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laura_a
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A 2nd year stat/probability question about P(A|B) I'm sooo close!
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]
P(A|B) = P(A n B) / P(B)
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..?
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..?