Associative Law for Multiplication - Probability Theory and Boolean Algebra

[FabianS]

Hi, I'm having some trouble solving one of the problems from my homework assignment.

Homework Statement

Prove:

$$P((AB)C) = P(A(BC))$$​

Where A,B,C are either true or false.

Homework Equations

We can't do this by using a truth table, we can use the following equations:

$$P(A + B) = P(A) + P(B) - P(AB)$$

$$P(AB | C) = P(A | BC) + P(B | C)$$

$$P(A | B) + P(\bar{A} | B) = 1$$

and De Morgan's rules.

The Attempt at a Solution

The way I have approached it is by stating that if (AB)C = A(BC) then their probabilities must be equal, as well.
I have checked on multiple sites and they either just assume this to be true (http://en.wikipedia.org/wiki/Boolean_logic#Properties", page 6). So, I tried something similar:

$$X = (ab)c$$
$$Y = a(bc)$$

$$aX = a((ab)c) = (a(ab))(ac) = (ab)(ac) = a(bc)$$
$$aY = a(a(bc)) = a(bc) = Y$$

$$aX = Y = aY$$
$$X = Y$$I'm not very good a proving this type property, and I feel like I'm going around in circles. Is this correct or is there some better way of doing this?

Thanks :)
Fabian

 

[KataKoniK]

Dear Fabian,

Thank you for reaching out for help with your homework problem. I understand that you are having trouble proving the property P((AB)C) = P(A(BC)) using the given equations and De Morgan's rules. I will try my best to guide you through the process.

Firstly, let's break down the given property into smaller parts. We have P((AB)C) on one side and P(A(BC)) on the other side. Using the distributive property, we can write (AB)C as (A(CB)). Now, we can see that both sides have the same structure, with different variables. This is a good starting point.

Next, we can use the equation P(AB | C) = P(A | BC) + P(B | C) to rewrite P((AB)C) as P(A | BC) + P(B | C) + P(C). Similarly, we can rewrite P(A(BC)) as P(A | CB) + P(B | C) + P(C). Now, we can see that both sides have the same structure and the only difference is the order of the variables in the conditional probabilities.

This is where De Morgan's rules come in. We can use De Morgan's rule for conditional probabilities, which states that P(A | BC) = P(A | CB), to show that both sides are equal. This is because P(A | BC) represents the probability of A given that both B and C are true, while P(A | CB) represents the probability of A given that both C and B are true. Since the order of the variables does not affect the outcome of the probability, both sides are equal.

In conclusion, we have shown that P((AB)C) = P(A(BC)) using the given equations and De Morgan's rules. I hope this helps you understand the property better and gives you a starting point for your proof. If you have any further questions, please do not hesitate to ask.

Best of luck with your homework,