# Thread: Probability of three dependent events

1. ## Probability of three dependent events

We got an assignment to find such three dependent events that satisfy this equation: P(A*B*C)=P(A)*P(B)*P(C). I was thinking to solve this with use of Bayes network: A->B<-C, so P(A*B*C)=P(A)*P(B|A*C)*P(C), and with Bayes' theorem I got P(B|AC)=P(AC|B)*P(B)/P(AC)=P(B), so the first equation is true. Is this approach correct?

2. ## Re: Probability of three dependent events

Hey truleX.

For this statement is P(A*B*C) = P(A and B and C)? I think that independence follows from P(A and B) = P(A)P(B) and using some induction P([A and B] and C) = P([A and B])P(C) = P(A)P(B)P(C) shows that if this is the case then A,B,C are all independent from each other (where P(A|B,C) = P(A), P(B|A,C) = P(B) and P(C|A,B) = P(C)).

If this is not the case what does P(A*B*C) mean?

3. ## Re: Probability of three dependent events

Yes, * means and, + means or. I'm not sure if you understood my problem. I am dealing with three _dependent_, ie non independent events that satisfy the equation above, so I can't use rules for probability of independent events.

4. ## Re: Probability of three dependent events

Isn't the independence condition an if and only if statement? I am under the impression that independence and the multiplication rule are equivalent statements and the only way this can hold is if the events are independent.

I can show you I obtain this, but I'll wait for your response.

5. ## Re: Probability of three dependent events

Originally Posted by truleX
We got an assignment to find such three dependent events that satisfy this equation: P(A*B*C)=P(A)*P(B)*P(C). I was thinking to solve this with use of Bayes network: A->B<-C, so P(A*B*C)=P(A)*P(B|A*C)*P(C), and with Bayes' theorem I got P(B|AC)=P(AC|B)*P(B)/P(AC)=P(B), so the first equation is true. Is this approach correct?
Here is an example due to Glyn George.

6. ## Re: Probability of three dependent events

Thank you Plato, this is exactly what I needed.