Question

Suppose you own three computers, each of which independently lasts for an exponentially distributed time with a mean of 4 years. Let X(t) be the number of these computers that have broken down after t years.

(a) What is the probability that all three computers are still working after six months?

(b) On average, how long will it take before there is only one working computer left?

Answer 1

This is a simple question related to the exponential distribution of the working computers which were 3 initially.

The mean life is given as 4 years.

The probability density function (pdf) for an exponential distribution is given by

f(x) = me^–mx where m=1/ (called as decay factor) : x= contineous random variable.)

clearly

m =1/4 =0.25

and the pdf will be

f(x) = 0.25 e^–0.25x

keeping values we can get the pdf as follows.

Now let us focus on the questions

It asks us to find the probability of all three laptops working after 6 years.

we shall find the probability of one laptop working first.

We need to find the probability of the laptop working after 6 years.

or P(x>6) ;P denotes probability

For an exponential distribution the

P(X<x) is given by

P(X < x) = 1 –e^–mx (read as probability of duration X being less than given duration of x)

Hence P(X>x) =1-P(X<x) (sum of probabiliites should be equal to 1)

or, P(X > x) = 1-(1 –e^–mx)

or, P(X > x) = e^–mx

Keeping values,

P(X > 6) = e^–0.25*6

or, P(X > 6) = 0.223

Thus the probability that a laptop survives 6 years is 0.223.

But in our case all three laptops survive thus

this probability will be given by the product of probabilitiies of survival of all three laptops

or P(X1 > 6)*P(X2 > 6)*P(X3 > 6) (X1,X2,X3 represent three laptops)

since probabilites of survival of each will be same (they have same mean value)

thus overall probabilty of survival of all 3 laptops will be

=(0.223)*(0.223)8(0.223)

=0.0111 (1.11 % probability)

b) It asks us to find the duration over which 2 computers will fail and only one will be left.

Clearly in other words, it is asking us to find the time over which 66.67 % of the computers fail .

let T1 = time after which 1st computer fails

T2 =time over which the second fails

thus T =T1+T2 will be the time over which the two computers fails

the expected value of t will hence be given by

E[T] =E[T1+T2]

or, E[T]=E[T1]+E[T2]

But for an exponential function the expected time of decay is given by the decay factor m=1/

Thus, E[T]=E[T1]+E[T2]

or, E[T]=1/+1/

or,E[T] = 2/

0r, E[T] =2/4 =0.50 years