Here is a problem that I just cannot figure out how to answer:
$z_1, z_2, \cdots, z_7$ are all independent, and are all normal random variables with $N(0,1)$. What is the probability of this occurring? $$ P(Z_1^2 + Z_2^2 + Z_3^2) \leq 4.943(Z_4^2 + Z_5^2 + Z_6^2 + Z_7^2) $$
Expanding on my comment: since a $\chi_\nu^2$ distribution has PDF $\frac{1}{2^{\nu/2}\Gamma(\nu/2)}x^{\nu/2-1}\exp\frac{-x}{2}$ for $x\ge0$, the PDF of $X:=A/B$ with $A:=\sum_{i=1}^3Z_i^2,\,B:=\sum_{i=4}^7Z_i^2$ is$$\frac{x^{1/2}}{4\sqrt{2\pi}}\int_0^\infty b^{5/2}\exp\frac{-b(1+x)}{2}db=\frac{15}{4}x^{1/2}(1+x)^{-7/2}$$for $x\ge0$. With $x=\tan^2t$, you can verify as a sanity check that this PDF integrates to $1$:$$\int_0^\infty x^{1/2}(1+x)^{-7/2}dx=\int_0^{\pi/2}2\sin^2t\cos^3tdt=\operatorname{B}(3/2,\,2)=\frac{\Gamma(3/2)\Gamma(2)}{\Gamma(7/2)}=\frac{4}{15}.$$Finally,$$\int_0^{4.943}\frac{15}{4}x^{1/2}(1+x)^{-7/2}dx\approx0.949991.$$
Edit: as @kimchilover notes, this is equivalent to an $F(3,\,4)$-distributed variable being $\le4.943\times\frac34$, so an $F$-distribution table gives the probability as $\approx0.95$.