   Chapter 15, Problem 52RE

Chapter
Section
Textbook Problem

A lamp has three bulbs, each of a type with average lifetime 800 hours. If we model the probability of failure of a bulb by an exponential density function with mean 800, find the probability that all three bulbs fail within a total of 1000 hours.

To determine

To find: The probability of all the bulb fail within 1000 hours.

Explanation

Given:

The lifetime of each bulb has an exponential density function and each bulb has an average lifetime of 800 hours.

Calculation:

The exponential density function is,

f(t)={0                 , if  t>01800et800  , if t0

Let X, Y and Z be the independent random variable which denotes the three bulbs of the lamp. Therefore, the exponential density function of each bulb is,

f(x)={0                 , if  x>01800ex800  , if x0f(y)={0                 , if  y>01800ey800  , if y0f(z)={0                 , if  z>01800ez800  , if z0

The probability that all the bulbs fails within 1000 hours is denoted by, P(X1000,Y1000,z1000) and the joint density function is,

f(x,y,z)=f(x)f(y)f(z)={                              0                                  , if  x,y,z<0(1800ex800)(1800ey800)(1800ez800)  , if x,y,z0={              0                  , if  x,y,z<0(1(800)3ex800y800z800)  , if x,y,z0={            0               , if  x,y<01(800)3e1800(x+y+z)  , if x,y0

Since, x+y+z1000, y varies from 0 to 1000x and z varies from 0 to 1000xy. Therefore, the value of P(X+Y+Z1000) is computed as follows.

P(X+Y+Z1000)=10001000x1000xy1(800)3e1800(x+y+z)dzdydx=[0001(800)3e1800(x+y+z)dzdydx+0100001000x01000xy1(800)3e1800(x+y+z)dzdydx]=0+0100001000x01000xy1(800)3e1800(x+y+z)dzdydx=0100001000x01000xy1(800)3e1800(x+y+z)dzdydx

Integrate with respect to z.

0100001000x01000xy1(800)3e1800(x+y)dzdydx=1(800)30100001000x01000xye1800xe1800ye1800zdzdydx=1(800)30100001000xe1800xe1800y[e1800z(1)800]01000xydydx=(800)(800)30100001000xe1800xe1800y[e1800(1000x

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