   Chapter 15.4, Problem 30E

Chapter
Section
Textbook Problem

(a) A lamp has two bulbs, each of a type with average lifetime 1000 hours. Assuming that we can model the probability of failure of a bulb by an exponential density function with mean µ = 1000, find the probability that both of the lamp’s bulbs fail within 1000 hours.(b) Another lamp has just one bulb of the same type as in part (a). If one bulb bums out and is replaced by a bulb of the same type, find the probability that the two bulbs fail within a total of 1000 hours.

(a)

To determine

To find: The probability of both 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 1000 hours.

Formula used:

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (1)

Calculation:

The exponential density function is,

f(t)={0                 , if  t>011000et1000  , if t0

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

f(x)={0                 , if  x>011000ex1000  , if x0f(y)={0                 , if  y>011000ey1000  , if y0

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

f(x,y)=f(x)f(y)={                   0                        , if  x,y<0(11000ex1000)(11000ey1000)  , if x,y0={              0                  , if  x,y<0(1(1000)2ex1000y1000)  , if x,y0={            0               , if  x,y<01(1000)2e11000(x+y)  , if x,y0

Obtain P(X1000,Y1000)

(b)

To determine

To find: The probability of two bulbs fail within a total of 1000 hours.

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