Conditional Probabilities I am faced with this question:
$10$% of all email you receive is spam. Your spam filter is $90$% reliable, that is, $90$% of the mails it marks as spam are indeed spam and $90$% of spam mails are correctly labelled as spam. If you see a mail marked spam by your filter, what is the probability that it is really spam?
This question was posed in a Chennai Mathematical Institute exam. This is how I'm trying it.
$10$% of all email you receive is spam
$$
P(\text{spam}) = 0.1
$$
$90$% of the mails it marks as spam are indeed spam
$$
P(\text{spam|marked as spam}) = 0.9
$$
$90$% of spam mails are correctly labelled as spam
$$
P(\text{marked as spam|spam}) = 0.9
$$
By Bayes' Theorem, we have
$$
P(\text{spam|marked as spam}) = \frac{P(\text{marked as spam|spam}) P(\text{spam})}{P(\text{marked as spam})}
$$
$$
0.9 = \frac{0.9 * 0.1}{P(\text{marked as spam})}
$$
Which is incorrect. I am not able to figure out what exactly I did wrong. The official solution goes like this

Out of 100 mails, 10 are spam. The filter will label 9 or 10 spam as
  spam and 9 of 90 non-spam as spam. So 18 are labelled spam, of which 9
  are actually spam.

Can somebody show the right way to solve this using conditional probabilities?
 A: There seems to be an error either in the question or your understanding of it. If the question indeed says 

90% of the mails it marks as spam are indeed spam and 90% of spam mails are correctly labelled as spam,

this means
\begin{equation}
P(\textrm{spam} \mid \textrm{marked}) =0.9, P(\textrm{marked} \mid \textrm{spam}) = 0.9,
\end{equation}
and the question doesn't make much sense as the answer is stated directly in the question, while the official solution is wrong. However, I suspect the intended premises actually were something like

It marks 90 % of spam mails as spam, and 90 % of non-spam mails as not spam,

which translates to conditional probability statements as follows:
\begin{equation}
P(\textrm{marked} \mid \textrm{spam}) = 0.9, P(\neg\textrm{marked} \mid \neg \textrm{spam}) = 0.9.
\end{equation}
In this case you can match the intended solution by using Bayes' theorem to compute
\begin{equation}
P(\textrm{spam} \mid \textrm{marked}) = \ldots = 0.5. 
\end{equation}
A: You say that '...90% of the mails marked is indeed spam'. That should probably be changed into '...90% of the mails that is spam is indeed marked'.
On base of$P\left[\text{spam}\right]=0.1$, $P\left[\text{marked}\mid\text{spam}\right]=0.9$ and $P\left[\text{marked}\mid\text{no spam}\right]=0.1$ we find: 
$$P\left[\text{spam}\mid\text{marked}\right]=\frac{P\left[\text{spam}\wedge\text{marked}\right]}{P\left[\text{marked}\right]}=\frac{P\left[\text{marked}\mid\text{spam}\right]P\left[\text{spam}\right]}{P\left[\text{marked}\mid\text{spam}\right]P\left[\text{spam}\right]+P\left[\text{marked}\mid\text{no spam}\right]P\left[\text{no spam}\right]}=\frac{0.9\times0.1}{0.9\times0.1+0.1\times0.9}=\frac{1}{2}$$
A: Why did you decide ${P(marked\ as\ spam)} = 0.1$ is incorrect?
In fact, this makes perfect sense..
Out of 100 mails, the classifier is expected to mark 10 as spam, 9 of which are actual spam and there's another spam mail which successfully bypassed the filter.

In your question you wrote 

"and 9 of 90 non-spam as spam."

Which is not the expected case, as ${P(marked\ as\ spam | not\ spam)} \neq 0.1$.
