please see the attached file.

Suppose and denote the lifetimes of the four components of the minicomputer. Here it is given that the lifetimes of the components are mutually independent exponentially distributed random variables with j-th component has mean lifetime 1/j, j = 1,2,3,4.
Thus the survival functions are , , and
a) The probability that no component of a minicomputer fails during the first 0.2 years of operation is given by

= , since the lifetimes of the components are mutually independent.
=
= 0.135335
b) At any time 't' the probability that component 1 of a minicomputer is the first component of that minicomputer to fail is given by

=
=
=
Let S denote the lifetime of the minicomputer and denote the survival function of the minicomputer system. Here it is given that the system ( minicomputer) remains functioning as long as at least 2 of the 4 components are functioning. Thus the given system is a 2 out of 4 system. Thus the survival function of the minicomputer system is given by

= +
+ +
+ +
+ +
+ +
+
= + +
+ + +
+ + +
+ +
=
c) The expected lifetime of the minicomputer is given by

=
= =
= 0.501587 = (1/2) years approximately
d) The variance of the lifetime of the minicomputer is given by
( This is the standard formula for variance for positive random variables)
But =
= = 0.358968
Thus =
= 0.107378
e) Let N(t) denote the number of renewal in the interval [0, t]. Then { N(t), t ≥ 0} will be a Poisson process if the interarrival times ( the time between successive replacement of the minicomputer ), the random variable S must be exponentially distributed. That is the survival function of S must be of the form . But here the distribution of S is not exponentially distributed as its survival function is of the form , which is not the survival function of an Exponential random variable. Thus { N(t), t ≥ 0} is not a Poisson process.
f) Even though { N(t), t ≥ 0} is not a Poisson process, it is a renewal process since the interarrival times are identically distributed with the survival function .
By the theory of renewal process for sufficiently large 't'
, where is the cumulative distribution function of a Standard Normal random variable ( For the Proof refer Ross( 1996, page 109)
Ross, S.M Stochastic Processes, 2nd Ed. Wiley, New York )
Now the probability that at least 50 computers have to be replaced during the first 10 years of operation of the alarm system is given by

where = 0.501587 and = 0.327686
Substituting these values we get
=1-1 = 0