1. ## Proof

How could this be done?

Show that the statement P(A,B|C) = P(A|C) P(B|C) is equivalent to both P(A|B,C| = P(A|C) and P(B|A,C) + P(B|C).

2. In the last statement, I think you mean an equal's sign instead of a plus sign yes?

3. Yes my apologies I am very tired. It should read

Show that the statement P(A,B|C) = P(A|C) P(B|C) is equivalent to both P(A|B,C| = P(A|C) and P(B|A,C) = P(B|C).

4. For problems such as these, I find it easier to assign actual "things" to these letters. So lets do that -

Lets let A be the event that Laura goes to school, B be the event the Kyle goes to school, and C be the event that it is raining. A and B are independent events since (unless the two live with each other or some such nonsense), the probability that Laura goes to school does not depend on Kyle going to school. So P(A,B|C) is just P(A|C)P(B|C) (whatever these actual probabilities are) since the two events are independent. Therefore in your second bit, the P(A|B,C) - Laura going to school given that is raining, is not dependent on the fact that B has occured (Kyle going to school).

Now do they actually want you to show this algebraically?

5. Yes, I am very confused as well. Not too sure where to begin.