Sum of Conditional Probabilities of two discrete random values

Question

While studying probability theory, I came across this truth statement. In my lecture notes, it has been depicted as False. However, according to my reasoning, it is supposed to be true. Could someone share their opinions on this?

Truth Statement:

Answer 1

This statement looks like a flawed version the law of total probability (see https://en.wikipedia.org/wiki/Law_of_total_probability). The first mistake is that the probability that $Y = y_{i}$ is missing. The second mistake is that the sum does not equal 1, but it equals the probability of $X=x_{i}$, so $ \mathbb{P}(X_{i}) = \sum\limits_{j=1}^{k} \mathbb{P}(X = x_{i} \mid Y = y_{j}) \mathbb{P}(Y = y_{i}), $ for all $i$.

Another possible way to correct the statement is as follows $ \sum\limits_{i=1}^{m} \mathbb{P}(X_{i}) = \sum\limits_{i=1}^{m} \sum\limits_{j=1}^{k} \mathbb{P}(X = x_{i} \mid Y = y_{j}) \mathbb{P}(Y = y_{i}) = 1,$ but then we lose the for all $i$ part of the equation.

A counterexample to the original statement is the following. Consider a deterministic random variable $X$, with $\mathbb{P}(X=0)=1$. Furthermore assume that $Y$ can take two values, 0 and 1, with probability 1/2 and that $X$ and $Y$ are independent. Then $\mathbb{P}(X=0 \mid Y = 0) + \mathbb{P}(X=0 \mid Y = 1) = 1 + 1 = 2$ (using the independence of $X$ and $Y$), but $\mathbb{P}(X=0 \mid Y = 0) \mathbb{P}(Y=0) + \mathbb{P}(X=0 \mid Y = 1)\mathbb{P}(Y=1) = 1/2 + 1/2 = 1$.