Proving P(B)>=P(A) using Kolmogrov's axioms

[amai]

Homework Statement

Suppose A is a subset of B. Using the three axioms to establish P(B)≥P(A)

Homework Equations

Axioms...
1. For any event A, P(A)\geq0
2.P(S)=1
3.If events A1,A2... are mutually exclusive, then P(\cupAi)=\sumP(Ai)

The Attempt at a Solution

From the venn diagram I drew, i have...

B=A+\overline{A}B
P(B)=P(A\cup\overline{A}B)
P(B)=P(A)+P(\overline{A}B)

According to axiom 1, P(\overline{A}B)≥0 ... but doesn't that mean i just proved P(B)≤P(A)?

 

[jbunniii]

Does your notation \overline{A}B mean the relative complement of $A$ in $B$, in other words $B \setminus A = B \cap A^c$?
If so, then I agree with your proof.

 

[amai]

Yes, that is what i meant. Maybe the solution is wrong then... i guess i'll have to ask my professor.

 

[jbunniii]

Yes, that is what i meant. Maybe the solution is wrong then... i guess i'll have to ask my professor.

Sorry, I misread your question:

but doesn't that mean i just proved P(B)≤P(A)?

No, you proved $P(B) \geq P(A)$, just like the question asked. Your obtained this result:
$$P(B) = P(A) + P(B \setminus A)$$
which means that $P(B) \geq P(A)$, since you have to add the nonnegative number $P(B \setminus A)$ to $P(A)$ in order to obtain $P(B)$.