Conditional probability question somebody please prove or disprove 
$P(x|y)=P(x|z)/P(y|z)$
I have tried to plug in numbers with example in Wikipedia, for example, 
$A$ is outcome number of dice 1
$B$ is outcome number of dice 2
$X$: event of $A=2$
$Y$: event of $A+B \leq 5$
$Z$: event of $B=2$
$P(x|y) = 3/10$ ,
$P(x|z) = 1/6$ , and
$P(y|z)= 3/6$
Question: Then the above equation does not Hold, am I missing anything?
 A: This is relatively easy to disprove. Take, for example the following events:


*

*$X$: A coin flip yields heads ($P(X) = 1/2$)

*$Y$: The same coin flip yields tails ($P(Y) = 1/2$)

*$Z$: The sky is blue (or some other event with probability one) ($P(Z) = 1$)


$P(X|Y) = 0$, as the two events are mutually exclusive. However, $P(X|Z) = P(Y|Z) = 1/2$, as $Z$ is independent of $X$ and $Y$. Therefore, $\frac{P(X|Z)}{P(Y|Z)} = 1 \ne P(X|Y)$.
The example you gave yourself also disproves the statement, but is perhaps less clear.
Another way of seeing that the statement is untrue is to notice that if true, it would imply that $P(X|Y) = 1/P(Y|X)$, which cannot be true in general as both of the conditional probabilities must be between zero and one.
A: The proposed equation can be writen
$$P(X | Z ) = P( X | Y ) P ( Y | Z )$$
We could think that X Y Z are events ocurring respectively tomorrow, today and yesterday.
To see in which particular case this could be true (and why it is not true in general):
$$P(X) = P (X \wedge Y) + P(X \wedge\bar{Y})$$
and also (conditioning on event Z)
$$P(X | Z ) = P (X \wedge Y | Z ) + P(X \wedge\bar{Y} | Z) = P(X | Y \; Z ) P (Y | Z) + P(X | \bar{Y} \; Z ) P (\bar{Y} | Z) $$
This is true in general.
Now, suppose (first particular assumption) that the probability that the event happens tomorrow given that it happened (or not) today doesn't depend of whether it happened yesterday (as a Markovian process). Then we'd have:
$$P(X | Z ) = P(X | Y ) P (Y | Z) + P(X | \bar{Y} ) P (\bar{Y} | Z) $$
Suppose further (second particular assumption) that the event can only happen tomorrow if it happened today.
Then, $ P(X | \bar{Y} )=0$ and the original equation would be true.
Example: 
Z = Event "the product lasted at least 1 year"
Y = Event "the product lasted at least 2 years"
X = Event "the product lasted at least 3 years"
