Example — Influenza vaccine

Example — Influenza vaccine

Giorgio has to decide whether to get vaccinated against influenza. We know that the vaccine has partial coverage. Let us define:

• $A$ = “Giorgio gets vaccinated”;
• $B$ = “Giorgio catches influenza”.

From the statistical data: $P(A) = \tfrac{1}{5}$, $P(B) = \tfrac{1}{3}$, and among the vaccinated the proportion of those who fall ill is $P(B\mid A) = \tfrac{1}{6}$.

Does the vaccine protect? If $A$ and $B$ were independent, getting vaccinated would make no difference. Let us compare $P(B\mid A)$ with $P(B)$:

$P(B\mid A) = \tfrac{1}{6} \approx 0,167 \quad < \quad P(B) = \tfrac{1}{3} \approx 0,333.$

Getting vaccinated halves the probability of falling ill: the events $A$ and $B$ are negatively correlated. By the law of compound probability:

$P(A\cap B) = P(A)\cdot P(B\mid A) = \tfrac{1}{5}\cdot\tfrac{1}{6} = \tfrac{1}{30}.$

So the correlation is:

$\operatorname{corr}(A,B) = \frac{1/30}{(1/5)(1/3)} - 1 = \frac{1/30}{1/15} - 1 = \tfrac{1}{2}-1 = \boxed{-\tfrac{1}{2}}.$

The negative correlation confirms the protective effect of the vaccine.

Let us fill in the table. With $P(A\cap B) = \tfrac{1}{30}$ we deduce all the cells:

	$A$ (vaccinated)	$\overline{A}$ (not vacc.)	Total
$B$ (influenza)	$\tfrac{1}{30}$	$\tfrac{9}{30}$	$\tfrac{10}{30}=\tfrac{1}{3}$
$\overline{B}$ (healthy)	$\tfrac{5}{30}$	$\tfrac{15}{30}$	$\tfrac{20}{30}=\tfrac{2}{3}$
Total	$\tfrac{6}{30}=\tfrac{1}{5}$	$\tfrac{24}{30}=\tfrac{4}{5}$	$1$

From here we can answer other questions, for example: among those who do not get vaccinated, what is the probability of falling ill?

$P(B\mid \overline{A}) = \frac{P(\overline{A}\cap B)}{P(\overline{A})} = \frac{9/30}{24/30} = \boxed{\tfrac{3}{8} = 37,5\%}.$

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Topics: Probability
Concepts: Correlation · Conditional probability · Contingency table
Methods: Conditional probability · Contingency table
Skills: Probability calculation · Modelling