Prove using three basic probability axioms

[hassman]

Homework Statement

Prove that $P(A \cap B)≥1-P(\bar{A})-P(\bar{B})$

for all $A, B \subseteq S$using only these axioms:
1) $0 \leq P(A) \leq 1$, for any event $A \subseteq S$
2) $P(S) = 1$
3) $P(A \cup B) = P(A) + P(B)$ if and only if $P(A \cap B) = 0$

Homework Equations

None.

The Attempt at a Solution

My proof:

First rearrange the shizzle:

$P(\bar{A}) + P(A \cap B) + P(\bar{B}) \geq 1$

Now using the fact that the first two terms are disjoint, use axiom 3 to obtain:

$P(\bar{A} \cup (A \cap B)) + P(\bar{B}) \geq 1$

Next note that the first term is equals to P(S), hence we get:

$P(S) + P(\bar{B}) \geq 1$

which holds for all $B \subseteq S$, because $P(S) = 1$ and $P(\bar{B}) \geq 0$ for all B.

Is this proof correct?

 

[daveb]

Homework Statement

Prove that $P(A \cap B)≥1-P(\bar{A})-P(\bar{B})$

for all $A, B \subseteq S$


using only these axioms:
1) $0 \leq P(A) \leq 1$, for any event $A \subseteq S$
2) $P(S) = 1$
3) $P(A \cup B) = P(A) + P(B)$ if and only if $P(A \cap B) = 0$

Homework Equations

None.

The Attempt at a Solution

My proof:

First rearrange the shizzle:

$P(\bar{A}) + P(A \cap B) + P(\bar{B}) \geq 1$

Now using the fact that the first two terms are disjoint, use axiom 3 to obtain:

$P(\bar{A} \cup (A \cap B)) + P(\bar{B}) \geq 1$

Next note that the first term is equals to P(S), hence we get:

$P(S) + P(\bar{B}) \geq 1$

which holds for all $B \subseteq S$, because $P(S) = 1$ and $P(\bar{B}) \geq 0$ for all B.

Is this proof correct?

In proof you can't start out claiming what you are trying to prove - you can only claim it isn't true and arriving at a contradiction. So you would say

Suppose $P(A \cap B)<1-P(\bar{A})-P(\bar{B})$ then try to arrive at a contradiction.

 

[hassman]

Got it! Thanks!