# Getting probablity of an event given a conditional probality

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)?

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

## 1 Answer

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*}