# Understanding conditional probability and Bayes' theorem

1. Nov 6, 2015

### Incand

I'm having trouble understanding an example supposed to motivate Bayes' theorem.

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.)

2. Nov 6, 2015

### PeroK

I'm not sure I can answer your question the way you posed it. What you've done seems too complicated. I would start by doing a probability tree:

Alcohol (30%)
- Speed (18%) (this being 60% of 30%)
- No Speed (12%)

No Alcohol (70%)
- Speed (22%) (as the total for speed is 40%)
- No speed (48%)

P(Alcohol|Speed) = 18/40 = 0.45

Or:

$P(A|S) = P(A \cap S)/P(S) = 18/40$

Or:

$P(A|S) = P(A)P(S|A)/P(S) = (0.3)(0.6)/(0.4) = 0.45$

3. Nov 6, 2015

### Incand

This is the answer you would get if using (1) with $P[E] = 0.4$. However according to the text this is also wrong (Again I only copied the text in the example in the book.) According to the book I should use that
$P[A|E] = \frac{P[E \cap A]}{P[E]} = \frac{P[E|A]P[A]}{P[E \cap A]+P[E\cap A']} = \frac{0.6\cdot 0.3}{P[E|A]P[A] + P[E|A']P[A']} = \frac{0.6\cdot 0.3}{0.6\cdot 0.3+0.1\cdot 0.7} = \frac{0.6\cdot 0.3}{0.25} = 0.72$ which is the end result the book gets. I pretty much don't understand why I can't do exactly what you did in your calculations. At least I'm not the only one being confused.

4. Nov 6, 2015

### PeroK

I've just spotted that "otherwise 10%", which makes no sense. They might be talking about alcohol and speed in general (not only relating to accidents) and they might be looking at individual drivers in an accident. Either (or both) of these would complicate matters.

I have, therefore, little idea what is intended by these figures. I wouldn't waste time if the question is unlcear!

5. Nov 6, 2015

### Incand

I believe the $10%$ refers to that when there is an accident and alcohol isn't involved there's only $10%$ probability that speeding was involved. I also got confused about if it matters that there's two drivers or not. However I suspect we both victims of the base rate fallacy somehow.
I try to ask our lecturer on monday about the question if he have any better idea on how to interpret this example. For now I probably better of doing as you say and move on.