Proving P(A)≥P(B): Probability Proof

[kuahji]

We have three events A, B, and C such that
P(A|C)$\geq$P(B|C) and P(A|C')$\geq$P(B|C')

Prove that P(A)$\geq$P(B).

First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

From above,
P(AC)/P(C)$\geq$P(BC)/P(C) and since P(C)$\geq$0,
P(AC)$\geq$P(BC).

Similarly from above,
P(AC')$\geq$P(BC').

Here is where I am stuck. I tried to expand P(AC)=P(A)+P(C)-P(AUC) with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.

 

[vela]

How did you get P(A|C)=P(AB)/P(C)? Why is B there?

 

[kuahji]

How did you get P(A|C)=P(AB)/P(C)? Why is B there?

Typo, fixed it above.

 

[drawar]

If I'm not mistaken then P(A and C) + P(A and not C) = P(A).

 

[HallsofIvy]

We have three events A, B, and C such that
P(A|C)$\geq$P(B|C) and P(A|C')$\geq$P(B|C')

Prove that P(A)$\geq$P(B).

First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

From above,
P(AC)/P(C)$\geq$P(BC)/P(C) and since P(C)$\geq$0,
P(AC)$\geq$P(BC).

Similarly from above,
P(AC')$\geq$P(BC').

Here is where I am stuck. I tried to expand P(AC)=P(A)+P(C)-P(AUC)

You have this backwards. It should be $P(A\cup C)= P(A)+ P(C)- P(A\cap C)$

with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.

 

[kuahji]

Okay, just figured it out. Thanks everyone for the assistance.