# Probability involving 3 standard normal random variables

If $X,Y,Z$ are independent standard normal random variables, compute $P(3X+2Y<6Z-7)$.

One way is to evaluate $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{(6z-2y-7)/3}\frac{\exp(-(x^2+y^2+z^2)/2)}{2\pi\sqrt{2\pi}}dxdydz.$$ But I don't know how to calculate this.

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Another way is to use the fact that any linear combination of independent normal random variables is a normal random variable again. –  Raskolnikov Apr 21 '12 at 16:16
I was thinking something along that line. So 3X+2Y~N(0,5) and 6Z-7~N(-7,6) isn't it? Then we have to calculate $$\int_{-\infty}^{\infty}\int_{\infty}^a\frac{e^{-b^2/50}}{5\sqrt{2\pi}}\frac{e^‌​{-(a+7)^2/72}}{6\sqrt{2\pi}}dbda$$ I can't figure out this one either. –  graidym Apr 21 '12 at 16:23
Hint: The probability is $P[3X+2Y-6Z<-7]$. –  David Mitra Apr 21 '12 at 16:49
I think I got it. 3X+2Y-6Z~N(0,-1), so $P(3X+2Y-6Z<-7)=P(A>7)$ where $A$ is the standard normal random variable. So the answer is $1-\Phi(7)$. Is it correct? –  graidym Apr 21 '12 at 16:58
Wait, the variance can't be negative, can it? So it should be $3X+2Y-6Z~N(0,11)$? –  graidym Apr 21 '12 at 16:59

$$\mu=\mu_1+\mu_2$$
$$\sigma^2=\sigma_1^2+\sigma_2^2$$
You can use this to solve your problem without integrals, other than one evaluation of the error function. Actually, you don't even need that, since the $7$ turns out to give you a special result for which you probably already know the result of applying the error function.