# Probability 2 - Lightbulbs

• Jan 28th 2009, 01:56 PM
mistykz
Probability 2 - Lightbulbs
Assume that a new light bulb will burn out after t hours, where t is chosen from [0,infinity) with an exponential density of f(t)= L*e^(-Lt)
In this context, L is often called the failure rate of the bulb.

a) assume that L = 0.01, and find the probability that the bulb will not burn out before T hours. This probability is often called the reliability of the bulb.
b) for what T is the reliability of the bulb = 1/2?

I set up part a as f(t) = 0.01e^(-0.01t), which is obvious, but I'm not sure where to go from here. Since this is the equation for the failure rate...would the probability of it NOT burning out be 0.99e^(-0.99t)? I am somewhat confused, my professor did not explain this too well.
• Jan 28th 2009, 04:01 PM
mr fantastic
Quote:

Originally Posted by mistykz
Assume that a new light bulb will burn out after t hours, where t is chosen from [0,infinity) with an exponential density of f(t)= L*e^(-Lt)
In this context, L is often called the failure rate of the bulb.

a) assume that L = 0.01, and find the probability that the bulb will not burn out before T hours. This probability is often called the reliability of the bulb.
b) for what T is the reliability of the bulb = 1/2?

I set up part a as f(t) = 0.01e^(-0.01t), which is obvious, but I'm not sure where to go from here. Since this is the equation for the failure rate...would the probability of it NOT burning out be 0.99e^(-0.99t)? I am somewhat confused, my professor did not explain this too well.

(b) Solve for $T$:

$\Pr(\text{Lifetime of bulb} > T) = \frac{1}{2}$, that is, $\int_T^{+\infty} 0.01 e^{-0.01 t} \, dt = \frac{1}{2}$.