Bayes's theorem problem issues

Question

Problem Statement:

Assume that 40% of all interstate highway accidents involve excessive speed by at least one of the drivers (event $E$) and that 30% involve alcohol use by at least one driver (event $A$). If alcohol is involved there is a 60% chance that excessive speed is also involved; otherwise, this probability is only 10%. An accident involves speeding. What is the probability that alcohol is involved?

Given probabilities:

P(E) = 0.40

P(A) = 0.30

P(E|A) = 0.60

P(E|A') = 0.10

To find $P(A|E)$, we use the definition of conditional probability:

$P(A|E) = \frac{P(E \cap A)}{P(E)}$

We need to find $P(E \cap A)$ and $P(E)$. Using the multiplication rule:

$P(E \cap A) &= P(E|A)P(A)$

$P(E \cap A') &= P(E|A')P(A')$

Subdividing event $E$:

$E = (E \cap A) \cup (E \cap A')$ Thus, $P(E) = P(E \cap A) + P(E \cap A')$

Substituting back:

$P(A|E) = \frac{P(E|A)P(A)}{P(E|A)P(A) + P(E|A')P(A')}$

Substituting the given probabilities:

$P(A|E) &= \frac{(0.60)(0.30)}{(0.60)(0.30) + (0.10)(0.70)}$

$= \frac{0.18}{0.18 + 0.07}$

$= \frac{0.18}{0.25}$

$= 0.72$

If excessive speed was involved in an accident, there is a 72% chance that alcohol was also involved.

I think the problem statement is inconsistent since P(E)=0.40 is given but nit used in the Bayes' formula P(E)=P(E|A)P(A) + P(E|A')P(A')=0.25. This clearly contradicts the statement P(E)=0.40. How can I fix the problem statement?

Answer 1

The issue is that your statement does not obey the law of total probability, the marginal and conditional probabilities cannot both hold. You can get a solution directly from Bayes' theorem using only what's fixed by the statement, $P(A|E)=\frac{P(E|A)\ P(A)}{P(E)}=\frac{0.6\ \cdot\ 0.3}{0.4}=0.45$.

The issue arises when you try recalculating $P(E)$: the law of total probability requires that $P(E)=\sum_nP(E \cap A_n)$. As you've discovered this is incompatible with the stated values of $P(E|A')$ and $P(E|A)$. You'll see the same if you just try to create a contingency table of the levels of $E$ and $A$, conditioning on the marginal probabilities - they will not add up.

One way to fix the statement is that $P(E|A')$ should actually be $\frac{0.22}{0.7}\approx 0.314$, this makes $P(E)=P(E\cap A)+P(E\cap A')$ hold and leads to the same solution as above. The other option is that 110% of alcohol-related accidents involve speeding (i.e. this is not really an option), or that indeed the marginal probability of an accident involving excessive speed is only 25% which leads to your solution of $P(A|E)=0.72$.