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Derive

1. Let the joint distribution of the events $A$ and $B$ be

   $$\begin{array}{rcc}& B& \mathrm{\neg }B\\ A& s& t\\ \mathrm{\neg }A& u& v\end{array}.$$

2. Find the probabilities

   $$\begin{array}{rlrlr}P(A)& ={\displaystyle \frac{s+t}{s+t+u+v}}& & \text{using}& \begin{array}{cc}\boxed{{\displaystyle s}}& \boxed{{\displaystyle t}}\\ u& v\end{array}\\ \text{and}P(B)& ={\displaystyle \frac{s+u}{s+t+u+v}}& & \text{using}& \begin{array}{cc}\boxed{{\displaystyle s}}& t\\ \boxed{{\displaystyle u}}& v\end{array}.\end{array}$$

3. Find the conditional probabilities

   $$\begin{array}{rlrlr}P(A|B)& ={\displaystyle \frac{s}{s+u}}& & \text{using}& \begin{array}{cc}\boxed{{\displaystyle s}}& \bcancel{t}\\ u& \bcancel{v}\end{array}\\ \text{and}P(B|A)& ={\displaystyle \frac{s}{s+t}}& & \text{using}& \begin{array}{cc}\boxed{{\displaystyle s}}& t\\ \bcancel{u}& \bcancel{v}\end{array}.\end{array}$$

4. See the equality

   $$\begin{array}{rl}\mathrm{Pr}\left(A\right)\mathrm{Pr}(B|A)& \stackrel{?}{=}\mathrm{Pr}\left(B\right)\mathrm{Pr}(A|B)\\ \frac{s+t}{s+t+u+v}\cdot \frac{s}{s+t}& =\frac{s+u}{s+t+u+v}\cdot \frac{s}{s+u}\\ \frac{\cancel{(s+u)}s}{(s+t+u+v)\cancel{(s+u)}}& =\frac{\cancel{s+t)s}}{(s+t+u+v)\cancel{(s+t)}}\\ \frac{s}{s+t+u+v}& =\frac{s}{s+t+u+v}\\ 1& \check{=}1.\end{array}$$

5. Divide by $P(B)$ and swap sides,

   $$\begin{array}{rl}\mathrm{Pr}\left(A\right)\mathrm{Pr}(B|A)& =\mathrm{Pr}\left(B\right)\mathrm{Pr}(A|B)\\ \mathrm{Pr}(A|B)& =\frac{\mathrm{Pr}\left(A\right)\mathrm{Pr}(B|A)}{\mathrm{Pr}\left(B\right)}.\end{array}$$

   This is the Bayes’ theorem.