---

Derive

1. Let the joint distribution of the events \(A\) and \(B\) be

   \[\begin{equation*}
     \begin{array}{r|cc}
       & B & \lnot B \\ \hline
       A & s & t \\ \lnot A & u & v
     \end{array}.
   \end{equation*}
   \]

2. Find the probabilities

   \[\begin{alignat*}{3}
     P(A)
     & = \dfrac{s + t}{s + t + u + v}
     && \qquad \text{using}
     & \qquad \begin{array}{cc} \boxed{s} & \boxed{t} \\ u & v \end{array}
     \\[0.5ex]
     \text{and} \qquad P(B)
     & = \dfrac{s + u}{s + t + u + v}
     && \qquad \text{using}
     & \quad \begin{array}{cc} \boxed{s} & t \\ \boxed{u} & v \end{array}.
   \end{alignat*}
   \]

3. Find the conditional probabilities

   \[\begin{alignat*}{3}
     P(A|B) & = \dfrac{s}{s + u}
     && \qquad \text{using} \qquad
     & \begin{array}{cc} \boxed{s} & \bcancel{t} \\ u & \bcancel{v} \end{array}
     \\[0.5ex]
     \text{and} \qquad P(B|A)
     & = \dfrac{s}{s + t}
     && \qquad \text{using} \qquad
     & \begin{array}{cc} \boxed{s} & t \\ \bcancel{u} & \bcancel{v} \end{array}.
   \end{alignat*}
   \]

4. See the equality

   \[\begin{align*}
     \Pr(A) \Pr(B | A)
     & \stackrel{?}{=} \Pr(B) \Pr(A | B)
     \\[1ex]
     \frac{s + t}{s + t + u + v} \cdot \frac{s}{s + t}
     & = \frac{s + u}{s + t + u + v} \cdot \frac{s}{s + u}
     \\[1ex]
     \frac{\cancel{(s + u)} s}{(s + t + u + v) \cancel{(s + u)}}
     & = \frac{\cancel{s + t) s}}{(s + t + u + v) \cancel{(s + t)}}
     \\[1ex]
     \frac{s}{s + t + u + v} & = \frac{s}{s + t + u + v}
     \\[1ex]
     1 & \stackrel{\checkmark}{=} 1.
   \end{align*}
   \]

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

   \[\begin{align*}
     \Pr(A) \Pr(B | A) & = \Pr(B) \Pr(A | B) \\[0.5ex]
     \Pr(A | B) & = \frac{\Pr(A) \Pr(B | A)}{\Pr(B)}.
   \end{align*}
   \]

   This is the Bayes’ theorem.