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Problem Statement

The problem 6, selected page 72, from “Introduction to Probability” book

Assume that a new light bulb will burn out after t hours, where t is chosen from [0,∞) with an exponential density \[ f(t)=\lambda { e }^{ -\lambda t } \]

In this context, λ is often called the failure rate of the bulb.

  1. Assume that λ = 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.

  2. For what T is the reliability of the bulb = 1/2?

Solution (a)

If X is a random variable, then we essentially want to find the probability of X less than or equal to T, where T is the time of failure. Hence we have the following:

The probability that the bulb will burn our before T hours is \[ P(x\le T) \]

The probability that the bulb will not burn out after T hours is simply the compliment \[ 1-P(x\le T) \]

Since we are dealing with a continous exponential function of time, we should consider using an integral with integration limits 0 to T and integrand f(t)

\[ 1-\int _{ 0 }^{ T }{ \lambda { e }^{ -\lambda t } } dt \] We are given a value of lambda= 0.01 \[ 1-\int _{ 0 }^{ T }{ 0.01 { e }^{ -0.01 t } } dt\\ 1-0.01\int _{ 0 }^{ T }{ { e }^{ -0.01 t } } dt \] The technique at hand is u-substitution (reverse chain rule) https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-9/a/review-applying-u-substitution \[ u=-0.01t\\ du=-0.01dt\\ \frac { -du }{0.01 } =dt \] \[ 1-0.01\int _{ 0 }^{ T }{ { e }^{ u } } \frac { -du }{0.01 }\\ 1+\int _{ 0 }^{ T }{ { e }^{ u } } du\\ \] Transform upper and lower limits using our u-sub \[ Upper: u=-0.01t=-0.01T\\ lower: u=-0.01t=-0.01(0)=0 \] Replace newlimits and evaluate in terms of u \[ 1+\int _{ 0 }^{-0.01 T }{ { e }^{ u } } du\\ \] Apply fundemental theorem of calculus for definite integrals (part 2 of theorem) http://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html \[ 1+(e^{-0.01T}-e^{0})=\\ 1+e^{-0.01T}-1=\\ =e^{-0.01T} \]

Solution (b)

We need to perform some algebra and solve for T using our derived probability. We need to solve for T
We need to solve exponential equations http://www.sosmath.com/algebra/logs/log4/log46/log46.html \[ \frac { 1 }{2 }=e^{-0.01T} \] \[ ln(1/2)=ln(e^{-0.01T})\\ ln(1/2)=-0.01T\\ \frac { ln(1/2) }{-0.01 }=T \] Using quotient to subtraction property of ln \[ \frac { ln(1)-ln(2) }{-0.01 }=T\\ \frac { 0-ln(2) }{-0.01 }=T\\ \frac { -ln(2) }{-0.01 }=T\\ \frac { ln(2) }{0.01 }=T \]

## [1] 69.31472

If you calculate these values in a calculator or in R, then T is roughly 70, hence 70 hours.