Does the Probability holds the distributivity?

I am doing a basic probability problem from a textbook exercise. The question is as follows:

Consider two events A and B such that P(A) = 1/3 and P(B) = 1/2 and P(AB) = 1/8. Determine the value of P(BA^c). The way I did the problem is as, P(BA^c) = P(B(1-A)) --> step 1 = P(B - BA) --> step 2 = P(B) - P(BA) --> step 3 = (1/2) - (1/8) = 3/8.

Answer from the book is same as mine.Is my approach correct? Does the Probability holds the distributivity as I did it from step 2 to step 3?

Please correct me if I were wrong with my terminology.

Thanks

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You could also used $P(B) = P(B,A \cup A^c )=\cdots$. –  Shai Covo May 15 '11 at 13:49

A way to see that your method works is to observe that for any two events $A$ and $B$ the probability of $B$ is the probability of "$B$ and $A$" plus the probability of "$B$ and not-$A$." Subtracting the "$B$ and $A$" term from both sides gives what you got.
If your notation means what I think it means, and if your question asks what I think it asks, then it should be easy for you to find examples where $P(A-B)\ne P(A)-P(B)$. –  Gerry Myerson May 15 '11 at 23:56
You can consider the notation you're using as shorthand for operations on sets, where juxtaposition/multiplication represents intersection, subtraction represents the set-theoretic difference, and the $1$ represents the entire event space. Then the distributivity that you used (from step 1 to 2, not 2 to 3 as you wrote) is the distributivity of set intersection over set differences: $R\cap(S\setminus T)=(R\cap S)\setminus(R\cap T)$.