# Conditinal probability theorem

• February 21st 2011, 12:04 PM
Jame
Conditinal probability theorem
I have to either prove or disprove the following:

If $P(A) \leq P(B)$, then $P(A|C) \leq P(B|C)$ (Assuming $P(C) \geq 0$)

After trying to prove this generally a million times and failing, I decided to find a counter example, I'm not sure if this works though.

All the following events have to do with rolling one fair dice.
A is getting a "2"
B is getting an odd number
C is getting an even number

Now $P(A) \leq P(B)$

i.e. $\frac{1}{6} \leq \frac{3}{6}$

but I believe that $P(B|C) = 0$

i.e $P( B \cap C) = 0$

and $P(A|C) = \frac{1/6}{3/6} = \frac{1}{3}$ (i.e $P(A \cap C) = \frac{1}{6}$

So this contradicts the hypothesis, but I am not sure if I calculated the probabilities of $P(A \cap C)$ and $P(B \cap C)$ right.

• February 21st 2011, 12:14 PM
Plato
Quote:

Originally Posted by Jame
I have to either prove or disprove the following:
If $P(A) \leq P(B)$, then $P(A|C) \leq P(B|C)$ (Assuming $P(C) \geq 0$)
All the following events have to do with rolling one fair dice.
A is getting a "2"
B is getting an odd number
C is getting an even number

Now $P(A) \leq P(B)$
i.e. $\frac{1}{6} \leq \frac{3}{6}$
but I believe that $P(B|C) = 0$

i.e $P( B \cap C) = 0$

and $P(A|C) = \frac{1/6}{3/6} = \frac{1}{3}$ (i.e $P(A \cap C) = \frac{1}{6}$

So this contradicts the hypothesis, but I am not sure if I calculated the probabilities of $P(A \cap C)$ and $P(B \cap C)$ right.

That is correct.
• February 21st 2011, 12:35 PM
Jame
Woot excellent. Thank you for your help!