How to find the probability of one sample variance is two times larger than another?

Question

I have two normal distribution where $X\sim N(\mu_{x}, 40^{2})$ and $Y\sim N(\mu_{y}, 50^{2})$. 8 samples from X and 16 samples from Y is drawn. How to determine the probability that the variance of the first sample is more than twice that of the second?

At first I was thinking about Chi-square distribution, but it seems that it only works for checking one variance.

Any help would be appreciated.

Edit:
I now reached to something like $\frac{25S_{X}^{2}}{16S_{Y}^2}\sim F_{7, 15}$,
then I am trying to calculate $P(S_{X}^{2}>2S_{Y}^2)$ $$\begin{eqnarray} P(S_{X}^{2}>2S_{Y}^2) &=& P(\frac{S_{X}^{2}}{S_{Y}^2}>2) \\ &=& P(\frac{25S_{X}^{2}}{16S_{Y}^2}>2(\frac{25}{16})) \\ &=& P(\frac{25S_{X}^{2}}{16S_{Y}^2}>3.125)\\ &=& 1 - F_{7, 15}(3.125) \end{eqnarray}$$ Am I doing it correct? I am trying to lookup the value from the table but where's the alpha-value (significance value)

Answer 1

First of all I underline that X and Y must be independent.

Hint:

$U=\frac{7S_X^2}{40^2}\sim \chi_{(7)}^2$

$V=\frac{15S_Y^2}{50^2}\sim \chi_{(15)}^2$

...but you know that $\frac{\frac{U}{7}}{\frac{V}{15}}=\frac{15}{7}\frac{U}{V}$ follows a Snedercor's F Distribution....

...now I think you can go on by youself