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 Susing 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!