Proving identities using the axioms of probability

phosgene
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Homework Statement



If A and B are events, use the axioms of probability to show:

a) If A \subset B, then P(B \cap A^{C}) = P(B) - P(A)

b) P(A \cup B) = P(A) + P(B) - P(A \cap B)

Homework Equations



Axiom 1: P(x)\geq 0

Axiom 2: P(S) = 1, where S is the state space.

Axiom 3: If A_{1},A_{2},...,A_{n},... is any set of disjoint events, then:

P(\bigcup_{i} A_{i})=\sum_{i} P(A_{i})

The Attempt at a Solution



It's easy to see why they are true using venn diagrams. In the first case, since A is a subset of B, the probability of the intersection of B and A complement is just the probability of B minus the probability of A. With the second one, you just want to add the P(A) and P(B) and then subtract the probability of the intersection so that you're not adding the probability of those events twice.

I just have no idea how to do this symbolically. This is the first time I've ever had to try and prove something in probability..
 
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phosgene said:

Homework Statement



If A and B are events, use the axioms of probability to show:

a) If A \subset B, then P(B \cap A^{C}) = P(B) - P(A)<br /> <br /> b) P(A \cup B) = P(A) + P(B) - P(A \cap B)<br /> <br /> <h2>Homework Equations</h2><br /> <br /> Axiom 1: P(x)\geq 0<br /> <br /> Axiom 2: P(S) = 1, where S is the state space.<br /> <br /> Axiom 3: If A_{1},A_{2},...,A_{n},... is any set of disjoint events, then:<br /> <br /> P(\bigcup_{i} A_{i})=\sum_{i} P(A_{i})<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> It's easy to see why they are true using venn diagrams. In the first case, since A is a subset of B, the probability of the intersection of B and A complement is just the probability of B minus the probability of A. With the second one, you just want to add the P(A) and P(B) and then subtract the probability of the intersection so that you're not adding the probability of those events twice.<br /> <br /> I just have no idea how to do this symbolically. This is the first time I've ever had to try and prove something in probability..
<br /> <br /> For (a): ##B = (B \cap A^c) \cup \{\text{something else disjoint}\}##. What must be that "something else"? What do the basic probability axioms then give you?
 
I use axiom three and re-arrange the result to get the statement that I was trying to prove. Thanks :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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