since he went to N twice as much as S. it means the possibility of getting on to the N bus is 2 times the possibility of getting on to the S bus. we can say P(N)=2*P(S) (equ 1)
let's assume S comes G mins after N.
| G | | G | | G |
N S N S N S
P(S) = G/10 (equ 2) P(event)= its duration/total duration
P(N) = (10 - G)/10 (equ 3)
plug equ 2 and 3 into equ 1. and solve for G
