Hi there,
Does anyone know how to prove the following...
P(E1nE2nE3nE4)=P(E1)P(E2\E1)P(E3\E2nE1)P(E4\E3nE2n E1)
apparently it can be proved by induction
Any better ideas
Thanks
Let’s agree that notation wise $\displaystyle P(A \cap B) = P(AB)$. It makes it easier.
Basically we know that $\displaystyle P(AB) = P(A|B)P(B)$.
Consider: $\displaystyle P(ABC)$ and let $\displaystyle X=BC$.
The we know that
$\displaystyle \begin{array}{rcl} {P(ABC)} & = & {P(AX)} \\ {} & = & {P(A|X)P(X)} \\
{} & = & {P(A|BC)P(BC)} \\ {} & = & {P(A|BC)P(B|C)P(C)} \\ \end{array} $
That is idea you could use in an inductive proof.