Conditional Probability - A breathalyser

Question

A breathalyser is used to check whether a driver’s blood alcohol level is over the legal limit. We define these events:

$A:$ breathalyser shows driver is over the limit;

$B:$ the driver’s blood alcohol really is over the limit.

(a) Suppose $P(A|B) = 0.9$, $P(\bar A|\bar B) = 0.99.$ A driver is stopped at random and breathalysed. If a conviction is made on the basis of the breathalyser alone, determine the error rate (rate of wrongful convictions) $P(\bar B|A) = P$(driver is under the limit given that the breathalyser shows driver over the limit).

Got stuck on this one, any help/tips would be much appreciated. Thanks!

Answer 1

This problem has more than one solution.

Let us denote:

• $p_1=P(A\cap B)$
• $p_2=P(A\cap B^{\complement})$
• $p_3=P(A^{\complement}\cap B)$
• $p_4=P(A^{\complement}\cap B^{\complement})$

The events mentioned above are disjoint and covering.

We have the following equalities:

• $p_1+p_2+p_3+p_4=1$
• $0.9(p_1+p_3)=p_1$ based on $P(A\mid B)P(B)=P(A\cap B)$
• $0.99(p_2+p_4)=p_4$ based on $P(A^{\complement}\mid B^{\complement})P(B^{\complement})=P(A^{\complement}\cap B^{\complement})$

If $p$ and $q$ are nonnegative and satisfy $10p+100q=1$ then a solution is:

• $9p=P(A\cap B)$
• $q=P(A\cap B^{\complement})$
• $p=P(A^{\complement}\cap B)$
• $99q=P(A^{\complement}\cap B^{\complement})$

leading to: $$P(B^{\complement}\mid A)=\frac{q}{q+9p}$$