# What is the probability that an employee is suffering from anxiety disorder?

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?

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

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

and

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

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

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

Now to calculate your probability it is very natural

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