- #1

- 333

- 47

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

##P[E] = 0.4 \; \; P[A]=0.3 \; \; P[E|A] = 0.6##

##P[E'] = 0.6 \; \; P[A'] = 0.7 \; \; P[E|A'] = 0.1##

We are being asked to find ##P[A|E]##. From the definition of conditional probability

##P[A|E]=\frac{P[E\cap A]}{P[E]}## (1)

We have by the multiplication rule that ##P[E\cap A] = P[E|A]P[A]## and that

##E = (E\cap A) \cup (E \cap A')## and hence

##P[E] = P[E\cap A]+ P[E\cap A']##.

substituting this into the equation we have

##P[A|E] = \frac{P[E|A]P[A]}{P[E|A]P[A]+P[E|A']P[A']}##.

My question is pretty much why the ##P[E]## in (1) isn't the same as the one given earlier? I.E. why do I have to split up ##E## into two parts (or use Bayes' theorem instead of the definition for conditional probability directly.)