Probability that a random variable X belongs to the set of rational nos.

Question

The question is a Multiple choice question

Let $X$ be a random variable with the M.G.F. $$M_{X}(t) = \frac{6}{\pi^{2}}\sum_{n\ge1}\frac{e^{\frac{t^2}{2n}}}{n^2}\,,\;t\in R$$ Then $P(X \in Q)$, where $\,Q$ is the set of rational nos., equals

The options are

1. 0
2. 1/4
3. 1/2
4. 3/4

The correct option is option 1.) i.e. 0

I'm looking for an explanation as to how did they deduce the answer to zero.

Answer 1

Let $Y$ be a random variable with pmf $P(Y=n)=\frac{6}{n^2\pi^2}$ and $Z|Y=n\sim N(0,1/n).$ Then $Z$ has the pmf $M_X(t).$ So, $Z$ and $X$ have same distribution. $P(X\in \mathbb{Q})=P(Z\in \mathbb{Q})=\sum_{n\geq 1}P(Z\in \mathbb{Q}|Y=n)P(Y=n)=\sum_{n\geq 1}0\times P(Y=n)=0.$