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Independence is a very
useful property.

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Whenever it is true, we can
break up complicated

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situations into simpler ones.

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In particular, we can do
separate calculations for each

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piece of a given model and
then combine the results.

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We're going to look at an
application of this idea into

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the analysis of reliability of
a system that consists of

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independent units.

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So we have a system that
consists of a number, let's

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say, n, of units.

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And each one of the units
can be "up" or "down".

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And it's going to be "up" with
a certain probability pi.

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Furthermore, we will assume
that unit failures are

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independent.

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Intuitively, what we mean is
that failure of some of the

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units does not the affect the
probability that some of the

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other units will fail.

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If we want to be more formal,
we might proceed as follows.

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We could define an event
Ui to be the event that

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the ith unit is "up".

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And then make the assumption
that the events U1, U2, and so

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on up to Un, if we have n
units, are independent.

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Alternatively, we could define
events Fi, where event Fi is

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the event that the ith unit is
down, or that it has failed.

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And we could assume that the
events Fi are independent, but

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we do not really need a
separate assumption.

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As a consequence of the
assumption that the Ui's are

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independent, one can argue
that the Fi's are also

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independent.

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How do we know that
this is the case?

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If we were dealing with just two
units, then this is a fact

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that we have already proved
a little earlier.

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We did prove that if two events
are independent, then

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their complements are
also independent.

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Now that we're dealing with
multiple events here, a

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general number n,
how do we argue?

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One approach would be to be
formal and start from the

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definition of independence
of the U events.

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And that definition gives
us a number of formulas.

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Then manipulate those formulas
to prove the conditions that

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are required in order to check
that the events Fi are

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independent.

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This is certainly possible,
although it is a bit tedious.

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However, the approach we will
be taking in situations like

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this one is that we will use
the intuitive understanding

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that we have of what
independence means.

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So independence in this context
means that whether

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some units are "up" or down,
does not change the

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probabilities that some
of the other units

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will be "up" or down.

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And by taking that
interpretation, independence

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of the events that units are
"up" is essentially the same

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as independence of the
units [having]

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failed.

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So we take this implication for
granted and now we move to

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do some calculations for
specific systems.

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Consider a particular system
that consists of three

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components.

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And we will say that the system
is "up", if there

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exists a path from the left to
the right that consists of

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units that are "up".

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So in this case, for the system
to be "up", we need all

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three components to be "up"
and we proceed as follows.

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The probability that the
system is "up"--

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this is the event that the first
unit is "up", and the

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second unit is "up", and
the third unit is "up".

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And now we use independence to
argue that this is equal to

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the probability that the first
unit is "up" times the

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probability that the second
unit is "up" times the

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probability that the
third unit is "up".

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And in the notation that we have
introduced this is just

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p1 times p2 times p3.

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Now, let us consider
a different system.

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In this system, we will say
that the system is "up",

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again, if there exists a path
from the left to the right

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that consists of units
that are "up".

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In this particular case the
system will be "up", as long

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as at least one of those three
components are "up".

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We would like again to calculate
the probability that

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the system is "up".

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And the system will be "up",
as long as either unit 1 is

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"up", or unit 2 is "up",
or unit 3 is "up".

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How do we continue from here?

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We cannot use independence
readily, because independence

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refers to probabilities of
intersections of events,

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whereas here we have a union.

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How do we turn a union
into an intersection?

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This is what De Morgan's Laws
allow us to do, and involves

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taking complements.

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Instead of using formally De
Morgan's Laws, let's just

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argue directly.

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Let us look at this event.

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That unit 1 fails, and unit
2 fails, and unit 3 fails.

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What is the relation between
this event and the event that

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we have here.

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They're complements.

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Why is that?

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Either all units fail, which is
this event, or there exists

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at least one unit,
which is "up".

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So since this event is the
complement of that event, this

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means that their probabilities
must add to 1, and therefore

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we have this relation.

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And now we're in better shape,
because we can use the

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independence of the events F to
write this as 1 minus the

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product of the probabilities
that each one

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of the units fails.

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And with the notation that we
have introduced using the

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pi's, this is as follows.

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The probability that unit
1 fails is 1 minus the

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probability that it is "up".

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Similarly, for the second unit,
1 minus the probability

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that it is "up".

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And the same for
the third unit.

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So we have derived a formula
that tells us the reliability,

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the probability that a system of
this kind is "up" in terms

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of the probabilities of its
individual components.

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You will have an opportunity to
deal with more examples of

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this kind, a little more
complicated, in the problem

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that follows.

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And even more complicated, in
one of the problem-solving

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videos that we will have
available for you.