Since both trains leave at the same frequency of 10 minutes, probably the solution lies in the gap between the two consecutive trains of each line. Suppose the train $M$ (towards Mary) leaves at the minutes: $0, 10, 20,...$

And the train $G$ (towards Gwen) leaves at the minutes: $x, (x+ 10 ), (x+20)...$ and so on, where $(0 < x < 10)$

Suppose Spiderman randomly reaches the station at some time $t + 10\cdot n$ minutes where $n$ is an arbitrary integer and $t$ is the fraction that determines his fate, $0 < t \le 10$.

Note that if $0 < t \le x$ means that he just missed $M$ and now has to wait for $G$. While $x <t \le 10$ means the opposite.

Thus, the probability that he takes the train $G$ is:

$P(G) = P(0 < t \le x)$

$0.1 = \dfrac{x-0}{10} \implies x = 1$

The train heading to Mary Jane's location leaves precisely 1 minute before the train heading to Gwen Stacy's location. Therefore, 9 out of 10 times, spiderman reaches the station when the next train is leaving towards Mary.