Toss a coin twice. Set the first toss as event A and the second is event B.A and B is independent. Therefore, P(A & B) = P(A)P(B) = 1/4 != 0. So the intersection of A and B is not NULL set. My question is what the common part between A and B is?

you need to formulate the thing a bit better... an event is A=tail not A...you have for example: P(A=tail and B=head)=P(A=tail)*P(B=head)=1/2*1/2=1/4

I like your question actually, don't let anyone tell you otherwise...Here is the story...(if you don't understand, feel free to ask, or if you want to think hard, read *the* book on Probability for able undergraduates..."Probability with Martingales" by Williams, from CUP)"call your 2 coins X_1 and X_2 , then each is a random variable with P(X_i = H) = P(X_i = T) = 0.5 "It is important to understand what this statement means:Each X_i is a FUNCTION from a space of possible outcomes,S, to the discrete 2 member space {H,T}X_i : S -> {H,T}so X_i (s) = H or T , where s is a member of S.Now, the "Probability" is another function, with "certain" properties, which measures how big "certain" subsets of S are (we call these special subsets "events") as a number between 0 and 1, i.e. not all subsets of S need be measurable by P, but some are.P : Certain Subsets of S -> [0,1]Now the main thing... the probabliity of X_i = H , say, is P(s in S such that X_i(s) = H)you might wonder what would happen if this subset, {s in S such that X_i(s) = H} is not one of those "certain" subsets that is measurable by P...well, this is where we cheat and say that by definition, this is true, i.e. we call X_i random variables that are "P-measurable".Now, the exact definition of independence in this setting is a little more tricky, but you can sort of think of it as you were initially.To answer your question, P(X_1=H ,X_2 = H) = P(s in S such that X_1(s) = H and X_2 = H) = 0.25 (via independence as you have shown)the event (set) which we are measuring is {s in S such that X_1(s) = H and X_2 = H} and this has non-zero measure, it is the subset of S that leads to X_i going to H , i=1,2. If this is all a little confusing, don?t worry, this is hard stuff...I think your question shows to me that you are thinking about probability in a measure theoretic way which is why I?ve mentioned all this stuff.

QuoteOriginally posted by: sjian99Toss a coin twice. Set the first toss as event A and the second is event B.A and B is independent. Therefore, P(A & B) = P(A)P(B) = 1/4 != 0. So the intersection of A and B is not NULL set. My question is what the common part between A and B is?Let A = H and B = T.Then S(A) = {(H,T), (H,H)}, S(B) = {(H,T), (T,T)}S(A).Int.S(B) = {(H,T)}This is the "common part".