To find the probability whether $X$ is rational Let $X$ be a random variable with the moment generating function
$$_() =\frac{6}{\pi^2}\sum_{n\geq 1}\frac{e^{\frac{t^2}{2n}}}{n^2}$$
Then $(X\in \mathbb{Q})$, where  is the set of rational numbers, equals
$(A) \space\space 0\space\space\space\space\space\space (B) \space\space \frac{1}{4}\space\space\space\space\space\space  (C)\space\space \frac{1}{2}\space\space\space\space\space\space (D) \space\space \frac{3}{4}\space\space\space\space\space\space $
I was thinking whether I can replace this sum by an integral , because I can't think of a distribution whose mgf looks like this.
Please Help!
 A: We can define a random variable with this distribution as follows.


*

*Let $P$ be an integer valued random variable with
$$\mathbf P[P = n] = \frac{6}{(\pi n)^2}.$$

*Define
$$Y = \mathbf{N}(0, P^{-1}),$$
that is let $Y$ be normally distributed with mean $0$ and variance $P^{-1}$, where $P$ is random as defined above.


(Aside: the letter $P$ was chosen to denote precision, which is the inverse of variance).
I claim that $Y$ has the moment generating function defined above. To see this we use the fact that for a normal random variable $X \sim \mathbf N(0,\sigma^2)$ the moment generating function is
$$M_X(t) = \exp\left( \frac12 \sigma^2 t^2\right),$$
and proceed by conditioning on $P = n$, so we have
\begin{align*}
M_Y(t)& = \mathbf E\left[ \exp(t Y)\right] \\
& \sum_{n=1}^\infty \mathbf E \left[ \exp\left(t \, \mathbf{N}(0,P^{-1}\right) \, | \, P = n \right] \mathbf P[P = n] \\
& = \frac{6}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2}  \mathbf E \left[ \exp\left(t \, \mathbf{N}(0,n^{-1}\right)  \right] \\
& = \frac{6}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2}e^{t^2/(2n)}
\end{align*}

With this in mind, you can now solve the problem, again relying on conditioning on  $P = n$, i.e. considering
\begin{align*}
\mathbf P[ Y \in \mathbb Q ] & = \sum_{n=1}^\infty \mathbf P\left[ \mathbf N(0,P^{-1}) \in \mathbb Q \, | \, P = n\right] \mathbf P[ N = n] \\
& = \frac{6}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \mathbf P\left[ \mathbf N(0,n^{-1}) \in \mathbb Q \right] \\
& = \frac{6}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} 0 \\
& = 0,
\end{align*}
where we used the fact that for any finite $\sigma^2 > 0$, we have $\mathbf P[ \mathbf N(0,\sigma^2) \in \mathbb Q] = 0$, since continuous distributions have measure $0$ on countable sets.

As a final remark, the general trick we rely on here is that given moment generating function $M_1,\ldots, M_n$ and positive values $p_1 + \cdots p_n = 1$, then the moment generating function
$$ M = p_1 M_1 + \cdots p_n M_n$$
describes a variable where you first choose $k$ according to the probabilities $(p_i)_{i=1}^n$, i.e $\mathbf P[k = i ] = p_i$, and then sample from the variable with moment generating function $M_k$.
This, combined with recognising that the terms in the above resemble the MGF of normal variables lead to the solution. 
