Problem 1a
For a certain brand of light bulbs, the time (in thousands of hours) until the instant a light bulb burns out follows an exponential distribution. From past experience, we know that 
Problem 1b
In a tech support call center for a brand of laser printers, the length of time (in minutes) for a technical support call follows an exponential distribution. From past experience, 60% of the calls last 3 minutes or less. What proportion of the calls can be expected to last more than 15 minutes?
Solution is found below.
Solution of Problem 1a
Let 
![\displaystyle \begin{aligned}P(X \le 1)&=1-e^{-\alpha} \\&=1-\biggl[e^{-0.5 \alpha}\biggr]^2 \\&=1-\biggl[\frac{3}{4}\biggr]^2 \\&=\frac{7}{16} \end{aligned}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7DP%28X+%5Cle+1%29%26%3D1-e%5E%7B-%5Calpha%7D+%5C%5C%26%3D1-%5Cbiggl%5Be%5E%7B-0.5+%5Calpha%7D%5Cbiggr%5D%5E2+%5C%5C%26%3D1-%5Cbiggl%5B%5Cfrac%7B3%7D%7B4%7D%5Cbiggr%5D%5E2+%5C%5C%26%3D%5Cfrac%7B7%7D%7B16%7D+%5Cend%7Baligned%7D&bg=ffffff&fg=333333&s=0 "\displaystyle \begin{aligned}P(X \le 1)&=1-e^{-\alpha} \\&=1-\biggl[e^{-0.5 \alpha}\biggr]^2 \\&=1-\biggl[\frac{3}{4}\biggr]^2 \\&=\frac{7}{16} \end{aligned}")
Answer to Problem 1b
Dan Ma
When I first attempted problem 1a, I solved for the exponential parameter (I use lambda). This requires one to use ln as the inverse and then plug this value back in to the equation a second time for t=1. Your solution is far more elegant– don’t solve for lambda, use the whole expression as a factor! Much quicker than the “textbook” way.
I then used your method for the second problem. Elegant indeed.