Introduction to Probability and Statistics for Engineers and Scientists

178 Chapter 5: Special Random Variables

However, if the lifetime distributionFis not exponential, then the relevant probability is

P{lifetime>t+ 5 |lifetime>t}=

1 −F(t+5)

1 −F(t)

wheretis the number of miles that the battery had been in use prior to the start of the
trip. Therefore, if the distribution is not exponential, additional information is needed
(namely,t) before the desired probability can be calculated. ■

For another illustration of the memoryless property, consider the following example.

EXAMPLE 5.6b A crew of workers has 3 interchangeable machines, of which 2 must be
working for the crew to do its job. When in use, each machine will function for an expo-
nentially distributed time having parameterλbefore breaking down. The workers decide
to initially use machines A and B and keep machine C in reserve to replace whichever of
A or B breaks down first. They will then be able to continue working until one of the
remaining machines breaks down. When the crew is forced to stop working because only
one of the machines has not yet broken down, what is the probability that the still operable
machine is machine C?

SOLUTION This can be easily answered, without any need for computations, by invoking
the memoryless property of the exponential distribution. The argument is as follows:
Consider the moment at which machine C is first put in use. At that time either A or
B would have just broken down and the other one — call it machine 0 — will still be
functioning. Now even though 0 would have already been functioning for some time,
by the memoryless property of the exponential distribution, it follows that its remaining
lifetime has the same distribution as that of a machine that is just being put into use. Thus,
the remaining lifetimes of machine 0 and machine C have the same distribution and so,
by symmetry, the probability that 0 will fail before C is^12. ■

The following proposition presents another property of the exponential distribution.

PROPOSITION 5.6.1 IfX 1 ,X 2 ,...,Xnare independent exponential random variables hav-
ing respective parametersλ 1 ,λ 2 ,...,λn, then min (X 1 ,X 2 ,...,Xn) is exponential with
parameter

∑n

t= 1 λi.

Proof

Since the smallest value of a set of numbers is greater thanxif and only if all values are
greater thanx, we have

P{min(X 1 ,X 2 ,...,Xn)>x}=P{X 1 >x,X 2 >x,...,Xn>x}

=

∏n

i= 1

P{Xi>x} by independence