In a company $7$ percent of the employees are suffering from anxiety disorder, $70$ percent of people who suffer from anxiety disorder don't show up one day at work and $10$ percent of people who don't have anxiety disorder don't show up one day at work.

A doctor justifies the absence from work of $5$ percent of people suffering from anxiety and of $30$ percent of people that don't have anxiety disorder.

An employee who skipped work was justified by a doctor. What is the probability that he was suffering from anxiety disorder?

My attempt let $A$ be the event of suffering from anxiety disorder, let $B^c$ be the event of not going to work one day and $C$ be the event that a doctor justifies ones absence.

So I have $$P(A)=7/100$$ $$P(B^c|A)=70/100$$ $$P(B^c | A^c)=10/100$$ $$P(C|A)=5/100$$ $$P(C|A^c)= 30/100$$

So I am looking for $$P(A|B^c \cap C)= \frac{P (A \cap B^c \cap C)}{P(B^c \cap C)} $$

I am stuck at this point since I can't find a useful formula to compute $ \frac{P (A \cap B^c \cap C)}{P(B^c \cap C)}$, any suggestions?


2 Answers 2


The problem is that it is not clear how $B$ and $C$ depend on each other. I think that the doctor's justification is understood to be $\textit{after}$ the absence of the employee (so $P(C\cap B)=0$), so the last two probabilities should be $P(C|A\cap B^c)=0.05$ and $P(C|A^c\cap B^c)=0.3$. Under this assumption we have

$P(A\cap B \cap C)=P(C|A\cap B^c)P(A\cap B^c)=P(C|A\cap B^c)P(B^c|A)P(A)$


$P(B^c\cap C)=P(C|B^c)P(B^c)=[P(C|B^c\cap A)P(A|B^c)+P(C|B^c\cap A^c)P(A^c|B^c)]\, P(B^c)$,

and the probabilities on the right handside can be computed from the information in the text.

  • $\begingroup$ Did you mean to write $P(C|A^c \cap B^c)=0.3$? $\endgroup$
    – user746545
    May 25, 2021 at 8:33
  • $\begingroup$ Yes, of course. I editted the answer. $\endgroup$
    – ym94
    May 25, 2021 at 9:55

In these cases, a tabular approach will get you the solution in a very easy way

These are your data in a tabular form

enter image description here

Now to calculate your probability it is very natural

$$\mathbb{P}[\text{Anxiety }|\text{Justified}]=\frac{0.05\times4.9}{0.05\times4.9+0.3\times9.3}=8.07\%$$


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