Conditional probability for two collections of items The random variable $X$ with mean $\mu$ and standard deviation $\sigma$ is distributed symmetrically about its mean, but not necessarily normally distributed. I also know that $P(X\leq \mu+\sigma)=a$ and $P(X\leq \mu+\frac{1}{2}\sigma)=b$, where $a$ and $b$ are unknown constants.
Suppose I have some items, $60\%$ of which satisfy condition $A$ and the remaining $40\%$ satisfy condition $B$. 
I know that items satisfying $A$ have mass modelled by $X$, with mean mass $500$ and standard deviation $10$, and items satisfying $B$ have mass modelled by $X$, with mean mass $495$ and standard deviation $10$. 
Let $Y$ the mass of an item picked at random. I want to find $P(Y>500 | Y\leq 505)$ in terms of $a$ and $b$. I know how to solve the problem - my question is not how to solve it.
When using 
$$P(Y>500|Y\leq 505)=\frac{P(500<Y\leq 505)}{P(Y\leq 505)},$$

I am struggling to understand why it is that I must compute $P(500<Y\le  505)$ and $P(Y\le 505)$ for all items together. That is, why it is not OK
  to compute $P(Y>500|Y\leq 505)$ just for items satisfying $A$ (and call
  the result $p_A$), compute $P(Y>500|Y\leq 505)$ just for items satisfying
  $B$ (and call the result $p_B$), and then do $0.6p_A+0.4p_B$?

 A: A simpler example may help to make this clearer.
Say items satisfying $A$ almost certainly have mass $510$ but with some small non-zero probability they have mass $503$; items satisfying $B$ always have mass $480$.
Now $P(500\lt Y\le505)\simeq0$ and $P(Y\le505)\simeq 0.4$, so $P(Y>500\mid Y\le505)\simeq0$.
On the other hand, for items satisfying $A$ we have $P(500\lt X\le505\mid X\le505)=1$, and for items satisfying $B$ we have $P(500\lt X\le505\mid X\le505)=0$, so if you used your alternative calculation you'd get $P(Y>500\mid Y\le505)\simeq0.6$.
$Y\le505$ actually tells you that the item almost certainly satisfies $B$, and in your alternative calculation you're not properly using that information; you calculate as if there were still an $0.6$ probability that the item satisfies $A$.
Of course your case is less clear cut and the information about $A$ or $B$ contained in $Y\le505$ is incomplete, but there is some such information and you're calculating as if you don't have it.
If the events of the item satisfying $A$ or $B$ happened to be independent of the event $Y\le505$, then your alternative calculation would work; but in your case they're clearly not.
