Waiting for the bus (II)

Consider the senario when there is two buses A and B coming every 6 and 10 minutes respectively. Let the time taken to wait for bus A and B after arriving at the bus stop be $T_{A}$ and $T_{B}$ respectively. $0\leq T_{A}\leq 6$ , $0\leq T_{B}\leq 10$

Then, the waiting time $T=min(T_{A},T_{B})$ . Graphing $T = \left\{\begin{matrix} T_{A} & T_{A}<T_{B}\\ T_{B} & T_{B}<T_{A} \end{matrix}\right.$ , we get and .

Let $l_{T}=$ length of a contour line (or the level curve) for some value of $T$ . Then, the probability distribution $f_{T}\propto l_{T}$ .

Normalising $f_{T}$ gives $\frac{l_{T}}{\int_{0}^{T_{A}}l_{T}\: dT}=\frac{l_{T}}{T_{A}T_{B}}$

The expected waiting time $E(T)=\int_{0}^{T_{A}}Tf_{T}\: dT=\frac{\int_{0}^{T_{A}}Tl_{T}dT}{T_{A}T_{B}}$

The volume of the solid represented $=\int_{0}^{T_{A}}Tl_{T}dT=\int_{0}^{T_{A}}T(T+b)dT$ by observation where $b=T_{B}-T_{A}$ . $\therefore E(T)=\frac{\int_{0}^{T_{A}}T(T+b)dT}{T_{A}T_{B}}$

Generalising this to higher dimensions gives $E(T)=\frac{\int_{0}^{T_{A}}T(T+b)(T+c)...dT}{T_{A}T_{B}T_{C}...}, T_{A}=min(T_{A}, T_{B}, T_{C}...)$ where $\\b=T_{B}-T_{A} \\c=T_{C}-T_{A} \\\begin{matrix} & & . & \\ & & .& \\ & & .& \end{matrix}$

Solving the above question, $E(T)=\frac{\int_{0}^{5}T(T+1)(T+2)(T+3)dT}{5\times6\times7\times8}=\boxed{\frac{2515}{2016}}$