The life expectancy of two headlights follow an exponential distribution with a MTBF of 1500 hours. Headlights were installed at the same time, are identical models, and are on at the same time. What is the probability that both headlights will burn out in 950 hours or less of operation?
My first thought is to use the joint exponential distribution:
f(x,y)=(e^(-y/1500))/1500y, but I am having trouble determining the upper and lower limits of the integration. Is this the right approach and how do I determine the limits?

Question

The life expectancy of two headlights follow an exponential distribution with a MTBF of 1500 hours.  Headlights were installed at the same time, are identical models, and are on at the same time.  What is the probability that both headlights will  burn out in 950 hours or less of operation?
My first thought is to use the joint exponential distribution:
f(x,y)=(e^(-y/1500))/1500y, but I am having trouble determining the upper and lower limits of the integration.  Is this the right approach and how do I determine the limits?

kropot72 · Answer

The exponential distribution is related to the Poisson process in the following way. If events occur in accordance with a Poisson process, then the time until the first occurence of the event has an exponential distribution.
Therefore if the number of occurences of an event in time t is given by the Poisson distribution then:
$$P(X _{t}=x)=\frac{(vt)^{x}}{x!}e ^{-vt}$$
Where
$$X _{t}=number\ of\ occurrences\ during\ time\ interval\ t$$
and

v = mean occurrence rate = 1/MTBF
$$P(X _{t}=2)=\frac{(\frac{950}{1500})^{2}}{2!}e ^{-\frac{950}{1500}}$$