Sourse: Tanenbaum
So, Success at first attempt = probability of zero frames = $e^{-G}$
Given, it generate 50 requests/sec i.e 50 requests in $10^{3}$ ms
i.e in $10^{3}$ ms ${\Rightarrow }$ 50 frames generated
So, in 40 ms ${\Rightarrow }$ $\frac{40 \times 50 }{10^{3}} = 2$ frames generated
So G = 2
So, Success at first attempt = $e^{-G}$ = $e^{-2}$