1.3 Calculation of Probability 1.3.2 Venn Diagrams 1.3.4 Problems

1.3.3 Bayes’s Theorem

Again you have your two events A and B but they no longer need to be independent. Then you can say the following

P(AB)=P(A)P(B|A).

(1.4)

What’s the meaning of this? $P(B|A)$ means ”the probability of B given A”. It’s called a ”conditional probability”. So this says the probability of having both A and B is (the probability of A) times (the probability of B given you’ve also got A).

Although this might seem unintuitive, it’s pretty simple if you think about it using Venn diagrams we just talked about. Let’s try to understand the case of two intersecting circles. What’s $P(B|A)$ ? That is the probability of B given A. In terms of darts, this means that we’re assuming that your dart fell inside circle A . What fraction of those darts were also inside B ? Well, that’s just just the area of $A B$ divided by the area of A , that is $P(B|A)=P(AB)/P(A)$ . Hey that’s just a rearrangement of Bayes’s Theorem, eqn 1.4.

Cracking a lock

Suppose you want to pick one of those combo locks you see on bicycles. You have 4 dials you can twirl and you have to guess the right combination. You don’t have any other information and so you try randomly twirling the lock. Since there are 10,000 possibilities, the probability that you’ll get it right on the first guess is 1/10,000. Not really worth it, considering the bicycle owner is 7 foot 1, with a tattoo saying ”Moma”.

But now suppose you overhear a conversation where you only catch the last digit of the combination, which is 6. Now what’s the probability you’ll get it right on the first try? You only have to guess three digits now, so that should be 1/1000.

Let’s see how this works from Bayes’s Theorem. Call B the correct combination. Call A all combinations ending with digit 6. What’s AB? That’s the last digit being 6 intersected with the correct combination. Well the last digit for the correct combination is 6, so $AB=B$ . What’s P(A), that is the probability of getting a last digit of 6? That’s 1/10. So Bayes’s Theorem says $P(B|A)=P(AB)/P(A)=P(B)/P(A)=(1/10000)/(1/10)=1/1000$ .

In this case, it was just easier to solve the problem the first way I did, but in lots of cases, Bayes’s Theorem is a lot easier.

Now this is obviously not the best way to pick a lock. If you’re interested in a life of crime, you’re studying the wrong thing. Go become a lawyer.

Back to Independence

Now we can better understand independence using Bayes’s Theorem. Eqn. 1.4 is particularly simple when A and B are independent. In this case the probability of B given A , $P(B|A)$ is just $P(B)$ because the probability of B is independent of the value of A . That’s the whole definition of independence. So in that case we have $P(AB)=P(A)P(B)$ .