Exam 1

Suppose that the queuing time, $T$ (in minutes), at a customer service desk has a probability given by

$$f(t)=\{\begin{array}{ll}\frac{1}{64}t(16-t^2)& 0\le t\le 4\\ 0& \text{elsewhere}\end{array}$$

What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute?