1. ## Proving Probabilities

I am supposed to show that if P(B) is not equal to 0, then P(A|B) is greater than or equal to 0, P(B|B) = 1, and P(A1 union A2 union A3...|B) = P(A1|B) + P(A2|B) + P(A3|B) + ... for any sequence of mutually exculsive events A1, A2, A3...

It doesn't seem like it should be so hard... in fact, it makes perfect sense that P(B|B) = 1... but I really don't know how to show it. Please help!

2. Apply the definition of conditional probability:

$\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)}$.

So

$\displaystyle P(B|B) = \frac{P(B \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$

Also we know $\displaystyle P(B) \geq 0 \text{ and } P(A \cap B) \geq 0$ so the non-negativity of the conditional probability of A given B is easy to show.

Now consider $\displaystyle P(A_1 \cup A_2 ... \cup A_m |B)$ by definition is $\displaystyle \frac{P((A_1 \cup A_2 ... \cup A_m ) \cap B)}{P(B)} = \frac{P((A_1 \cap B) \cup (A_2 \cap B) ... \cup (A_m \cap B)}{P(B)}$
by basic set theory,
$\displaystyle = \frac{P(A_1 \cap B)}{P(B)} + ... + \frac{P(A_m \cap B)}{P(B)}$
since we assume mutual exclusivity
$\displaystyle = P(A_1|B) + ... + P(A_m|B)$