Suppose I have a random variable $X \in \{1,2,3\}$ with pmf,
$$ p(x) = [ p_1, \ p_2, \ p_3] $$
such that $p_1 +p_2 +p_3 =1$.
I would like to find the probability $P(X=x | x \ne2)$
That is,
$$ P(X=x |x \ne 2) = \frac{P(X=x,x\ne2)}{P(x\ne2)} $$
Now, $P(x\ne2) = P(x=1,x=3) =1-p_2$
But I am having difficulty calculating $P(X=x,x\ne2)$.
I believe some how that $p_2$ is distributed to the events $x=1$ and $x=3$. But not sure how it is distributed.
I wrote the conditional probability matrix for events $x\ne2$ and $x=2$
$$ \begin{array}{l|ll} P(X|X\ne 2) &1 &2 &3\\ \hline 1 &? &0 &?\\ 0 &0 &1 &0 \end{array} $$
Since $P(X|X=2) = [0,\ 1,\ 0]$
I have read that when events such as $X$ are partitions of the sample space that it must be $\sum_x P(X=x|x\ne2) =1$. So row's must sum to one in the conditional probability matrix.
Any help would be much appreciated. Thank you.
$\displaystyle \small P(X=x |x \ne 2) = \frac{P \big((X=x) \cap (x \ne 2)\big)}{P(x\ne2)}$
$\small P(x \ne 2) = 1 - p_2 = p_1 + p_3$
$\small P ((X=1) \cap (x \ne 2)) = P(X = 1) = p_1$
$\small P ((X=3) \cap (x \ne 2)) = P(X = 3) = p_3$
So,
$\displaystyle \small P(X=1 |x \ne 2) = \frac{p_1}{p_1 + p_3}$
$\displaystyle \small P(X=3 |x \ne 2) = \frac{p_3}{p_1 + p_3}$