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Hello and welcome to the lecture in the course
Probability Methods in Civil Engineering.
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Today, we will cover the Probability of Events,
which is very useful for the different
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applications in the problems related to Civil
Engineering.
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..
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In this lecture, we will first touch a few
basic concepts, that is, equality of events
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and
concept of fields, which are useful, particularly
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when we deal with most of the problems
in the Civil Engineering. Then, we will touch
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the countable and non-countable space,
followed by, we will go to the conditional
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probability; and with the help of this, we
will
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try to explain the total probability and related
theorem, Theorem of Total Probability;
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and, after that, we will cover this Bayes’
theorem and rule. And finally, we will see
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some
of the application problems for applying to
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this particular concept, and we will go one
after another, starting from this equality
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of events.
.
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.This equality of events says, that the two
events A and B are called equal, if they consist
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of same elements; the event A and B are called
equal with probability 1, this equal with
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probability 1 is important because, then we
can say that, all these elements of both the
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events are same, and that is, if the set consisting
all outcome, those are in A or in B, but
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not in A intersection B has 0 probability;
that is, a probability of these which is equal
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to
this one, as I discussed in the previous class,
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this probability should be equal to 0.
.
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This second point, if I want to explain graphically,
then it looks like this. Suppose, that
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this is one sample space in which there are
two events, one is A; this one is your A and
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another one is your B. Now, what it says is
that, if these A and B are equal with
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probability 1, if the set consist of all the
outcomes; those are in A as well as in B,
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but not
in the A intersection B.
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So, what we mean is that, these two areas,
one is this in A or in B, but not in their
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intersection. So, the probability of these
two events, probability of this area should
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be
equal to 0; so, this is exactly what is meant.
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So, these two area is nothing but your A
intersection B prime, which is union with
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A prime, A complementary with intersection
B. So, this is the area, and if we say that
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there is no such element in this area, then
that
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means, what we are trying to say is that,
the probability of this event, if it is 0,
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then we
can say that this A and this B, these two
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events areas are equal.
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..
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So, coming back to this point again, that
this event A and B are equal with probability
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1,
if the set consisting all outcomes those are
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in A or in B but not in A intersection B,
has
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zero probability, that is, probability of
that …. ; this is the thing I explained,
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which is
obviously equal to… this one also; so, these
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two events are referring to the same event,
whose probability…; if this probability
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is 0, then we can say that this event A and
B are
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equal. Thus, the event A and B are equal with
probability 1, if and only if, the probability
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of A is equals to probability of B is equals
to probability of A intersection B. This is
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important, once again, if we just refer to
that particular Venn diagram here …
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.
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.That is, if this probability of this A and
this probability of B is equal to the probability
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of
their intersection, which is nothing, but
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basically, we are just pulling these two events
to
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be on the same event, that is, this, as well
as this, means, this is your A as well … this
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itself is your B; so then only we can say
that, probability of A is equals to probability
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of
B equals to probability of A intersection
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B. All these three elements of this equation
is
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important because, if we do not consider these,
if we say that probability of A equals to
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probability of B, that does not mean that
these two events are same, so, this must be
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there.
For example, if we take the example of throwing
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of one dice, and if we say that the
probability of getting 1 or probability of
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getting 2, both are same; or, if I say the
probability of getting an even outcome and
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probability of getting an odd outcome, one
is
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event A, another one is event B, the probability
of these two events are same; but we
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cannot say that these two events are same.
So, this one, this intersection, this is the
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last
part, this probability of A intersection B
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is important to declare, that these event
A and B
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are equal.
.
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Thus, it is stated that the event A and B
are equal with probability 1, if and only
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if their
individual probability is equal to their intersection.
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If only, what we discussed just now;
if only, we say that probability A is equals
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to B, then A and B are equal in probability;
but no conclusion can be drawn about the probability
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of A intersection B. The example
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.that we are telling, that getting a dice
- throwing a dice and getting the even number
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and
odd number. So, these two events are equal
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in probability, but there, the probability
of
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their intersection is 0. So, that is why the
statement states clearly that, if only probability
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of A is equals to probability of B, then A
and B equal in probability; but no conclusion
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can be drawn about the probability of A intersection
B; A and B might be mutually
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exclusive.
.
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Then, a concept is important, which is known
as field. The definition of fields says, that
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a field F is a non-empty set of events, non-empty
subset of events, called a class of event.
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Now, a class of event is again another definition,
where this class means that, we are
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considering …, instead of considering each
and every event of a sample space, we are
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considering only particular subset of the
whole sample space; and that particular subset
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is
generally denoted by the class of event.
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So, this field, what we are now trying to
understand, this field F is a non-empty subset
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of
the event, this is called a class of event,
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in such a way, this F is defined in such a
way
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that, if any event A belongs to F, then its
complementary also belongs to F.
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If one event A belongs to F and another event
B belongs to F, then their union also
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belongs to F, so, these are the two bare minimum
criteria to define one field, which is a
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non-empty subset; so, based on these two,
there are other properties, as well, which
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can
also be drawn. The other properties of this
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field, that states that, if one event belongs
to
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.that field and another event belongs to that
field, then their intersection also will be
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in
that field.
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Also, if the complementary of one event belongs
to F, and the complementary of another
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event belongs to F, then, we can say the union
of their complementary, that is,
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complementary A union complementary B also
belongs to F. And, the complementary of
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the union of individual complementary, that
is, A complementary union B
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complementary (full thing), their complementary,
which is nothing but equal to A
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intersection B, also belongs to F.
Last one, since F a nonempty, sorry for this
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mistake, F is a nonempty set and contains
at
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least one event A. Also, it also contains
that A complementary, that is, it contains
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at least
one event which is denoted as A here; so,
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it also contains that A complementary. Thus,
the A union A complementary which is nothing
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but the full sample space, so, that also
belongs to that field; and, A intersection
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A complementary, which is nothing but a null
event; so, that also belongs to that F. So,
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this S is nothing but almost a certain event;
and this is almost, this is the impossible
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event, that is, a null set; these two are
also, these
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two extremes are also belongs to the field.
.
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Next is countable space. Sometimes, if the
space, S contains N outcomes and N is a finite
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number, then the probabilities of all outcomes
can be expressed in terms of the
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probabilities of the elementary event, probability
of ai equals to pi.
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.So, if there are N finite, N countable or
countable events are there in the space, then
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their
probability can be defined by the individual
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events. For example, here it is shown that
probability of ai
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is equals to p . However, it should follow
the axioms that, the
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probability of each and every event should
be greater than equal to zero and their
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summation should be 1; which is directly following
from the axioms of the probability
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that is discussed in previous classes.
If A is an event having m elementary elements,
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ai ; A can be written as the union of the
elementary events ai then the probability
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of A is nothing but the summation of their
individual probabilities, which is nothing
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but, p1 plus p2 plus up to pm . So, there
are m
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elementary events are there; if we just add
up, we will get the probability of that event
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A.
This is also true; even if the set S comprised
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of an infinite, but countable number of
elements a1 , a2 in this, and so on. So, even
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though I am talking about this countable
space, it is true, when the S comprised of
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an infinite but countable number of elements
in
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such a way, that a1 , a2 in this way, and
so on.
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.
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So, a contrast to this countable space, which
is more important, particularly for these
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applications in this Civil Engineering, where
most of the cases we will see, which is the
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non-countable space. In many cases, we have
seen that the total sample space cannot be
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defined just in terms of a few elementary
events, rather it should be expressed in terms
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of
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.a non-countable set. For example, if we take
the example of the real line, then whatever
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the number that lies on this real line, it
consist of this full sample space.
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Now, for such cases, how to define the probability,
that is, now our, goal to understand.
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So, it is not only for the real line which
is a one dimensional feature, the concept
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can be
extended to any n dimensional space. So, it
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can be the two dimensional, where it refers
to
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the areas, or three dimensional which is referring
to the volume; and in this way, it can
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be extended, the concept can be extended to
any n dimensional space. Now, here we will
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discuss about the one dimensional, that is,
the real line. So, if S is set of all real
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numbers, its subset might be considered as
a set of points on the real line. It is generally
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impossible to assign the probabilities for
all subsets of the S to satisfy the axioms.
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It is
true for any n dimensional space, just now
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what I have discussed.
So, the probability space on the real line
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can be constructed considering all the events
at
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any intervals, where x lies between x1 and
x2 , and x1 and x2 can be of any real number
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on the real line and their countable unions
and intersection. So, in this case,
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particularly… So, what I mean is a real
line, that is the one dimensional case, the
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probability is assigned to the event, x less
than equal to x i , so, xi is any number on
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this
real line. So, if we just define the probability
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of x less than equals to x i , then this is
sufficient to explain the entire set of this
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probability for the entire sample space. The
probability of any other event, all other
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events can be determined with the help of
the
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probability axioms.
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..
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Now, probability masses. Now the probability
P(Ai) of an event Ai, can be interpreted as
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the mass of the corresponding figure of its
Venn diagram here; whatever you can see
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here, that, this is a Venn diagram that is
shown. Now, if these dots are the outcome
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of the
experiment and all these dots consist of the
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sample space, now the probability can be
treated as a concentrated mass to these points,
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to this outcome.
Now here, if I just extend this one to this
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continuous field; suppose that, instead of
being
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an elementary event, if the set consists of
a continuous event, then what will happen;
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we
will just discuss in a minute.
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..
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The second thing is that, for these cases,
where it is an elementary event, if a sample
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space S consist of this finite number of outcomes
a1, a2 up to an, and this A1, A2… An are
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the elementary events, then by that, this
ai corresponds to the Ai, then this probability
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of
all these events should be equal to 1 which
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directly follows from the axiom of
probability.
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Now, what just now I was telling was that,
instead of being this discrete point, if it
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is a
continuous point; this is important in the
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sense that, what we can…; in that case,
what
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we can imagine that, this probability is a
mass and where, in this field, that can be
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expressed in terms of the density. Now, if
I take one elemental area of that particular
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sample space, then the total mass, that is,
the total area multiplied by the density;
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the
total mass will be give you the probability
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for that particular event.
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..
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Now, in the previous class also we have discussed
about the concept of this conditional
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probability. So, here, just to recall the
fact that, if there are two events, one is
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A and
another one is B, and these events are taken
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in such a way that, the probability of A is,
sorry for this mistake, this should be greater
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than 0. So, if we say that the probability
of
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A greater than 0, not 1, this is greater than
0, then, the probability of B given that A
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has
already occurred … so, this is expressed
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in terms of the probability B on condition
A, so,
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this is known as the conditional probability
which differs from the probability of this
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B,
in the sense that, when there is no other
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information available, this is simply the
probability of one event B, so, probability
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of B.
Now, if we say that A has already occurred,
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now, one information is available. So, based
on the available information, the probability
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of the other event may or may not change;
and this is known as the conditional probability,
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which is denoted like this, B on
condition A, which is derived as that probability
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B on condition A which is equal to the
probability A intersection B divided by probability
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of A.
So, here, so, if we see, if we refer to this
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Venn diagram, then what is the probability
of
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B? Then, we will just concentrate on the event
B which is shown by this circle. Now, if
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we say that A has already occurred, then we
know that our sample space, our total
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feasible space is now within this; this zone
which is denoted as the event A. Now, the
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success of this one; so, what is the probability
of B? So, the success area is highlighted
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in
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.these area, this is red which is the intersection
of these two events; that is A intersection
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B. That is why, this is the area, where the
success lies to declare the probability of
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00:20:14,330 --> 00:20:17,210
B.
Now, the total feasible space as we have seen
209
00:20:17,210 --> 00:20:22,179
that, as A has already occurred, so, the
probability of A comes here; so, it says that
210
00:20:22,179 --> 00:20:27,690
conditional probability of B, given that A
has already occurred, is equal to probability
211
00:20:27,690 --> 00:20:30,940
of A intersection B divided by probability
of
212
00:20:30,940 --> 00:20:39,759
A. We will use this relationship to form our
Bayes’ rule which is very important, so
213
00:20:39,759 --> 00:20:43,019
far
as the application is concerned, we will refer
214
00:20:43,019 --> 00:20:50,599
to this one. Before that, we will try to
understand how this probability of particular
215
00:20:50,599 --> 00:20:54,380
event can be derived in terms of the other
events.
216
00:20:54,380 --> 00:20:55,380
.
217
00:20:55,380 --> 00:21:03,679
However, still, before that discussion, we
will discuss about the conditional probability,
218
00:21:03,679 --> 00:21:10,940
when we are talking about more than two events.
So, for any three events, that is A1, A2
219
00:21:10,940 --> 00:21:18,110
and A3, the probability that all of them occurred
is the same as the probability of A1
220
00:21:18,110 --> 00:21:26,159
times probability of A2, given A1 has occurred,
times; probability of A 3 given that both
221
00:21:26,159 --> 00:21:33,399
A1 and A2 has occurred.
So, if we know that there are three events
222
00:21:33,399 --> 00:21:38,509
A1, A2, A3; if we say that - what is the
probability of the simultaneous occurrence
223
00:21:38,509 --> 00:21:44,629
of all these three events; this can be expressed
in terms of probability of A1 multiplied by
224
00:21:44,629 --> 00:21:52,019
probability of A2 on condition A1, probability
of A3 on condition A1 and A2 both has occurred.
225
00:21:52,019 --> 00:21:55,159
This is just follows from this two event
226
00:21:55,159 --> 00:22:00,279
.case, from this conditional probability case,
that is, probability of A on condition B is
227
00:22:00,279 --> 00:22:06,500
nothing, but equal to probability of A multiplied
by probability of B on condition A.
228
00:22:06,500 --> 00:22:11,260
Now, extending the same thing, we are getting
here, for these three events, that is
229
00:22:11,260 --> 00:22:17,059
probability of A1, A2, A3. First we take the
first event, that is probability of A1 multiplied
230
00:22:17,059 --> 00:22:23,240
by probability of A2 on condition A1, probability
of A3 on condition A1 and A2, both has
231
00:22:23,240 --> 00:22:26,769
occurred.
So, extending this same thing, this can be
232
00:22:26,769 --> 00:22:33,880
generalized for this n numbers of events also;
so, probability of A1 intersection A2 intersection
233
00:22:33,880 --> 00:22:40,539
A3 intersection A4 up to, if we go ahead
like this, then we will say that probability
234
00:22:40,539 --> 00:22:45,850
of A1 multiplied by probability of A2 on
condition A1, probability of A3 on condition
235
00:22:45,850 --> 00:22:52,169
A1 and A2 multiplied by probability of A4
on condition A1 A2 A3, all three has occurred;
236
00:22:52,169 --> 00:22:57,019
and this will go on in the same way, as it
is
237
00:22:57,019 --> 00:23:06,690
going on for this n numbers of different events.
.
238
00:23:06,690 --> 00:23:13,559
Next is the concept of the Total probability.
Sometimes, the occurrence of one event A
239
00:23:13,559 --> 00:23:19,349
cannot be determined directly. It depends
on the occurrence of the other events such
240
00:23:19,349 --> 00:23:24,169
as
say B1, B2 upto Bn which are mutually exclusive
241
00:23:24,169 --> 00:23:30,749
and collectively exhaustive.
So, now, I just spoke these condition here,
242
00:23:30,749 --> 00:23:35,119
which are mutually exclusive and collectively
exhaustive; which is generally leading to
243
00:23:35,119 --> 00:23:40,909
the Theorem of Total probability. So, even
if I
244
00:23:40,909 --> 00:23:45,690
do not know the probability of a particular
event, but from the experience if you know
245
00:23:45,690 --> 00:23:49,980
.the probability of the other events, then
this probability can be expressed in terms
246
00:23:49,980 --> 00:23:53,159
of this
one. To calculate that probability of this
247
00:23:53,159 --> 00:23:56,730
event A, the weightage of the probabilities
of
248
00:23:56,730 --> 00:24:07,409
the events B1, B2 and Bn are generally used;
and this approach is known as the Theorem
249
00:24:07,409 --> 00:24:09,059
of Total probability.
.
250
00:24:09,059 --> 00:24:21,309
So, this Theorem of Total Probability says,
that if any event A must result in one of
251
00:24:21,309 --> 00:24:25,059
the
mutually exclusive and collectively exhaustive
252
00:24:25,059 --> 00:24:30,559
events A1 to An; I take a minute to
explain once again; the mutually exclusive
253
00:24:30,559 --> 00:24:33,960
and collectively exhaustive - this means,
the
254
00:24:33,960 --> 00:24:39,490
occurrence of one event, for example, this
A1 to An, what is shown here, the occurrence
255
00:24:39,490 --> 00:24:46,409
of any one event implies the non-occurrence
of all other events. This is meant by the
256
00:24:46,409 --> 00:24:51,210
mutually exclusive.
And, collectively exhaustive means, the probability
257
00:24:51,210 --> 00:24:55,320
of A1 plus probability of A2 plus, up
to in this way, probability of An should be
258
00:24:55,320 --> 00:25:03,289
equal to 1. So, if we see here, if this full
rectangle is your sample space, then these
259
00:25:03,289 --> 00:25:10,640
are the events which are non-overlapping to
each other, this A1, A2, A3, A4, up to An.
260
00:25:10,640 --> 00:25:17,599
So, these events are known as mutually
exclusive and collectively exhaustive. So,
261
00:25:17,599 --> 00:25:21,509
if these are the events, then the probability
of
262
00:25:21,509 --> 00:25:31,269
another event A which is overlapping with
all these events can be expressed as,
263
00:25:31,269 --> 00:25:39,090
probability of A is equals to the probability
of A1 multiplied by probability A on
264
00:25:39,090 --> 00:25:47,970
condition A1 plus probability of A2 on condition
probability of A on condition A2, in this
265
00:25:47,970 --> 00:25:53,139
.way it will go up to probability of An multiplied
by probability of A on condition An; so,
266
00:25:53,139 --> 00:25:59,289
this theorem is known as the Theorem of Total
probability.
267
00:25:59,289 --> 00:26:07,039
So, the events, this probability of A1, probability
of A2, probability of An generally are
268
00:26:07,039 --> 00:26:13,619
known from the experience, and probability
of A1 on condition A on condition A1,
269
00:26:13,619 --> 00:26:19,529
probability A on condition A2 are also known
from the previous experience. When both
270
00:26:19,529 --> 00:26:24,619
these information is known to us, then, if
you want to know what is the total probability
271
00:26:24,619 --> 00:26:31,070
of the event A, then these probabilities,
these conditional probabilities are, are weighted
272
00:26:31,070 --> 00:26:41,200
to the individual probability of the individual
events A1, A2 up to An. This is the, basis
273
00:26:41,200 --> 00:26:43,839
of
this total probability theorem.
274
00:26:43,839 --> 00:26:44,839
.
275
00:26:44,839 --> 00:26:50,830
Now, if we see one example problem using this
total probability problem, this is a water
276
00:26:50,830 --> 00:26:56,470
supply problem; Municipality of a city uses
70 percent of its required water from a
277
00:26:56,470 --> 00:27:02,059
nearby river and remaining from the ground
water, that is, 30 percent is used from this
278
00:27:02,059 --> 00:27:10,919
groundwater. Now, there could be various reasons
for not getting sufficient water from
279
00:27:10,919 --> 00:27:16,719
the sources including the pump failure, non-availability
of the sufficient water and so on.
280
00:27:16,719 --> 00:27:24,580
So, the failures can be of, I can, I am just
dividing the failure of not supplying sufficient
281
00:27:24,580 --> 00:27:31,539
water into 2 parts.
One is related to the supply from the river,
282
00:27:31,539 --> 00:27:33,510
another one is related to the supply from
the
283
00:27:33,510 --> 00:27:39,360
groundwater. So, if the probability of the
shortage of water due to the system involved
284
00:27:39,360 --> 00:27:46,000
.with the river is 0.3, and that with the
ground water is 0.15, what is the probability
285
00:27:46,000 --> 00:27:50,099
of
insufficient supply of this water to the city?
286
00:27:50,099 --> 00:27:56,210
Now, here in this problem, we can see, that
this probability of insufficient supply of
287
00:27:56,210 --> 00:27:59,279
the
water is my total probability that I am looking
288
00:27:59,279 --> 00:28:06,519
for, which is depending on this; here it is
2, and that can depend on many factors.
289
00:28:06,519 --> 00:28:07,519
.
290
00:28:07,519 --> 00:28:14,380
So, now, if I just define the events like
this, that A event is the insufficient supply
291
00:28:14,380 --> 00:28:17,639
of the
water to the city and R is the water from
292
00:28:17,639 --> 00:28:24,139
the river and G is the water from the
groundwater. Now, from the problem, we have
293
00:28:24,139 --> 00:28:30,049
seen the probability of the water that we
get from the river is 0.7, from the groundwater
294
00:28:30,049 --> 00:28:32,899
is 0.3. Now, probability of the
295
00:28:32,899 --> 00:28:36,600
insufficient supply in the case of the river,
it is 0.3; and probability of insufficient
296
00:28:36,600 --> 00:28:42,600
water
in case of the ground water, it is 0.15 so,
297
00:28:42,600 --> 00:28:46,279
from the total probability theorem, that is
probability of the insufficient supply of
298
00:28:46,279 --> 00:28:50,179
the water to the city, this equals to probability
of
299
00:28:50,179 --> 00:28:57,149
R weighted to the probability of insufficient
supply in case of river, and similarly, for
300
00:28:57,149 --> 00:29:02,990
this groundwater so, which is equals to this
calculation which comes, that the total
301
00:29:02,990 --> 00:29:11,340
probability of insufficient supply of the
water to the city is 25.5 percent.
302
00:29:11,340 --> 00:29:12,340
..
303
00:29:12,340 --> 00:29:20,940
Now, another important concept is known as
independence which is very relevant here to
304
00:29:20,940 --> 00:29:28,519
discuss is that, independent event, if the
probability of a event B occurring completely
305
00:29:28,519 --> 00:29:35,820
unaffected by the occurrence or non-occurrence
of the event A, then event A and B are
306
00:29:35,820 --> 00:29:42,070
said to be independent. Thus, if A and B are
independent, then it can be expressed as
307
00:29:42,070 --> 00:29:46,969
probability of B on condition A is equals
to probability of B.
308
00:29:46,969 --> 00:29:53,379
So, whether A has occurred or not, it does
not have anything to change the probability
309
00:29:53,379 --> 00:29:57,581
of
B, so, which is equivalent as the probability
310
00:29:57,581 --> 00:30:05,210
of A intersection B is equals to probability
of A multiplied by probability of B.
311
00:30:05,210 --> 00:30:06,210
..
312
00:30:06,210 --> 00:30:11,539
So, which is directly following from the conditional
probability which is, just we have
313
00:30:11,539 --> 00:30:22,830
seen, the probability of B on condition A
is equals to probability of A intersection
314
00:30:22,830 --> 00:30:30,360
B,
probability of A. Now, this is probability
315
00:30:30,360 --> 00:30:35,380
of B on condition A; if these two are
independent, then, this is nothing, but equal
316
00:30:35,380 --> 00:30:50,830
to probability of B. So, A intersection B
divided by probability of A which … So,
317
00:30:50,830 --> 00:31:03,309
this is generally the basis to declare in
mathematically that two variables, two random
318
00:31:03,309 --> 00:31:08,649
variables are independent which will be
used from the next class onwards.
319
00:31:08,649 --> 00:31:09,649
.
320
00:31:09,649 --> 00:31:14,570
.So, this is forming the mathematical basis
to declare two events to be independent; it
321
00:31:14,570 --> 00:31:19,209
is
said that, if and only if, the probability,
322
00:31:19,209 --> 00:31:25,919
their joint probability is equals to the
multiplication of their individual probability,
323
00:31:25,919 --> 00:31:30,319
then we say that these two events are
independent.
324
00:31:30,319 --> 00:31:31,319
.
325
00:31:31,319 --> 00:31:41,950
Bayes’ theorem or rule. If A1, A2, A3, …, An
are mutually exclusive events and
326
00:31:41,950 --> 00:31:46,049
collectively exhaustive; just now, we have
discussed about this mutually exclusive and
327
00:31:46,049 --> 00:31:56,330
collectively exhaustive; then for any event
Ak, k can range from 1 to n, what we can say
328
00:31:56,330 --> 00:32:03,149
that this probability of Ak on condition A
is equals to probability of Ak multiplied
329
00:32:03,149 --> 00:32:06,600
by
probability of A on condition Ak divided by
330
00:32:06,600 --> 00:32:13,679
summation of all these probabilities, which
probability of Aj multiplied by probability
331
00:32:13,679 --> 00:32:21,190
of A on condition Aj. This is known as Bayes’
theorem which we will just see that, this
332
00:32:21,190 --> 00:32:23,280
comes from this total probability theorem;
that
333
00:32:23,280 --> 00:32:31,049
is, if we know the probability of individual
event and after that we know the occurrence
334
00:32:31,049 --> 00:32:37,279
of one particular event A, which is shown
as a red ellipse here, then, what are the
335
00:32:37,279 --> 00:32:43,440
probability of these different sub-events
which are mutually exclusive and collectively
336
00:32:43,440 --> 00:32:50,129
exhaustive, is generally obtain from this
Bayes’ theorem or Bayes’ rule.
337
00:32:50,129 --> 00:32:51,129
..
338
00:32:51,129 --> 00:32:56,479
The proof of Bayes’ theorem can be explained
like this, that is, if I take any particular
339
00:32:56,479 --> 00:33:04,709
event Ak, k can be from 1 to n; So, Ak, a
particular event on condition that A has
340
00:33:04,709 --> 00:33:09,090
occurred can be expressed in terms of, just
now we have seen the conditional probability,
341
00:33:09,090 --> 00:33:15,429
so, this A intersection Ak divided by probability
of A. Now, again from this Joint
342
00:33:15,429 --> 00:33:20,570
probability, this probability A intersection
Ak can be expressed as probability of A on
343
00:33:20,570 --> 00:33:25,240
condition Ak multiplied by probability of
Ak.
344
00:33:25,240 --> 00:33:32,909
Now, from here if we just replace this one
to this part, then we see that probability
345
00:33:32,909 --> 00:33:38,239
of A k
on condition A is equals to probability of
346
00:33:38,239 --> 00:33:47,539
Ak multiplied by probability of A on condition
Ak divided by probability of A. Now, this
347
00:33:47,539 --> 00:33:50,179
probability of A can be expressed in terms
of
348
00:33:50,179 --> 00:33:55,070
this total probability theorem. Then, from
this total probability theorem, this probability
349
00:33:55,070 --> 00:34:02,250
of A can be expressed as this; for all this
k, it should be the summation of the probability
350
00:34:02,250 --> 00:34:09,780
of A k multiplied by probability of A on condition
probability of A k. Now, if we replace
351
00:34:09,780 --> 00:34:16,169
this one here, so, probability of a particular
event on condition A is equals to probability
352
00:34:16,169 --> 00:34:23,880
of Ak multiplied by probability of A on condition
Ak divided by summation of for all k
353
00:34:23,880 --> 00:34:28,240
probability of Ak multiplied by probability
of A on condition A k.
354
00:34:28,240 --> 00:34:35,399
Now, these probabilities, these individual
probabilities is generally having different
355
00:34:35,399 --> 00:34:42,500
meaning. First of all, this probability of
Ak; so, what we are doing is that, if we say
356
00:34:42,500 --> 00:34:44,899
that,
this left hand side is unknown and right hand
357
00:34:44,899 --> 00:34:50,929
side is known, then what we are trying to
do is that, we know the probability of Ak
358
00:34:50,929 --> 00:34:53,819
without any condition; what we are trying
to
359
00:34:53,819 --> 00:34:59,829
.understand, what we are trying to get is,
the probability of the same event here is
360
00:34:59,829 --> 00:35:03,140
A k,
here is also Ak; but here, the condition is
361
00:35:03,140 --> 00:35:09,240
that occurrence of the particular event is
known. So, this one, this probability of this
362
00:35:09,240 --> 00:35:14,549
individual events, which are mutually
exclusive and collectively exhaustive, these
363
00:35:14,549 --> 00:35:19,880
events are my prior knowledge.
Now, from this prior knowledge, due to the
364
00:35:19,880 --> 00:35:22,069
occurrence of the particular event, I want
to
365
00:35:22,069 --> 00:35:28,819
update the knowledge of the probability of
Ak. So, this is known as the posterior, so,
366
00:35:28,819 --> 00:35:35,390
probability of Ak is your prior and probability
of Ak on condition A is your posterior.
367
00:35:35,390 --> 00:35:43,480
Now, this probability of A on condition Ak,
that is, if we have the data available to
368
00:35:43,480 --> 00:35:46,200
us,
then these probabilities can be calculated
369
00:35:46,200 --> 00:35:55,420
from the previous experience and the previous
experiments. So, this is known as the likelihood
370
00:35:55,420 --> 00:36:03,029
of that particular event; so, this is known
as the likelihood. Now, this denominator part
371
00:36:03,029 --> 00:36:08,890
is coming from this Total probability
theorem, which is equals to the probability
372
00:36:08,890 --> 00:36:19,180
of A. So, as compared to these probabilities,
this can be treated as a constant. So, as
373
00:36:19,180 --> 00:36:21,619
this can be treated as a constant, if we take
this
374
00:36:21,619 --> 00:36:26,589
out, then this equality sign will convert
to proportional sign.
375
00:36:26,589 --> 00:36:27,589
.
376
00:36:27,589 --> 00:36:35,490
So, this, if we just want to discuss it, then
in the equation of this Bayes’ theorem,
377
00:36:35,490 --> 00:36:38,130
we
have seen that this probability of different
378
00:36:38,130 --> 00:36:43,960
events, that is probability of Ak are generally
the belief of the engineer from the previous
379
00:36:43,960 --> 00:36:46,520
experience. These probabilities are known
as
380
00:36:46,520 --> 00:36:54,030
prior. Whereas the probability of that particular
event on the condition that one event,
381
00:36:54,030 --> 00:36:59,000
one particular event has already occurred,
this is known as the posterior.
382
00:36:59,000 --> 00:37:05,440
.Similarly, the probabilities of this A on
condition Ak are known as a likelihood, which
383
00:37:05,440 --> 00:37:11,410
are obtained from the earlier experiments.
The denominator, there would be one space
384
00:37:11,410 --> 00:37:16,980
here; the denominator, that is the total probability
can be treated as a constant. Thus, the
385
00:37:16,980 --> 00:37:22,180
Bayes’ rule is also expressed as left hand
side that is your posterior, which is
386
00:37:22,180 --> 00:37:29,180
proportional to the prior multiplied by the
likelihood.
387
00:37:29,180 --> 00:37:30,180
.
388
00:37:30,180 --> 00:37:39,360
So with this, we will take one particular
example where a particular construction material
389
00:37:39,360 --> 00:37:49,890
is ordered from 3 different companies. So,
the company A delivers 600 units per day,
390
00:37:49,890 --> 00:37:52,940
out
of which, 3 percent do not satisfy the specific
391
00:37:52,940 --> 00:38:04,920
quality. Company B delivers 400 units and
out of which, 1 percent do not satisfy the
392
00:38:04,920 --> 00:38:14,019
specific quality. Now, the company C delivers
500 units per day, so, out of which 2 percent
393
00:38:14,019 --> 00:38:20,369
do not satisfy the specific quality. So, the
total units per day being supplied is, 1500
394
00:38:20,369 --> 00:38:24,240
units are being supplied by 3 different
companies.
395
00:38:24,240 --> 00:38:30,349
Now, we want to know at the construction site
that, what is the probability that 1 unit
396
00:38:30,349 --> 00:38:33,260
of
the material picked at random, this picked
397
00:38:33,260 --> 00:38:38,210
at random is important, so, I do not know;
without knowing, the knowledge that which
398
00:38:38,210 --> 00:38:43,369
company has supplied this one, if you do not
know that information, that is why it is picked
399
00:38:43,369 --> 00:38:52,430
at random, will not satisfy the specific
quality. So, to address this problem we have
400
00:38:52,430 --> 00:39:00,230
to use the theorem of Total probability; the
total probability of getting one particular
401
00:39:00,230 --> 00:39:03,099
unit which is defective or not satisfying
the
402
00:39:03,099 --> 00:39:04,099
specific quality.
403
00:39:04,099 --> 00:39:12,630
.Now, on the other hand, if a particular unit
is found to be substandard, then what is the
404
00:39:12,630 --> 00:39:21,660
probability that it has come from supplier
B? Now, here we are giving one condition that
405
00:39:21,660 --> 00:39:28,160
one unit is found to be substandard; that
is already fact. Now depending on that fact,
406
00:39:28,160 --> 00:39:33,740
depending on that event, what is the probability
that it is being supplied by B? So, we
407
00:39:33,740 --> 00:39:38,430
will see this two problem. The first one will
be solved by this Total probability theorem
408
00:39:38,430 --> 00:39:42,180
and the second one will be solved by this
Bayes’ rule.
409
00:39:42,180 --> 00:39:43,180
.
410
00:39:43,180 --> 00:39:50,859
So, to answer the first one, the substandard
unit may come either from the company A or
411
00:39:50,859 --> 00:39:58,089
B or C. Thus, the theorem of Total probability
should be applied to obtain the probability
412
00:39:58,089 --> 00:40:07,310
of the event E, that is selecting a substandard
unit at random. So, this E is denoted as the
413
00:40:07,310 --> 00:40:13,900
probability of the event, that is selecting
a substandard unit. So, probability of E which
414
00:40:13,900 --> 00:40:18,740
should be expressed in terms of that probability
of A, that is, it is supplied by …, what
415
00:40:18,740 --> 00:40:22,011
is
the probability of supplying A, probability
416
00:40:22,011 --> 00:40:30,390
of supplying B and probability of supplying
C and what are their chances of supplying
417
00:40:30,390 --> 00:40:41,110
the defective units.
So, this first one is 0.03 and their probability
418
00:40:41,110 --> 00:40:48,450
of supplying by event A is 600 by total
units being supplied in a day is 1500. And,
419
00:40:48,450 --> 00:40:54,730
the second one is 0.01 and the probability
it is
420
00:40:54,730 --> 00:41:02,190
being supplied by company B is 400 by 1500,
and the probability that it is being supplied
421
00:41:02,190 --> 00:41:09,049
by C, company C is 500 by 1500. So, if we
do this calculation, it comes that the total
422
00:41:09,049 --> 00:41:18,119
probability of getting a substandard unit
is 0.0213. So, this is the total probability
423
00:41:18,119 --> 00:41:19,119
that we
424
00:41:19,119 --> 00:41:24,359
.get. Now, if we just see the second question,
if the unit is defective, then what is the
425
00:41:24,359 --> 00:41:28,109
probability that it has come from company
B?
426
00:41:28,109 --> 00:41:29,109
.
427
00:41:29,109 --> 00:41:33,710
So, once it is known that the unit is substandard,
the probability of the unit being
428
00:41:33,710 --> 00:41:40,119
supplied by a particular company is not the
same as that when the information of the
429
00:41:40,119 --> 00:41:47,730
substandard unit was not known. So, what is
meant here is that, if this information was
430
00:41:47,730 --> 00:41:54,390
not available, then probability of supplying
the substandard unit by with the company B
431
00:41:54,390 --> 00:42:02,279
is known to us, which is nothing but your
0.01 as supplied in this, from the earlier
432
00:42:02,279 --> 00:42:06,380
experiences. Company B delivers this one,
out of which 1 percent do not satisfy the
433
00:42:06,380 --> 00:42:10,700
specific quality.
So, now we have to update that information,
434
00:42:10,700 --> 00:42:15,600
that is the probability of B on condition
that
435
00:42:15,600 --> 00:42:21,359
one substandard unit has come. So, this is
equals to probability of E on condition that
436
00:42:21,359 --> 00:42:26,820
probability of E on condition, it is supplied
by B multiplied by the probability of B, this
437
00:42:26,820 --> 00:42:32,180
one; which is now this A is again expressed
in terms of this total probability which is
438
00:42:32,180 --> 00:42:37,040
expressed in this form - Probability of E
on condition A multiplied by a probability
439
00:42:37,040 --> 00:42:40,460
A,
similarly from probability B, similarly from
440
00:42:40,460 --> 00:42:45,619
company C.
Now, if we just put this forms, then it comes
441
00:42:45,619 --> 00:42:52,310
that probability B is 400 by 1500. Similarly,
the total probability as we have seen in this
442
00:42:52,310 --> 00:43:03,820
last slide that this quantity comes to 0.0213
and this is 0.0027, a ratio gives that 0.215.
443
00:43:03,820 --> 00:43:04,820
..
444
00:43:04,820 --> 00:43:10,549
So, the first one, we have solved from this
Total probability; and second one, we have
445
00:43:10,549 --> 00:43:16,760
solved from this Bayes’ rule. We will take
another interesting problem which is taken
446
00:43:16,760 --> 00:43:25,019
from Kottegoda and Rosso, 2008, from that
book. where this is a basically a geotechnical
447
00:43:25,019 --> 00:43:32,420
problem, where the the design of foundation
for the tall structure, needs to know the
448
00:43:32,420 --> 00:43:37,769
depth of the soil above the bedrock which
is denoted as h.
449
00:43:37,769 --> 00:43:48,660
Now, four categories of h are denoted as B1;
so, there are four different categories of
450
00:43:48,660 --> 00:43:52,490
this
depth. The first one which is less than 5
451
00:43:52,490 --> 00:44:01,619
meter and second one B2 is 5 meter to 10 meter;
third one is 10 meter to 15 meter and B4 is
452
00:44:01,619 --> 00:44:08,019
greater than 15 meter. So, belief of the
geologist states that, the prior probabilities
453
00:44:08,019 --> 00:44:14,820
of these 4 events are as follows: the
probability of B1, that is the bedrock should
454
00:44:14,820 --> 00:44:20,920
be within the 5 meter depth, is equals to
60
455
00:44:20,920 --> 00:44:27,289
percent; probability of B2 is equals to 0.2;
probability of B3 is equals to 0.15 and
456
00:44:27,289 --> 00:44:33,920
probability of B4 is equals to 0.05.
Now, a seismic recorder is being used to measure
457
00:44:33,920 --> 00:44:41,010
this h. Now, obviously any instrument
that we will use; this type of measurement
458
00:44:41,010 --> 00:44:47,359
is generally not always perfect, this is also
having some certain percentage of error. So,
459
00:44:47,359 --> 00:44:54,430
this is coming from this earlier experiment,
where both the true depth, as well as the
460
00:44:54,430 --> 00:45:05,000
record both, are available to us. The
performance of the instrument is shown in
461
00:45:05,000 --> 00:45:09,089
the table in the next slide which is here.
462
00:45:09,089 --> 00:45:10,089
..
463
00:45:10,089 --> 00:45:16,779
So, this is the measured state which is Bi
and these are the true state which is Bj.
464
00:45:16,779 --> 00:45:20,420
So, if
the measured states generally say that, it
465
00:45:20,420 --> 00:45:26,430
is within the 5 meter, so, there is 90 percent
chance that it is actually in the first state;
466
00:45:26,430 --> 00:45:28,970
there is still 5 percent chance it is in the
second
467
00:45:28,970 --> 00:45:36,260
state, 3 percent chance it is in the third
state and 2 percent chance - this one. So
468
00:45:36,260 --> 00:45:39,019
similarly,
in this way for all these cases, if it is
469
00:45:39,019 --> 00:45:43,829
measured to this particular fact and if the
true data
470
00:45:43,829 --> 00:45:49,200
is also available, then we can complete this
particular table here. See this is quite…
471
00:45:49,200 --> 00:45:54,549
There are two important thing that should
be observed here; one is that, if the instrument
472
00:45:54,549 --> 00:46:03,440
is perfect, then we can say that this one,
if it is measured in the state one, then this
473
00:46:03,440 --> 00:46:06,200
should
be 100 percent probability and all other probability
474
00:46:06,200 --> 00:46:10,329
should be 0.
If it is two, this is also the true state
475
00:46:10,329 --> 00:46:13,130
will be two, so, this should be a perfect
one, that is,
476
00:46:13,130 --> 00:46:21,259
100 percent probability and other should be
0. But, as this instrument is not perfect,
477
00:46:21,259 --> 00:46:23,250
that
is why we are getting this distribution of
478
00:46:23,250 --> 00:46:29,440
these probabilities from the earlier experiments;
and obviously, this diagonal is heavy diagonal,
479
00:46:29,440 --> 00:46:38,040
that means, most of the time it generally
measures the true fact. Another thing, that
480
00:46:38,040 --> 00:46:44,450
is, if it is measured in state one or a particular
state, and there are the probabilities where
481
00:46:44,450 --> 00:46:50,220
this state will be; so, it should be exhaustive.
So, the state one whatever the probability
482
00:46:50,220 --> 00:46:57,880
is, if we just add up the probabilities in
a rowwise fashion or in a column-wise fashion,
483
00:46:57,880 --> 00:47:06,039
this should be equal to 1. So, this is denoting
that this is collectively exhaustive.
484
00:47:06,039 --> 00:47:11,660
.Now, the readings are obtained at a site,
so, now, the same instrument, once we know
485
00:47:11,660 --> 00:47:14,339
the
performance of this instrument, the same instrument
486
00:47:14,339 --> 00:47:22,020
is being used to know the depth of
the bedrock at a particular site. Now, the
487
00:47:22,020 --> 00:47:26,640
readings are obtained at a site and found
to be 7
488
00:47:26,640 --> 00:47:33,289
meter for the first reading which is denoted
as sample 1 and 8 meter for the second
489
00:47:33,289 --> 00:47:40,380
reading. Now, we have to calculate the probability
of the different event, that is B1, B2,
490
00:47:40,380 --> 00:47:44,529
B3 given the record obtained from the successive
readings.
491
00:47:44,529 --> 00:47:45,529
.
492
00:47:45,529 --> 00:47:52,269
So, if no reading is taken, then the probabilities
are listed here. So, probability of it being
493
00:47:52,269 --> 00:47:59,079
in the B1 state, that is, below 5 meter is
0.6, 60 percent, 20 percent, 15 percent, 5
494
00:47:59,079 --> 00:48:05,670
percent. Now, I got one sample which is 7
meter depth from this instrument which is
495
00:48:05,670 --> 00:48:11,980
obviously not perfect, erroneous instrument.
So, after getting that sample#1 from that
496
00:48:11,980 --> 00:48:17,570
instrument, what are the probability, how
these probabilities are being updated?
497
00:48:17,570 --> 00:48:18,570
..
498
00:48:18,570 --> 00:48:24,289
So, to know this fact, once we know this,
after this sample one, then another sample
499
00:48:24,289 --> 00:48:27,430
has
been taken which is sample#2. So, these probabilities
500
00:48:27,430 --> 00:48:32,630
will be updated after sample#1.
Again it will be, again updated after taking
501
00:48:32,630 --> 00:48:37,970
this sample#2. How these probabilities are
changing over these successive probabilities;
502
00:48:37,970 --> 00:48:42,349
that, we will see now in this problem.
.
503
00:48:42,349 --> 00:48:49,730
So, the sample#1 was found to be 7 meter which
corresponds to the B2. Now, the
504
00:48:49,730 --> 00:48:54,670
posterior probabilities of the actual states
are obtained from Bayes’ theorem, that is,
505
00:48:54,670 --> 00:48:57,279
what
is the probability of a particular state Bk?
506
00:48:57,279 --> 00:49:01,240
Now, here Bk means that B1, B2, B3, and B4
507
00:49:01,240 --> 00:49:08,400
.on condition that sample#1 is in B2. So,
this is ... directly from the Bayes’ rule,
508
00:49:08,400 --> 00:49:12,910
we can
state that, this is the probability of sample#1
509
00:49:12,910 --> 00:49:20,180
belonging to B2 on condition Bk multiplied
by Bk; and the total probability is the probability
510
00:49:20,180 --> 00:49:24,390
of sample#1 belongings to B2 on
condition Bk probability of Bk.
511
00:49:24,390 --> 00:49:33,180
Now, this total probability is calculated
from this one, that is if I just take that
512
00:49:33,180 --> 00:49:37,099
sample is
two, what is on condition that it is in actually
513
00:49:37,099 --> 00:49:45,049
in one and multiplied the probability one,
which is 0.07 multiplied by 0.6. Now, if you
514
00:49:45,049 --> 00:49:55,359
see this slide, this is your 0.07; it is in
measure state two. So, this is in 0.07 and
515
00:49:55,359 --> 00:49:57,880
the probability of the prior knowledge of
B1 is
516
00:49:57,880 --> 00:50:04,400
0.6; and, that is why this 0.07 and 0.6 are
multiplied. Similarly, taking the other
517
00:50:04,400 --> 00:50:09,730
probabilities from that table and their prior
probability 0.2, in this way, if you just
518
00:50:09,730 --> 00:50:15,049
add
up, we get the total probability 0.236. Now,
519
00:50:15,049 --> 00:50:21,630
putting this one, so, we will get the updated
probabilities for difference state which is
520
00:50:21,630 --> 00:50:22,749
as follows:
.
521
00:50:22,749 --> 00:50:30,819
So, probability that, it is the state one
is equals to 0.07 multiplied by 0.6 is divided
522
00:50:30,819 --> 00:50:35,130
by the
total probability 0.236. Now, this 0.07 comes
523
00:50:35,130 --> 00:50:41,450
from that table, just now I showed from
here, and this 0.6 comes from this prior knowledge.
524
00:50:41,450 --> 00:50:46,730
So, this is now after the sample#1,
the prior information was 0.6; it is updated
525
00:50:46,730 --> 00:50:52,150
to 0.178. So, the probability of the depth
of
526
00:50:52,150 --> 00:51:00,859
this strata in one is now reduced from 0.6
to 0.175.
527
00:51:00,859 --> 00:51:07,170
Similarly, the state which is in the B2 is
in the 0.2, which is going to be increased
528
00:51:07,170 --> 00:51:10,299
from
this 20 percent to 74.6 percent; similarly,
529
00:51:10,299 --> 00:51:13,730
other probabilities also. Now, after taking
the
530
00:51:13,730 --> 00:51:19,160
.sample#1, these probabilities are modified,
and this probability of it being in the second
531
00:51:19,160 --> 00:51:26,390
state has increase to almost 75 percent.
But, still there is 25 percent chance that,
532
00:51:26,390 --> 00:51:36,640
this depth of this bedrock may not be in these
strata two. So, another observation is collected,
533
00:51:36,640 --> 00:51:43,240
where it is again saying that it is the
depth is 8 meter, that is, it is again in
534
00:51:43,240 --> 00:51:46,970
that state two, that is the second stage.
So, after
535
00:51:46,970 --> 00:51:52,950
getting the second information again this
probabilities are being updated.
536
00:51:52,950 --> 00:51:53,950
.
537
00:51:53,950 --> 00:51:54,950
.
538
00:51:54,950 --> 00:51:59,940
.So, now what is the probability of Bk on
condition the sample#1 and #2, both are in
539
00:51:59,940 --> 00:52:03,160
B2.
So, this can be expressed in terms of this
540
00:52:03,160 --> 00:52:10,069
Bayes’ theorem and here the total probability,
when we are calculating; we are using that
541
00:52:10,069 --> 00:52:12,009
the performance of the …., this is from
that
542
00:52:12,009 --> 00:52:17,019
table, performance of the instrument, and
this is from the updated information after
543
00:52:17,019 --> 00:52:25,460
getting sample#1. So, if we add up this thing
it comes to be 0.675. Now, using that 0.675
544
00:52:25,460 --> 00:52:31,269
we are getting different probability for the
difference state. So, from 175 it is further
545
00:52:31,269 --> 00:52:36,671
reduced to that 1 percent, 1.8 percent, and
the probability that it is in state 2, it
546
00:52:36,671 --> 00:52:40,130
is
increased to 97 percent and similarity for
547
00:52:40,130 --> 00:52:44,310
the other states.
So, thus, it is noticed that after obtaining
548
00:52:44,310 --> 00:52:47,690
sample#2, the chance of true state being B2
is
549
00:52:47,690 --> 00:52:53,910
very high, which is 97.2 percent. Thus, with
the help of the Bayes’ theorem, the
550
00:52:53,910 --> 00:52:59,669
probability of unknown are improved with the
availability of more information.
551
00:52:59,669 --> 00:53:00,669
.
552
00:53:00,669 --> 00:53:05,299
So, in this class the concept of probability
events is discussed through different theorems
553
00:53:05,299 --> 00:53:13,440
and problems. The theorem of conditional probability
is defined, the probability of one
554
00:53:13,440 --> 00:53:17,940
event based on the occurrence of the other
events. Concept of total probability theorem
555
00:53:17,940 --> 00:53:23,900
and Bayes’ theorem is useful for revising
or updating with the availability of more
556
00:53:23,900 --> 00:53:30,970
information which is seen in the last example
and in the next class, we will cover the
557
00:53:30,970 --> 00:53:36,170
concept of random variable and this we will
see in the next class.
558
00:53:36,170 --> 00:53:37,170
Thank you.
559
00:53:37,170 --> 00:53:37,170
.