Guided example — Drug trial

Example — Drug trial: from the counts to the three representations

A medical trial reports: $35$ patients took the drug ($A$), of whom $10$ had a headache ($B$); $22$ patients did not take it ($\overline{A}$), of whom $15$ had a headache.

Step 1 — Counts $\to$ Table. The data are already in tabular form:

	$A$ (drug)	$\overline{A}$ (no drug)	Tot
$B$ (headache)	$10$	$15$	$25$
$\overline{B}$ (healthy)	$25$	$7$	$32$
Tot	$35$	$22$	$57$

Marginals in probability: $P(A)=35/57\approx 0,614$, $P(\overline{A})=22/57\approx 0,386$, $P(B)=25/57\approx 0,439$.

Step 2 — Table $\to$ Venn. The four joint probabilities are the fractions of the cells over the total:

$P(A\cap B)=\tfrac{10}{57},\quad P(A\cap\overline{B})=\tfrac{25}{57},\quad P(\overline{A}\cap B)=\tfrac{15}{57},\quad P(\overline{A}\cap\overline{B})=\tfrac{7}{57}.$

The Venn diagram with the absolute counts: in the intersection the $10$ patients with drug and headache, outside the ovals the $7$ without drug and healthy.

Step 3 — Table $\to$ Tree. We compute the conditionals for each row of $A$:

$P(B\mid A)=\tfrac{10}{35}=\tfrac{2}{7}\approx 0,286,\qquad P(B\mid\overline{A})=\tfrac{15}{22}\approx 0,682.$

So among those who take the drug only $28\%$ have a headache, among those who do not take it $68\%$: the drug is protective (negative correlation between $A$ and $B$).

The tree conditioned on $A$: on the branches the conditionals, on the leaves in red the products (the joint probabilities).

Check of the total probability (sum over the paths): $P(B) = \tfrac{35}{57}\cdot\tfrac{2}{7} + \tfrac{22}{57}\cdot\tfrac{15}{22} = \tfrac{10}{57}+\tfrac{15}{57}=\tfrac{25}{57}$ ✓, in agreement with the marginal of the table.

Seventh conversion: tree $\to$ inverted tree. We want the tree conditioned on $B$ ($B\to A$ instead of $A\to B$). We proceed with the three steps seen for the inversion of the tree:

$\text{joint probabilities already computed: }\tfrac{10}{57},\ \tfrac{25}{57},\ \tfrac{15}{57},\ \tfrac{7}{57};$

$\text{marginals: }P(B)=\tfrac{25}{57},\ P(\overline{B})=\tfrac{32}{57};$

$P(A\mid B)=\frac{10/57}{25/57}=\tfrac{10}{25}=\tfrac{2}{5},\qquad P(A\mid\overline{B})=\frac{25/57}{32/57}=\tfrac{25}{32}.$

The inverted tree has first level $B,\overline{B}$ with weights $25/57$ and $32/57$, second level $A,\overline{A}$ with $2/5$ and $25/32$ (and the complements).

Reading. Given that the patient had a headache, the probability that they had taken the drug is $P(A\mid B)=2/5=40\%$: the majority of those who have a headache did not take the drug (negative correlation, as expected).

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Topics: Probability
Concepts: Tree diagram · Venn diagram · Tree inversion · Moving between representations · Contingency table
Methods: Tree diagram · Moving between representations · Conditional probability · Contingency table
Skills: Probability calculation · Modelling