Let $\xi_1, \xi_2, \xi_3, \xi_4$ independent random variables with normal distribution with parameters (0, 1), (-1, 1), (0, 4) and (1, 4) respectively. So, I have a easy question, how to find $P(|2\xi_1 - 3\xi_2 + \xi_3 - \xi_4| < 13)$.
And in general, how to find probability that simple arithmetic combination(like above) of random variables is less of the given number?
The general technique is:
Note that $Z = X_1+X_2+X_3+X_4$ is a random variable.
Derive a closed form for it (generally by convolution)
Use the CDF to find $P(Z<13)$.
In this case, as the r.v.'s are all normal, then if $X\sim\mathcal{N}(\mu_X,\sigma_X^2)$ and $Y\sim\mathcal{N}(\mu_Y,\sigma_Y^2)$, then $Z = X+Y\sim\mathcal{N}(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2)$
To deal with $cX$ where $X\sim\mathcal{N}(\mu_X,\sigma^2_X)$, we just note that $E[cX] = cE[X]$, and $V[cX] = c^2V[X]$, so $cX\sim\mathcal{N}(c\mu_X,c^2\sigma_X^2)$.
Note that $E[2\xi_1] = 0\times 2$, $E[-3\xi_2] = 3$, $E[\xi_3] = 0$, and $E[-\xi_4] = -1$, so $E[Z] = 2$.
We also have that $V[2\xi_1] = 4\times 1, V[-3\xi_2] = 9\times 1, V[\xi_3] = 4, V[-\xi_4] = 4$.
So, we have that: $$Z\sim\mathcal{N}(2,21)$$
Then, all you need to do is compute $\mathbb{P}(Z < 13)$ for $Z\sim\mathcal{N}(2,21)$.