Examples

Bayes’ Theorem: COVID tests

Assume that 0.5% of people have COVID. Suppose a COVID test gives a (true) positive on 96% of patients who have COVID, but gives a (false) positive on 2% of patients who do not have COVID. Bob tests positive. What is the probability that Bob has COVID?

Solution

(1) Label events:

Event $A_{P}$ : Bob is actually positive for COVID
Event $A_{N}$ : Bob is actually negative; note $A_{N} = A_{P}^{c}$
Event $T_{P}$ : Bob tests positive
Event $T_{N}$ : Bob tests negative; note $T_{N} = T_{P}^{c}$

(2) Identify known data:

Know: $P [T_{P} | A_{P}] = 96 %$
Know: $P [T_{P} | A_{N}] = 2 %$
Know: $P [A_{P}] = 0.5 %$ and therefore $P [A_{N}] = 99.5 %$

We seek: $P [A_{P} | T_{P}]$

(3) Translate Bayes’ Theorem:

Set $A = T_{P}$ and $B = A_{P}$ in Bayes’:

P [A_{P} | T_{P}] = P [A_{P}] \cdot \frac{P [T_{P} | A_{P}]}{P [T_{P}]}

We know all values on the right except $P [T_{P}]$

(4) Denominator: apply Total Probability (Division into Cases):

Observe that $T_{P} \cap A_{P}$ and $T_{P} \cap A_{N}$ are exclusive events, and that:

T_{P} = T_{P} \cap A_{P} ⋃ T_{P} \cap A_{N} .

Therefore:

P [T_{P}] = P [A_{P}] \cdot P [T_{P} | A_{P}] + P [A_{N}] \cdot P [T_{P} | A_{N}]

Plug in data and compute:

≫ ≫ P [T_{P}] = \frac{5}{1000} \cdot \frac{96}{100} + \frac{995}{1000} \cdot \frac{2}{100} ≫ ≫ \approx 0.0247

(5) Plug in and compute:

\begin{matrix} P [A_{P} | T_{P}] = P [A_{P}] \cdot \frac{P [T_{P} | A_{P}]}{P [T_{P}]} \\ ≫ ≫ 0.96 \cdot \frac{0.005}{0.0247} ≫ ≫ \approx 19 % \end{matrix}

Intuition - COVID testing

Some people find this low number surprising. In order to repair your intuition, think about it like this: roughly 2.5% of tests are positive, with roughly 2% coming from false positives, and roughly 0.5% from true positives. Only $1 / 5$ of all the positive results are true ones!

(This rough approximation assumes that $96 % \approx 100 %$ .)

If two tests both come back positive, the odds of COVID are now 98%.

If only people with symptoms are tested, so that, say, 20% of those tested have COVID, that is, $P [A_{P} | T_{P}] = 20 %$ , then one positive test implies a COVID probability of 92%.

Inferring bin from marble

There are marbles in bins in a room:

Bin 1 holds 7 red and 5 green marbles.
Bin 2 holds 4 red and 3 green marbles.

Your friend goes in the room, shuts the door, and selects a random bin, then draws a random marble. (Equal odds for each bin, then equal odds for each marble in that bin.) He comes out and shows you a red marble.

What is the probability that this red marble was taken from Bin 1?

Solution

(1) Label events:

Event $B_{1}$ : friend chooses Bin 1.

Event $B_{2}$ : friend chooses Bin 2.

Event $R$ : friend draws a red marble.

Event $G$ : friend draws a green marble.

Answer will be $P [B_{1} | R]$ .

(2) Identify knowns:

Know $P [R | B_{1}] = 7 / 12$ .

Know $P [G | B_{1}] = 5 / 12$ .

Know $P [R | B_{2}] = 4 / 7$ .

Know $P [G | B_{2}] = 3 / 7$ .

Know $P [B_{1}] = P [B_{2}] = 1 / 2$ .

(3) Apply Bayes’ Theorem for $P [B_{1} | R]$ ::
$P [B_{1} | R] = P [R | B_{1}] \cdot \frac{P [B_{1}]}{P [R]}$
Division into Cases for the denominator:
$P [R] = P [B_{1}] \cdot P [R | B_{1}] + P [B_{2}] \cdot P [R | B_{2}]$

(3) Plug in data and compute:
$\begin{matrix} P [R] ≫ ≫ \frac{1}{2} \cdot \frac{7}{12} + \frac{1}{2} \cdot \frac{4}{7} ≫ ≫ \frac{97}{168} \\ P [B_{1} | R] ≫ ≫ \frac{7}{12} \cdot \frac{1 / 2}{97 / 168} ≫ ≫ \frac{49}{97} \end{matrix}$

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Examples

Bayes’ Theorem: COVID tests

Inferring bin from marble

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Table of Contents

Examples

Bayes’ Theorem: COVID tests

Inferring bin from marble

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