---

Law of total probability

Define

\[\begin{equation*}
  P(A) = P(A | B) \, P(B) + P(A | \lnot B) \, P(\lnot B).
\end{equation*}
\]

Prove

1. Define probabilities of events \(A\) and \(B\) as

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

2. Substitute \(s, t, u, v\) into the law

   \[\begin{equation*}
     \begin{array}{ccccccccc}
       P(A) & = & P(A | B) & \cdot & P(B) & + & P(A | \lnot B) & \cdot & P(\lnot B)
       \\[3ex]
       \begin{array}{cc} \boxed{s} & \boxed{t} \\ u & v \end{array}
       & = & \begin{array}{cc} \boxed{s} & \bcancel{t} \\ u & \bcancel{v} \end{array}
       & \cdot & \begin{array}{cc} \boxed{s} & t \\ \boxed{u} & v \end{array}
       & + & \begin{array}{cc} \bcancel{s} & \boxed{t} \\ \bcancel{u} & v \end{array}
       & \cdot & \begin{array}{cc} s & \boxed{t} \\ u & \boxed{v} \end{array}
       \\[3ex]
       \dfrac{s + t}{s + t + u + v}
       & = & \dfrac{s}{s + u} & \cdot & \dfrac{s + u}{s + t + u + v}
       & + & \dfrac{t}{t + v} & \cdot & \dfrac{t + v}{s + t + u + v}.
     \end{array}
   \end{equation*}
   \]

3. Verify the equality

   \[\begin{align*}
       \frac{s + t}{s + t + u + v}
       & \stackrel{?}{=}
       \frac{s}{s + u}
       \cdot
       \frac{s + u}{s + t + u + v}
       +
       \frac{t}{t + v}
       \cdot
       \frac{t + v}{s + t + u + v}
       \\[2ex]
       \frac{s + t}{s + t + u + v}
       & =
       \frac{s \cancel{(s + u)}}{\cancel{(s + u)} (s + t + u + v)}
       +
       \frac{t \cancel{(t + v)}}{\cancel{(t + v)} (s + t + u + v)}
       \\[2ex]
       \frac{s + t}{s + t + u + v}
       & =
       \frac{s}{s + t + u + v}
       +
       \frac{t}{s + t + u + v}
       \\[2ex]
       \frac{s + t}{s + t + u + v}
       & =
       \frac{s + t}{s + t + u + v}.
       \\[2ex]
       1
       & \stackrel{\checkmark}{=}
       1.
   \end{align*}
   \]

\(\blacksquare\)

Use

A person tests positive for a disease.

• The epidemiology of the disease is 1 in 5000 people.
• The sensitivity of the test is 99%.
• The specificity of the test is 100%.

What is the probability of the person having the disease?

Let

	\(t\)	denote the event “tests positive” and
	\(d\)	denote the event “has the disease”.

We know that

\[\begin{align*}
  P(d) & = 1 / 5000 = 0.0002, && \text{\(1\) in \(5000\) prevalence} \\
  P(t | d) & = 0.99, && \text{\(99\%\) true positives} \\
  P(\lnot t | \lnot d) & = 1. && \text{\(100\%\) true negatives}
\end{align*}
\]

We must find

\[\begin{equation*}
  P(d | t).
\end{equation*}
\]

N.B. We cannot use the Bayes’ Theorem right away because \(P(t)\) is unknown.

1. Find the probability of false positives

   \[\begin{equation*}
     P(t | \lnot d) = 1 - P(t | d) = 1 - 0.99 = 0.01.
   \end{equation*}
   \]

2. Find the probability of not having the disease

   \[\begin{equation*}
     P(\lnot d) = 1 - P(d) = 1 - 1 / 5000 = 0.9998.
   \end{equation*}
   \]

3. Find the probability of testing positive

   \[\begin{align*}
     P(t)
     & = P(t | d) \, P(d) + P(t | \lnot d) \, P(\lnot d) \\
     & = 0.99 \cdot 0.0002 + 0.01 \cdot 0.9998 \\
     & = 0.010196
   \end{align*}
   \]

   using the law of total probability.

4. Find the probability of having the disease given a positive test

   \[\begin{equation*}
     P(d | t)
     = \frac{P(d) \, P(t | d)}{P(t)}
     = \frac{0.0002 \cdot 0.99}{0.010196}
     \approx 0.0194193 \\
     = 2\%
   \end{equation*}
   \]

   using the Bayes’ theorem.

(Webb & Sidebotham, 2020)