Waiting for the bus (I)

Suppose that the last type $A$ bus arrived $T_A$ minutes before you arrive at the bus stop, and the last type $B$ bus arrived $T_B$ minutes before you arrive at the bus stop, and let $T$ be the time you have to wait for the next bus. Then $T_A$ is uniformly distributed on $[0,6]$ , while $T_B$ is uniformly distributed on $[0,10]$ . For any $o \le x \le 6$ we will only have $T \ge x$ if $T_A \le 6-x$ and $T_B \le 10-x$ . Thus $\begin{aligned} P[T \ge x] & = \; \tfrac{1}{6}(6-x)\times \tfrac{1}{10}(10-x) \\ P[T \le x] & = \; 1 - \tfrac{1}{60}(6-x)(10-x) \; = \; \tfrac{1}{60}(16x - x^2) \end{aligned}$ Differentiating gives the probability distribution function for $T$ as $f_T(x) \; = \; \left\{ \begin{array}{lll} \tfrac{1}{30}(8-x) & \hspace{1cm} & 0 \le x \le 6 \\ 0 & & \mathrm{o.w} \end{array}\right.$ and hence the expected waiting time is $E[T] \; = \; \int_0^6 x f_T(x)\,dx \; = \; \tfrac{12}{5}$ making the answer $12+5 = \boxed{17}$ .