How is the right term derived in this conditional probability statement?

Question

Does anyone know how to get the right term? It seems like the right term is actually $P(C, x_1, x_2, \ldots, x_n)$ and not $P(C \mid x_1, x_2, \ldots, x_n)$?

Answer 1

As Blaza said, a common devisor cancels, is all.

$\require{cancel}\begin{align}\dfrac{\mathsf P\big(C=c^1\mid \bigcap\limits_{i=1}^n (X_i=x_i)\big)}{\mathsf P\big(C=c^2\mid \bigcap\limits_{i=1}^n (X_i-x_i)\big)} ~ & = ~ \dfrac{\mathsf P\big((C=c^1)\cap\bigcap\limits_{i=1}^n(X_i=x_i)\big)/\cancel{\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\big)}}{\mathsf P\big((C=c^2)\cap\bigcap\limits_{i=1}^n(X_i=x_i)\big)/\cancel{\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\big)}} \\[1ex] &=~ \dfrac{\mathsf P(C=c^1)~\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\mid C=c^1\big)}{\mathsf P(C=c^2)~\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\mid C=c^1\big)}\\[2ex] &=~\dfrac{\mathsf P(C=c^1)}{\mathsf P(C=c^2)}\prod_{i=1}^n\dfrac{\mathsf P(X_i=x_i\mid C=c^1)}{\mathsf P(X_i=x_i\mid C=c^2)}\end{align}$

The last step relies on the conditional mutual independence of $(X_\star)$ when given a value for $C$.

Anything else?