on statistics and statistical research. $\endgroup$ – Jeremiah Dunivin Sep 6 '18 at 8:14
  • 2
    $\begingroup$ Probability is crucial to Statistics, for me this is a legit question. $\endgroup$ – Digio Sep 6 '18 at 9:47
2
$\begingroup$

Yes, you can.

Let $S$ denote a sample space. Let $A$ and $B$ denote two events of $S$ such that \begin{align*} P(A \, | \, B) &= \alpha \\[5pt] P\big( \, \overline{A} \, | \, \overline{B} \big) &= \beta \\[5pt] P(B) = \gamma &\neq 0 \end{align*}

where I am using $\overline{A}$ and $\overline{B}$ for the complement of $A$ and $B$, respectively. In other words, not $A$ and not $B$.

Then by the Law of Total Probability:

\begin{align*} P(A) &= P(A \cap B) + P\big( \, A \cap \overline{B} \, \big) \\[8pt] &= P(B) \, P\big(\,A \, | \, B\,\big) + P\big(\, \overline{B} \, \big) P\big(\, A \, | \, \overline{B}\,\big) \\[8pt] &= \gamma \alpha + (1-\gamma) \big[1-P\big(\, \overline{A} \, | \, \overline{B} \, \big)\big] \\[8pt] &= \alpha \gamma + (1-\gamma)(1-\beta) \end{align*}

$\endgroup$

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.