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We have three events A, B, and C such that

P(A|C)[itex]\geq[/itex]P(B|C) and P(A|C')[itex]\geq[/itex]P(B|C')

Prove that P(A)[itex]\geq[/itex]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)[itex]\geq[/itex]P(BC)/P(C) and since P(C)[itex]\geq[/itex]0,

P(AC)[itex]\geq[/itex]P(BC).

Similarly from above,

P(AC')[itex]\geq[/itex]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.

P(A|C)[itex]\geq[/itex]P(B|C) and P(A|C')[itex]\geq[/itex]P(B|C')

Prove that P(A)[itex]\geq[/itex]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)[itex]\geq[/itex]P(BC)/P(C) and since P(C)[itex]\geq[/itex]0,

P(AC)[itex]\geq[/itex]P(BC).

Similarly from above,

P(AC')[itex]\geq[/itex]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.

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