Thread: Proving for mutally disjoint events A and B

1. Proving for mutally disjoint events A and B

I dont really know how to do this question..mainly were to start.
Here it is exactly..

Just as the difference rule gives rise to a formula for the probability of the complement of an event, so the addition and inclusion/exclusion rules give rise to formulas for the probability of the union of mutually disjoint events and for a general union of (not necessarily mutually exclusive) events.

a. Prove that for mutually disjoint events A and B.

P(A U B) = P(A) + P(B)

b. Prove that for any events A and B

P(A U B) = P(A) + P(B) - P(A ∩ B)

These are just practice questions outta my book but it doesnt provide solutions for them. Can someone help me out please?

Thanks.

2. I am sorry to tell you this, but you may be out of luck on this one.
Without knowing you exact set of axioms it is impossible to proceed.
In a great many textbooks your part a) is actually an axiom so there is nothing to prove.

However, once we have part a) either by axiom or proof the part b) is easy.
Note that:
$A \cup B = \left( {A \cap B^c } \right) \cup B\;\& \;A = \left( {A \cap B} \right) \cup \left( {A \cap B^c } \right)$ this means:
$P\left( {A \cup B} \right) = P\left( {A \cap B^c } \right) + P\left( B \right)\;\& \;P\left( A \right) = P\left( {A \cap B} \right) + P\left( {A \cap B^c } \right)$.

You can finish?

3. Hmm the inclusion/exclusion rule says:

If A, B, and C are finite sets, then

N(A U B) = N(A) + N(B) - N(A ∩ B)

and..

N(A U B U C) = N(A) + N(B0 +N(C0 - N(A ∩ B) - B(A ∩ C) - N(B ∩ C) + N(A ∩ B ∩ C).

so..it almost seems that
N(A U B) = N(A) + N(B) is by def but then its missing the N(A ∩ B) part but it would make sense if a) was the axiom and yeah I am pretty sure I could manage to get b) out of that.
Part a) is now confusing me now though...I guess we cant do anything but just assume that a) is true and use that to prove b) ?