Going wrong with Bayes theorem

Question

Lets say we have following case:

Probability of bring a drunk driver = $0.10$

Probability of a drinking test coming positive = $0.30$

Probability of a drinking test coming negative, given the subject was not drunk = $0.90$

Then by Bayes theorem,

$$P(Not Drunk|Negative Test) = \frac{P(Negative Test|Not Drunk) \times P(Not Drunk) }{P(Negative Test)}.$$ Now, \begin{align} P(Negative Test|Not Drunk)& = 0.90\\ P(Not Drunk)& = 0.90\\ P(Negative test)& = 0.70 = (1 - Probability(Positive Test)) \end{align} Thus, $$P(Not Drunk|Negative Test) = (0.90 * 0.90) / 0.70 = 1.51.$$

As far as I understand, the probabilities shouldn't ever become more than 1 and above result is counterintuitive to me. Is this correct, if not where am I going wrong?

Answer 1

Your initial numbers are impossible.

To use your notation, and the Law of Total Probability:

$$.7=P(NegativeTest)=P(Negative\,|\,Drunk)P(Drunk)+P(Negative\,|\,NotDrunk)P(NotDrunk)$$

Thus $$.7=.9\times.9 + P(Negative\,|\,NotDrunk)\times .1\implies P(Negative\,|\,NotDrunk)=-\frac {.11}{.1}=-1.1<0$$

Informally: say you had $100$ people. Then you have $90$ sober and $10$ drunk. Of the ninety, you tell us that exactly $9$ tests negative...so $81$ of those test positive. Thus the number of positive tests is at least $81$. Yet you claim that only $70$ test positive.

Answer 2

The problem is that your initial numbers cannot happen. This is because of the law of total probability which tells us that $P(A)=P(A|H_1)\cdot P(H_1) + P(A|H_2)\cdot P(H_2)$ if $H_1$ and $H_2$ are two oposing hypotheses. In your case, setting $A$ to be "Test is negative", $H_1$ to be "Driver is drunk"and $H_2$ to be "Driver is not drunk", you get

$$0.7 = P(A) = P(A|H_1)\cdot P(H_1) + P(A|H_2)\cdot P(H_2) = P(A|H_1)\cdot 0.1 + 0.9\cdot 0.9 = \\=P(A|H_1)\cdot 0.1 + 0.81$$

meaning that $P(A|H_1)$ is negative.