Question

G A professor has two light bulbs in his garage. When both are burned out, they are replaced, and the next day starts with two working light bulbs. Suppose that when both are working, one of the two will go out with probability .02 (each has probability .01 and we ignore the possibility of losing two on the same day). However, when only one is there, it will burn out with probability .05.
(a) What is the long-run fraction of time that there is exactly one bulb working

Answer 1

Answer:

2/7 or 0.2857

Step-by-step explanation:

The expected time before the first bulb burns out (two bulbs working) is given by the inverse of the probability that a bulb will go out each day:

$E_1 = \frac{1}{0.02}=50\ days$

The expected time before the second bulb burns out (one bulb working), after the first bulb goes out, is given by the inverse of the probability that the second bulb will go out each day:

$E_2 = \frac{1}{0.05}=20\ days$

Therefore, the long-run fraction of time that there is exactly one bulb working is:

$t=\frac{E_2}{E_1+E_2}=\frac{20}{20+50}\\t=\frac{2}{7}=0.2857$

There is exactly one bulb working 2/7 or 0.2857 of the time.