Proving the Inequalities of Simple Probability: P(A∩B) ≤ P(A) ≤ P(AUB)

[TheHamburgler1]

Homework Statement

Prove the following:

P(A∩B) ≤ P(A) ≤ P(AUB)

Homework Equations

The Attempt at a Solution

I first attempt to show that P(A) is a subset of P(AUB) which therefore means it is ≤ P(AUB):

P(AUB)=P(A)+P(B)-P(A∩B), thus P(A) is a subset of P(AUB)...I think that works...

Any help would be great,

cheers

 

[lanedance]

be careful, probabilities are scalar numbers in the interval [0,1], whilst A & B are sets

so if you can show
$$A \subseteq B$$

it follows directly that
$$P(A) \leq P(B)$$

as every event in A is also in B

 

[TheHamburgler1]

be careful, probabilities are scalar numbers in the interval [0,1], whilst A & B are sets

so if you can show
$$A \subseteq B$$

it follows directly that
$$P(A) \leq P(B)$$

as every event in A is also in B

So what you're saying is that I have to show that A is a subset of AUB. This is done by definition that AUB=A+B therefore A is a subset of AUB and it follows (as you said) that P(A) is less than or equal to P(AUB). Correct? What would I do to prove the intersection is less than A?

cheers

 

[lanedance]

right, and once you have proved one of the inequalities, you could do the other similarly

or P(AUB)=P(A)+P(B)-P(A∩B) yields it directly