A 2nd year stat/probability question about P(A|B) I'm sooo close

[laura_a]

A 2nd year stat/probability question about P(A|B) I'm sooo close!

Homework Statement

Consider 2 events A and B such that P(A)=x and P(B)=y, given x,y >0 and x + y > 1

Prove that
P(A|B) >= 1 - [(1-x)/y]

Homework Equations

P(A|B) = P(A n B) / P(B)

The Attempt at a Solution

Now I know that P(A)= x so P(not a) = 1-x and P(B) = y and P(Not B) = 1-y

so the answer that I have to prove is actually the same as 1 - P(not a)/P(b)

So all I have to work out is how to go from

P(A|B) = P(A n B) / P(B)

to

1 - P(not a)/P(b)

And I'm going off the part where x + y > 1

What step am I missing? Am I using the right formula? There must be a trick or formula I can manipulate as this is a question at the start of a chapter so should be easyish..?

 

[Zone Ranger]

Homework Statement

Consider 2 events A and B such that P(A)=x and P(B)=y, given x,y >0 and x + y > 1

Prove that
P(A|B) >= 1 - [(1-x)/y]

Homework Equations

P(A|B) = P(A n B) / P(B)

The Attempt at a Solution

Now I know that P(A)= x so P(not a) = 1-x and P(B) = y and P(Not B) = 1-y

so the answer that I have to prove is actually the same as 1 - P(not a)/P(b)

So all I have to work out is how to go from

P(A|B) = P(A n B) / P(B)

to

1 - P(not a)/P(b)

And I'm going off the part where x + y > 1

What step am I missing? Am I using the right formula? There must be a trick or formula I can manipulate as this is a question at the start of a chapter so should be easyish..?

$$1\geq P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$1\geq x+y-P(A\cap B)$$

$$P(A\cap B)\geq x+y-1$$

$$\frac{P(A\cap B)}{P(B)}\geq \frac{x+y-1}{y}$$

$$P(A|B)\geq \frac{y-(1-x)}{y}$$

$$P(A|B)\geq 1-\frac{1-x}{y}$$