How to find $P(S_1 \cap S_2^C | K_i)$ given $P(S_1 \cap S_2 | K_i)$ From medical investigations it is known that the symptoms $S_1$ and $S_2$ can appear with three different diseases $K_1, K_2, K_3$. The conditional probabilities $a_{i,j}=P(S_j|K_i), i \in \{1,2,3\}, j \in \{1,2\}$ are given by the following matrix.
$$ A= (a_{i,j}) = \left(
\begin{array}{cc}
 0.8 & 0.3 \\
 0.2 & 0.9 \\
 0.4 & 0.6 \\
\end{array}
\right)$$
In the first part of the question I already calculated $P(S_j)$ and $P(K_i|S_j)$. For the second part of the question, I'm given the conditional probabilities $P(S_1 \cap S_2 |K_i)$ by the following vector $(0.2, 0.1, 0.3)$. Assuming that a patient shows symptoms $S_1$, but not $S_2$, what is the probability that he suffers from $K_1, K_2$ and $K_3$?
So I'm looking for $P(K_i|S_1 \cap S_2^C)$. I can get $P(S_1 \cap S_2^C)$ using $P(S_1 \cap S_2^C) = P(S_1) - P(S_1 \cap S_2)$, but I have problems getting the joint distribution $P(S_1 \cap S_2, K_i)$.
First I was trying to show that $S_1$ and $S_2$ are independent and use that to derive the joint probability, but they are not. Alternatively, is it true that the formula $P(S_1 \cap S_2^C) = P(S_1) - P(S_1 \cap S_2)$ remains true when conditioning on K, so that I have $P(S_1 \cap S_2^C | K_i) = P(S_1 | K_i) - P(S_1 \cap S_2 | K_i)$? Or can anybody help me by providing an alternative way to solve this?

The formula you mention remains true under conditioning, so can be
used:
$P\left(S_{1}\mid K_{i}\right)=\frac{P\left(S_{1}\cap K_{i}\right)}{P\left(K_{i}\right)}=\frac{P\left(S_{1}\cap S_{2}\cap K_{i}\right)}{P\left(K_{i}\right)}+\frac{P\left(S_{1}\cap S_{2}^{c}\cap K_{i}\right)}{P\left(K_{i}\right)}=P\left(S_{1}\cap S_{2}\mid K_{i}\right)+P\left(S_{1}\cap S_{2}^{c}\mid K_{i}\right)$
