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(N) = (10 - G)/10 (equ 3)

  • P(event)= its duration/total duration


plug equ 2 and 3 into equ 1. and solve for G

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood