Question

A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of these bulbs by an exponential density function with mean μ = 1400, find the probability that both of the lamp's bulbs fail within 1500 hours. (Round your answer to four decimal places.)

Answer 1

The probability of failure of both the bulbs is 0.4323.

Explanation:

For an exponential distribution the distribution is given by

`f(x,\lambda )=\int_(0)^(x )\lambda e^(-\lambda x)dx`

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

`P(X)=\int_(0)^(x )\lambda _(X)e^{-\lambda _(X)x}dx`

Similarly for the bulb Y the distribution function is given by

`P(Y)=\int_(0)^(y )\lambda _(Y)e^{-\lambda _(Y)y}dy`

Thus the probability for both the bulbs to fail within 1500 hours is

`P(E)=\int_(0)^(1500)\int_(0)^(1500)(1)/(1400)e^{(-x)/(1400)}\cdot (1)/(1400)e^{(-y)/(1400)}dxdy\\\\P(E)=(1)/(1400^2)(\int_(0)^(1500)\int_(0)^(1500)e^{(-x)/(1400)}\cdot e^{(-y)/(1400)}dxdy)\\\\P(E)=(1)/(1400^2)(\int_(0)^(1500)e^{(-x)/(1400)}dx)\cdot (\int_(0)^(1500)e^{(-y)/(1400)}dy)\\\\P(E)=(1)/(1400^(2))* 920.473* 920.473\\\\\therefore P(E)=0.4323`