| Line 3: | Line 3: | ||
let's assume S comes G mins after N. | let's assume S comes G mins after N. | ||
| − | + | ||
| G | | G | | G | | | G | | G | | G | | ||
N S N S N S | N S N S N S | ||
| − | P(S) = G/10 (equ 2) | + | P(S) = G/10 (equ 2) |
| − | + | ||
P(N) = (10 - G)/10 (equ 3) | 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 | plug equ 2 and 3 into equ 1. and solve for G | ||
Latest revision as of 19:34, 9 September 2008
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
