Prove P(A/B and C) = P(A/C) if A and B are independent

Answer 1

Independent means P(A/B) = P(A) and P(B/A) = P(B). Knowing whether one event happened doesn't change the probability of the other even happening. The waiter says they have specials on pork and steak. You say you'll have the steak. The waiter then says, oh, and we also have a special on chicken. You then say in that case, I'll have the pork. The waiter then decides you're a little loony. An irrelevant alternative should have no effect on a choice like that, unless you're missing a few marbles. (this is related to Arrow's Paradox which treats it as a rational axiom (with no need of proof) that irrelevant alternatives do not affect choice or probability.)

Bayes Theorem is P(A/B)P(B) = P(B/A)P(A) used usually when A and B are not independent. If A and B are independent Bayes Theorem collapses to a more obvious tautology, P(A)P(B) = P(B)P(A)

P(A/B & C) = P(A/C) because B is totally irrelevant to calculating the probability of A

if C also were independent, then P(A/B&C) = P(A/C) = P(A). Just delete the irrelevant events, the independent events.