use the law of total probabilty to prove that

a. if P (A l B) = P (A l B,) then A and B are independent

b. if P (A l C ) > P (B l C) and P (A l C' ) > P( B l C' ), then P (A) > P (B)

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- Jan 30th 2010, 01:21 AMstressedoutlaw of total probability
use the law of total probabilty to prove that

a. if P (A l B) = P (A l B,) then A and B are independent

b. if P (A l C ) > P (B l C) and P (A l C' ) > P( B l C' ), then P (A) > P (B) - Jan 30th 2010, 01:47 AMMoo
Hello,

The law of total probability says that $\displaystyle P(A)=P(A|B)P(B)+P(A|B')P(B')$

So here, $\displaystyle P(A)=P(A|B)[P(B)+P(B')]=P(A|B)$ since B and B' are complementary events.

Provided that $\displaystyle P(B)\neq 0$, it follows that $\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}=P(A) \Rightarrow P(A\cap B)=P(A)P(B)$

For the second one... :

$\displaystyle P(A)=P(A|C)P(C)+P(A|C')P(C')$

But from the two inequalities, we can say that $\displaystyle P(A)>P(B|C)P(C)+P(B|C')P(C')$, and the RHS is exactly $\displaystyle P(B)$