Looking at @Ingix's answer at this link, is it possible to have the following event manipulation?
X = find in A day 2,
Y = not find in A day 1,
Z = in A
$\mathbf P(X∣Y\cap Z) = \frac{\mathbf P(X\cap Y∣Z)}{P(Y)}$
Basically, I want to account for the event in A that he left out. However, I am not sure it would be actually like this, when Y and Z not not known to depend on each other.
$\mathbf P(X∣Y\cap Z) = \frac{\mathbf P(X\cap Y∣Z)}{P(Y\mid Z)}$
As per similar question found here:
By switching the equality above, we have:
The answer should be:
$\mathbf P(X∣Y\cap Z) = \frac{\mathbf P(X\cap Y∣Z)}{P(Y\mid Z)}$