# Finding $P(X_1+X_2 > 1.9X_3)$ where $X_1$, $X_2$, and $X_3$ are independent, normal distributed random variables

Suppose that $X_1$, $X_2$, and $X_3$ are independent, normal distributed with same mean and standard deviation. How do I find $P(X_1+X_2 > 1.9X_3)$?

What I did so far:

$$P(X_1+X_2 > 1.9X_3) = 1- P\left(\dfrac{X_1+X_2}{X_3}\le1.9\right)$$

Adding $X_1$ and $X_2$ together will yield a new random variable with different normal distribution:

$$X_1+X_2\sim N(\mu_{X_1}+\mu_{X_2},\sigma^2_{X_1}+\sigma^2_{X_2})$$ Then how do I calculate a random variable with ratio distribution of $\dfrac{X_1+X_2}{X_3}$? Cauchy distribution would not work here because the means are not zero. Is there other way to do this?

The key observation you need to make is that sums of independent Normal variables are themselves Normal.

If we define $Z = X_1 + X_2 - 1.9 X_3$ with $X_1,X_2,X_3 \sim N(\mu, \sigma^2)$ then

$$Z \sim N((1 + 1 - 1.9) \mu, (1^2 + 1^2 + (-1.9)^2)\sigma^2),$$

i.e. $Z \sim N(0.1\mu, 5.61\sigma^2)$.

And then

$$\mathbf P [ X_1 + X_2 > 1.9X_3] = \mathbf P[ Z > 0] = 1 - \Phi \left( \frac{-0.1\mu}{\sqrt{5.61\sigma^2}} \right)= \Phi \left( \frac{0.1\mu}{\sqrt{5.61\sigma^2}} \right),$$ where $\Phi$ is the CDF of the standard Normal Distribution.

$X_1+X_2-(1.9)X_3$ is normal with mean $(0.1)m$ and variance $(5.61) \sigma^{2}$ where $m$ and $\sigma ^{2}$ are the mean and variance of the $X_i$'s. Write $Y$ for $X_1+X_2-(1.9)X_3$. What you want to find is $P\{Y>0\}$. The ansqer id $1-\Phi (-\frac {(0.1)m)} {\sigma\sqrt (5.61)}$