1
00:00:18,130 --> 00:00:19,130
.
2
00:00:19,130 --> 00:00:25,140
Hello there and welcome to the course Probability
methods in Civil Engineering.
3
00:00:25,140 --> 00:00:29,039
In
today’s class we are going through this
4
00:00:29,039 --> 00:00:33,070
module 2, which is Random Events and today
is
5
00:00:33,070 --> 00:00:39,240
the third lecture in which we will cover the
part called Axioms of Probability.
6
00:00:39,240 --> 00:00:47,420
So, these axioms basically are the fundamentals
to the Probability theory and anything,
7
00:00:47,420 --> 00:00:54,940
any conclusion, that we generally draw in
the Theory of Probability, these are generally,
8
00:00:54,940 --> 00:01:02,000
directly or indirectly, somehow it is based
on these axioms of probability.
9
00:01:02,000 --> 00:01:05,909
So, to know
the Theory of Probability, understanding these
10
00:01:05,909 --> 00:01:10,100
axioms is very important.
11
00:01:10,100 --> 00:01:11,100
..
12
00:01:11,100 --> 00:01:20,120
So, it is important that we should first know
the probability of an event A, in the
13
00:01:20,120 --> 00:01:24,430
previous classes, we have covered these concepts
of event and sample space.
14
00:01:24,430 --> 00:01:30,530
So, here
when we are going to assign probability to
15
00:01:30,530 --> 00:01:33,729
some event, we generally follow certain
norms.
16
00:01:33,729 --> 00:01:41,100
So, here probability of an event A, which
is denoted as P(A) is assigned in such a way
17
00:01:41,100 --> 00:01:44,150
that it satisfies certain conditions.
18
00:01:44,150 --> 00:01:48,310
These conditions are known as axioms of probability.
19
00:01:48,310 --> 00:01:55,380
There are three such axioms that we will go
through one after another.
20
00:01:55,380 --> 00:02:00,270
In the development of this probability, as
we just discussed, in the development of this
21
00:02:00,270 --> 00:02:07,579
probability theory all conclusions are directly
or indirectly based on these axioms.
22
00:02:07,579 --> 00:02:08,579
..
23
00:02:08,579 --> 00:02:16,709
So, the first axiom says that, for any event
A, which belongs to the sample space S, as
24
00:02:16,709 --> 00:02:24,750
shown in this figure, which should lie between
the numbers, real numbers 0 to 1.
25
00:02:24,750 --> 00:02:29,970
So, it
should be greater than or equal to 0 and it
26
00:02:29,970 --> 00:02:32,440
should be less than or equal to 1.
27
00:02:32,440 --> 00:02:40,370
So, this axiom states that, the probability
are real numbers in the interval from 0 to
28
00:02:40,370 --> 00:02:41,370
1.
29
00:02:41,370 --> 00:02:49,580
Obviously, here one point should be, should
automatically be known, that if a particular
30
00:02:49,580 --> 00:02:58,390
event that has no point, if the event is a
null set, then obviously the probability of
31
00:02:58,390 --> 00:03:00,570
that
event equals to 1.
32
00:03:00,570 --> 00:03:04,630
On the other hand, if the event consists of
the full set, which is
33
00:03:04,630 --> 00:03:10,960
nothing, but the axiom two and here we will
go for this axiom two that, the probability
34
00:03:10,960 --> 00:03:15,590
of
the full sample space or the certain event,
35
00:03:15,590 --> 00:03:20,480
which is denoted as S the sample space, the
probability of S equals to 1.
36
00:03:20,480 --> 00:03:21,480
..
37
00:03:21,480 --> 00:03:29,690
So, if the event consists of all the sample
space, that means any one of this event, if
38
00:03:29,690 --> 00:03:31,610
this
event is a certain event.
39
00:03:31,610 --> 00:03:36,830
So, that is why the total probability, which
is probability of the
40
00:03:36,830 --> 00:03:41,770
full sample space, which is equal to 1.
41
00:03:41,770 --> 00:03:50,570
This axiom states that the sample space as
a whole is assigned, sorry for the spelling
42
00:03:50,570 --> 00:03:56,840
mistake, this will be axiom, this axiom states
that, sample space as a whole is assigned
43
00:03:56,840 --> 00:04:02,090
to
a probability 1 since S contains all the possible
44
00:04:02,090 --> 00:04:06,090
outcomes, the S is a certain event.
45
00:04:06,090 --> 00:04:11,930
So,
this is the second axiom of the probabilities.
46
00:04:11,930 --> 00:04:12,930
.
47
00:04:12,930 --> 00:04:21,729
.The third axiom states that, for any two
mutually exclusive event, now, in the last
48
00:04:21,729 --> 00:04:23,940
class,
we discussed about this mutually exclusive
49
00:04:23,940 --> 00:04:25,010
event.
50
00:04:25,010 --> 00:04:34,340
The two mutually exclusive events
means the two event, if there is nothing in
51
00:04:34,340 --> 00:04:40,130
common; so, then the occurrence of one event
automatically implies the non-occurrence of
52
00:04:40,130 --> 00:04:45,150
the other event; then this two events are
known as mutually exclusive event.
53
00:04:45,150 --> 00:04:51,320
So, this axiom three states that, for any
two mutually exclusive events A and B, that
54
00:04:51,320 --> 00:04:54,940
is A
intersection B is a null set.
55
00:04:54,940 --> 00:04:59,660
Then probability of A union B is the summation
of their
56
00:04:59,660 --> 00:05:04,600
individual probabilities, that is probability
of A plus probability of B.
57
00:05:04,600 --> 00:05:13,640
So, this axiom states that, the probability
of the union of two events is the summation
58
00:05:13,640 --> 00:05:18,190
of
the probability of the individual event, if
59
00:05:18,190 --> 00:05:21,000
the events have no outcome in common.
60
00:05:21,000 --> 00:05:24,910
So, if
the events have no outcome in common, that
61
00:05:24,910 --> 00:05:33,660
means, that there is no space, there is no
overlap between these two events, as it is
62
00:05:33,660 --> 00:05:35,430
shown through this Venn diagram.
63
00:05:35,430 --> 00:05:40,680
That is the
intersection between two events, is a null
64
00:05:40,680 --> 00:05:41,680
set.
65
00:05:41,680 --> 00:05:42,680
.
66
00:05:42,680 --> 00:05:47,910
Now, based on this three axioms, there are
few elementary properties are there.
67
00:05:47,910 --> 00:05:52,770
Those
are very useful to draw a certain conclusion
68
00:05:52,770 --> 00:05:56,730
from these axioms that we will go one after
another.
69
00:05:56,730 --> 00:06:02,990
The first elementary property that we can
draw from this one, is just the extension
70
00:06:02,990 --> 00:06:06,210
of the
last axiom that we have seen.
71
00:06:06,210 --> 00:06:10,241
If A1 , A2 and in this way, these are mutually
exclusive
72
00:06:10,241 --> 00:06:17,050
.events, then following the third axiom, that
is probability of the union of these events
73
00:06:17,050 --> 00:06:20,060
is
simply the summation of the probability of
74
00:06:20,060 --> 00:06:22,080
all such events.
75
00:06:22,080 --> 00:06:30,510
So, this is basically an extension of the
third axiom, to include any number of mutually
76
00:06:30,510 --> 00:06:31,950
exclusive events.
77
00:06:31,950 --> 00:06:34,919
This is known as the property of infinite
additivity.
78
00:06:34,919 --> 00:06:44,480
So, for a sample space, if we see, if we can
have the different events then and if we say
79
00:06:44,480 --> 00:06:50,620
that these events are not overlapping to each
other, then if we want to know what is the
80
00:06:50,620 --> 00:06:57,110
total probability of occurring any of these
events, then that can be achieved, that can
81
00:06:57,110 --> 00:06:58,510
be
obtained by
82
00:06:58,510 --> 00:07:02,790
simply summing up the probabilities of the
individuals events, of
83
00:07:02,790 --> 00:07:10,270
summation of those probabilities for the individual
events.
84
00:07:10,270 --> 00:07:11,270
.
85
00:07:11,270 --> 00:07:17,500
Second elementary property says, if A1 is
one event which is belongs to A2 , that means,
86
00:07:17,500 --> 00:07:24,340
if we refer to this particular Venn diagram,
you see that, if this A1 is a subset of this
87
00:07:24,340 --> 00:07:32,340
bigger set A2 , which is denoted here as that
A1 belongs to A2 , then the probability A1
88
00:07:32,340 --> 00:07:40,500
is less than equals to probability A2 and
probability of A2 minus A1 , it is equals
89
00:07:40,500 --> 00:07:44,440
to
probability of A2 , sorry, probability of
90
00:07:44,440 --> 00:07:54,610
A2 minus probability of A1 .
So, in this Venn diagram to represent this
91
00:07:54,610 --> 00:08:00,949
one, this probability of A2 minus A1 is
nothing but, this dark red area which is nothing,
92
00:08:00,949 --> 00:08:08,540
but this probability of A2 is the total
probability as shown in this, inside the bigger
93
00:08:08,540 --> 00:08:11,830
circle minus the probability of this small
94
00:08:11,830 --> 00:08:18,759
.subset of this A2 , which is quite straight
forward, from this Venn diagram
95
00:08:18,759 --> 00:08:19,849
representation.
96
00:08:19,849 --> 00:08:20,849
.
97
00:08:20,849 --> 00:08:21,849
.
98
00:08:21,849 --> 00:08:29,230
The third property is, if that A complement,
that is A bar, is the complement of the event
99
00:08:29,230 --> 00:08:37,630
A, then the probability of A complement is
equal to 1 minus probability of A. Here one
100
00:08:37,630 --> 00:08:45,170
thing is that, the event A and its complement,
union of these two events consist of the
101
00:08:45,170 --> 00:08:46,170
full set.
102
00:08:46,170 --> 00:08:56,650
Now, if you see here, if this is your, if
this is your full sample space, and if this
103
00:08:56,650 --> 00:09:02,120
is
your event A, then the event A and the event,
104
00:09:02,120 --> 00:09:06,360
this area is your event A complement that
105
00:09:06,360 --> 00:09:16,060
.means, the union of these two event is nothing
but, the full set, which is the full sample
106
00:09:16,060 --> 00:09:20,041
space S.
Now, from the second axiom we know that the
107
00:09:20,041 --> 00:09:23,540
probability of S is equal to one.
108
00:09:23,540 --> 00:09:31,830
So, this
S, we can break by probability of A plus probability
109
00:09:31,830 --> 00:09:33,260
of A complement.
110
00:09:33,260 --> 00:09:37,620
So, that we will
get this equals to 1.
111
00:09:37,620 --> 00:09:43,190
So, this probability of A complement is equals
to 1 minus probability
112
00:09:43,190 --> 00:09:48,350
of A.
So, from this second axiom, we got this particular
113
00:09:48,350 --> 00:09:57,860
conclusion which is shown here, that
probability of A complement is equal to the
114
00:09:57,860 --> 00:10:04,140
total probability which is 1 minus
probability of A.
115
00:10:04,140 --> 00:10:05,140
.
116
00:10:05,140 --> 00:10:14,320
Then, the next property says that, if we say,
it is just in the sense, it is just opposite
117
00:10:14,320 --> 00:10:15,940
to the
property.
118
00:10:15,940 --> 00:10:23,240
Here if we say that one event which is the
union of different n numbers of
119
00:10:23,240 --> 00:10:31,240
mutually exclusive events, as it is written
here, if A is the union of A1 to An or A1
120
00:10:31,240 --> 00:10:36,110
to An
are the mutual exclusive events, then the
121
00:10:36,110 --> 00:10:39,870
probability of the total event that is A is
nothing
122
00:10:39,870 --> 00:10:44,709
but the summation of probability of A1 , A2
up to probability of An
123
00:10:44,709 --> 00:10:45,709
..
124
00:10:45,709 --> 00:10:51,850
which is again the extension of those, that
we just discussed, that third axiom.
125
00:10:51,850 --> 00:10:54,810
That in the
third axiom, we have seen that, that probability
126
00:10:54,810 --> 00:11:05,510
of A union B is equals to your
probability of A plus probability of B.
127
00:11:05,510 --> 00:11:11,529
That means, now, if I say that, probability
of A plus probability of B, now, in other
128
00:11:11,529 --> 00:11:14,740
side
we are just saying that A is the split up
129
00:11:14,740 --> 00:11:20,430
of, say for example, A1 and A2, which are
mutually exclusively of course, then probability
130
00:11:20,430 --> 00:11:31,970
of A1 union A2 is equals to probability
of A1 plus probability of A2 .
131
00:11:31,970 --> 00:11:38,670
And this A1 union A2 is nothing but probability
of A which is equal to probability of A1
132
00:11:38,670 --> 00:11:41,220
plus probability of A2 .
133
00:11:41,220 --> 00:11:42,220
..
134
00:11:42,220 --> 00:11:52,180
Now, just by going on this induction, if I
just say that, this A is the union of, so,
135
00:11:52,180 --> 00:12:02,940
I can
write that A equals to the union of this events
136
00:12:02,940 --> 00:12:10,540
up to An , then obviously, the probability
of A should be equals to probability of A1
137
00:12:10,540 --> 00:12:15,750
plus probability of A2 and in this way, it
will
138
00:12:15,750 --> 00:12:25,980
go up to probability of An , which is here
shown in this diagram, that, if these are
139
00:12:25,980 --> 00:12:27,730
the
partition, partitioned in such a way that,
140
00:12:27,730 --> 00:12:29,930
these are mutually exclusive events, then
the
141
00:12:29,930 --> 00:12:36,819
probability of A should be equals to the summation
of their individual probability.
142
00:12:36,819 --> 00:12:37,819
.
143
00:12:37,819 --> 00:12:44,990
.So now, if I replace this A, so this is also
going to this, if I just replace this A in
144
00:12:44,990 --> 00:12:48,990
terms of
this full set, full, full sample space and
145
00:12:48,990 --> 00:12:52,649
this full sample space is partitioned like
this and
146
00:12:52,649 --> 00:13:00,081
it is collectively, if it is exhaustive, then,
we can say that, probability of A 1 plus
147
00:13:00,081 --> 00:13:05,610
probability of A2 up to probability of An
is equals to probability of S, which is nothing
148
00:13:05,610 --> 00:13:07,220
but equal to 1.
149
00:13:07,220 --> 00:13:14,670
So, if for, if the full sample space is partitioned
into A1 , A2 up to An , which are
150
00:13:14,670 --> 00:13:19,960
mutually exclusive and collectively exhaustive,
then the summation of these probabilities
151
00:13:19,960 --> 00:13:21,520
are equals to 1.
152
00:13:21,520 --> 00:13:30,950
Then, so, this one also we discussed, one
in the last lecture also.
153
00:13:30,950 --> 00:13:31,950
.
154
00:13:31,950 --> 00:13:38,180
For any two events in the sample space, this
is the fifth property now.
155
00:13:38,180 --> 00:13:42,040
For any two events
A and B, inside this sample space which is
156
00:13:42,040 --> 00:13:49,300
shown in this Venn diagram, one is that black
shaded circle and one is the red circle.
157
00:13:49,300 --> 00:13:50,339
These are two events.
158
00:13:50,339 --> 00:13:57,800
Now, any one event,the
probability of the one event can be expressed
159
00:13:57,800 --> 00:14:00,040
in terms of this expression.
160
00:14:00,040 --> 00:14:01,040
..
161
00:14:01,040 --> 00:14:08,010
Now, if we just see here, that is, this is
your full sample space.
162
00:14:08,010 --> 00:14:17,600
Now, we are having the
two events A and B. Now, if we want to know
163
00:14:17,600 --> 00:14:23,200
what is the probability of A, this is equal
to, what we are doing is that, this probability
164
00:14:23,200 --> 00:14:30,010
of A, we are just making it the summation
of two zones, one is that, this red shaded
165
00:14:30,010 --> 00:14:37,660
zone and another one is this particular, this
black shaded zone.
166
00:14:37,660 --> 00:14:42,850
Now, what is this, that your red shaded zone?
167
00:14:42,850 --> 00:14:48,279
This red shaded zone is the probability of
A intersection probability of B. So, this
168
00:14:48,279 --> 00:14:49,279
much done.
169
00:14:49,279 --> 00:14:51,560
Now, what is this black shaded zone?
170
00:14:51,560 --> 00:14:56,230
This black shaded join again is the intersection
of
171
00:14:56,230 --> 00:14:57,550
two events.
172
00:14:57,550 --> 00:14:59,430
One is that B complement.
173
00:14:59,430 --> 00:15:03,769
So, outside this B area, the full area is
the B
174
00:15:03,769 --> 00:15:08,959
complement intersection with A, so, which
is nothing but this black shaded area.
175
00:15:08,959 --> 00:15:13,680
So,
which is your B complement intersection A.
176
00:15:13,680 --> 00:15:14,680
..
177
00:15:14,680 --> 00:15:16,680
So, these are the two events.
178
00:15:16,680 --> 00:15:21,840
So, this, this event corresponds to this red
zone and this one,
179
00:15:21,840 --> 00:15:23,880
this probability corresponds to this zone.
180
00:15:23,880 --> 00:15:26,670
So, this is, these two summation is nothing,
but
181
00:15:26,670 --> 00:15:39,170
your probability of A. Here, if you see that,
this probability, if you see the monitor here,
182
00:15:39,170 --> 00:15:43,920
you see that, this probability of A is equal
to probability of A intersection B, which
183
00:15:43,920 --> 00:15:48,480
is
nothing, but this, this zone and this A intersection
184
00:15:48,480 --> 00:15:50,930
B complementary, which is nothing
but, this area.
185
00:15:50,930 --> 00:15:57,050
So, one particular, the probability of any
one event can be expressed in terms of the
186
00:15:57,050 --> 00:16:06,200
summation of its partition, which is in terms
of this two events in the sample space.
187
00:16:06,200 --> 00:16:07,200
..
188
00:16:07,200 --> 00:16:17,761
The next property tells us that, if A1 and
A2 are any two events, then the probability
189
00:16:17,761 --> 00:16:24,720
of
A1 union A2 is equals to probability of A1
190
00:16:24,720 --> 00:16:31,339
plus probability of A2 minus probability of
A1 intersection A2 . This is one, this important
191
00:16:31,339 --> 00:16:36,810
in the sense, you see as you, if you
compare this property with respect to the
192
00:16:36,810 --> 00:16:43,010
third axiom of this probability, then we see
that, here it is the any two events, we are
193
00:16:43,010 --> 00:16:47,380
not putting the constraint of this mutually
exclusive event here.
194
00:16:47,380 --> 00:16:51,320
So, here this A1 and A2 need not be mutually
exclusive.
195
00:16:51,320 --> 00:16:57,690
So, if
these are any two events, then this is the
196
00:16:57,690 --> 00:16:59,060
expression that holds.
197
00:16:59,060 --> 00:17:00,060
.
198
00:17:00,060 --> 00:17:15,679
.If you now, see here, that if, in this area
there are two events again, what we are trying
199
00:17:15,679 --> 00:17:19,000
to
get is, that is the union, if this is your
200
00:17:19,000 --> 00:17:23,380
A and if this is your B, then what we are
trying to
201
00:17:23,380 --> 00:17:31,180
get is the probability of A union probability
of B. So, what is this probability of A union
202
00:17:31,180 --> 00:17:32,990
B is this green area.
203
00:17:32,990 --> 00:17:38,850
If I just draw it in the, along the side of
this circles, this union
204
00:17:38,850 --> 00:17:44,050
concept was given in the last class, so, this
is your union of two events.
205
00:17:44,050 --> 00:17:48,800
Now, this two events, so, how can I express
this one?
206
00:17:48,800 --> 00:17:56,460
First thing, from this thing, I can
write probability of A that means, I am just
207
00:17:56,460 --> 00:18:03,380
writing it, I am just shading it, the area,
corresponding area here, in this Venn diagram,
208
00:18:03,380 --> 00:18:04,380
which is this.
209
00:18:04,380 --> 00:18:08,570
This is your probability of
A. Now, I am adding it up, probability of
210
00:18:08,570 --> 00:18:15,400
B, which is again, I am just shading it in
this,
211
00:18:15,400 --> 00:18:17,190
this is the area.
212
00:18:17,190 --> 00:18:25,240
So, if you see, while adding up this two process,
while adding up these two probabilities,
213
00:18:25,240 --> 00:18:27,910
we are adding this area twice.
214
00:18:27,910 --> 00:18:31,220
One, which is there in probability of A and
also in
215
00:18:31,220 --> 00:18:35,360
probability of B, that is why this area has
to be deducted.
216
00:18:35,360 --> 00:18:42,660
So, this is minus probability of
A intersection probability of B.
217
00:18:42,660 --> 00:18:43,660
.
218
00:18:43,660 --> 00:18:44,660
..
219
00:18:44,660 --> 00:18:45,660
.
220
00:18:45,660 --> 00:18:49,150
So, this is the graphical representation of
this one, this particular equation.
221
00:18:49,150 --> 00:18:54,010
Now, this
equation can also be proven in terms of this,
222
00:18:54,010 --> 00:18:58,330
what we have just now, what you have seen,
the representation of one event in terms of
223
00:18:58,330 --> 00:19:01,880
the event, in terms of two different event
in
224
00:19:01,880 --> 00:19:03,210
the sample space.
225
00:19:03,210 --> 00:19:10,730
So, if we see that proof, it states like these,
that probability of A union
226
00:19:10,730 --> 00:19:15,020
B can be expressed in this way.
227
00:19:15,020 --> 00:19:18,620
That probability of A intersection B plus
this two.
228
00:19:18,620 --> 00:19:23,570
Now
again, if you refer to this diagram, that
229
00:19:23,570 --> 00:19:28,710
this probability of, I just take a separate
sheet
230
00:19:28,710 --> 00:19:31,059
again.
231
00:19:31,059 --> 00:19:32,059
..
232
00:19:32,059 --> 00:19:46,220
So, now, this one, for this is A and this
is your B. So, probability of A union probability
233
00:19:46,220 --> 00:19:54,059
of B, which is equals to, I am just separating
it out into three different places, one is
234
00:19:54,059 --> 00:20:02,620
the,
first one is this place, second one is this
235
00:20:02,620 --> 00:20:10,260
area and third one is here, this particular
thing.
236
00:20:10,260 --> 00:20:17,320
So, then, what we are getting is, this, the
middle one, that is the intersection part,
237
00:20:17,320 --> 00:20:26,950
probability of A intersection B plus so, this
first area, the way that we can write is,
238
00:20:26,950 --> 00:20:29,780
that,
this is nothing but the intersection between
239
00:20:29,780 --> 00:20:36,350
the B complement that is outside B and
intersection with the event A. So, this will
240
00:20:36,350 --> 00:20:38,510
give you this red shaded area.
241
00:20:38,510 --> 00:20:44,320
So, which I can write as this probability
of A intersection B complementary.
242
00:20:44,320 --> 00:20:50,299
So, this is,
this one corresponds to this red shaded area
243
00:20:50,299 --> 00:20:59,350
plus similarly, if I want to write this black
shaded area which is probability of A complementary
244
00:20:59,350 --> 00:21:10,210
intersection B. Now, these events
can be now, these events can be written that,
245
00:21:10,210 --> 00:21:21,120
probability of A intersection B plus
probability of A intersection B prime.
246
00:21:21,120 --> 00:21:26,131
I am just clubbing these two parts together
plus, I
247
00:21:26,131 --> 00:21:36,419
am adding one more part, which is probability
of A intersection B, plus, I am taking this
248
00:21:36,419 --> 00:21:44,880
term, which is probability of A complementary
B intersection with B. So, I have added
249
00:21:44,880 --> 00:21:45,880
one part here.
250
00:21:45,880 --> 00:21:50,330
So, I have to delete this part to make this
equality, probability of A
251
00:21:50,330 --> 00:21:51,830
intersection B.
252
00:21:51,830 --> 00:21:52,830
..
253
00:21:52,830 --> 00:21:53,830
.
254
00:21:53,830 --> 00:21:59,770
Now, this one if you just match and this one
if you just match with our fifth property
255
00:21:59,770 --> 00:22:03,530
that
we have seen in the monitor , if you see,
256
00:22:03,530 --> 00:22:08,110
just, if you just see in this one, then in
the fifth
257
00:22:08,110 --> 00:22:14,770
of property, then what we can do here, if
you see it here, so, this we can write as
258
00:22:14,770 --> 00:22:22,390
probability of A and similarly, this we can
write as probability of B and obviously, this
259
00:22:22,390 --> 00:22:35,820
minus probability of A intersection B which
is obviously, this A union B.
260
00:22:35,820 --> 00:22:36,820
..
261
00:22:36,820 --> 00:22:42,180
So, this proof is given here on this slide
also.
262
00:22:42,180 --> 00:22:51,350
These are the breakup of three probabilities
which is shown in the Venn diagram there and
263
00:22:51,350 --> 00:22:57,180
there are some algebraic calculation to get
this proof.
264
00:22:57,180 --> 00:22:58,180
.
265
00:22:58,180 --> 00:23:02,620
The next property, rather, I should say that
this is the extension of, instead of having
266
00:23:02,620 --> 00:23:06,580
these two events, if we have more than two
events, if we have the three events, then
267
00:23:06,580 --> 00:23:09,809
this
will be simply the extension of this thing,
268
00:23:09,809 --> 00:23:13,760
which can be easily shown in terms of this
Venn diagram.
269
00:23:13,760 --> 00:23:21,900
.Here, what we are saying that, that if these
A1 , A2 and A3 are any three events in the
270
00:23:21,900 --> 00:23:27,860
sample space, then what is the probability
of A1 union A2 union A3 ?
271
00:23:27,860 --> 00:23:34,250
So, similarly, what we should do, we should
add up these probability, probability of A1
272
00:23:34,250 --> 00:23:37,460
,
which corresponds to this circle here, you
273
00:23:37,460 --> 00:23:44,770
can see, probability of A2 , this one and
probability of A3 which is this.
274
00:23:44,770 --> 00:23:50,980
Now, after doing all, after adding all this
probabilities,
275
00:23:50,980 --> 00:23:56,539
what we are doing that, we are adding individual
intersection, the pair-wise intersection
276
00:23:56,539 --> 00:23:59,750
of these probabilities twice.
277
00:23:59,750 --> 00:24:02,440
So, we have to deduct that part.
278
00:24:02,440 --> 00:24:09,789
So, we are deducting this
first pair, A1 , A2 , minus A2 A3 minus A3
279
00:24:09,789 --> 00:24:14,049
A1 .
Now, while doing this deduction from this
280
00:24:14,049 --> 00:24:17,210
one, we can see that this particular area
has
281
00:24:17,210 --> 00:24:20,669
been deducted once more.
282
00:24:20,669 --> 00:24:24,090
So, we have to add this particular area, which
is the
283
00:24:24,090 --> 00:24:26,450
intersection of all three, all three events.
284
00:24:26,450 --> 00:24:29,870
So, have to add this part, that is probability
of
285
00:24:29,870 --> 00:24:33,600
A1 A2 A3 intersection.
286
00:24:33,600 --> 00:24:37,340
So, this is just a simple extension of this
earlier thing.
287
00:24:37,340 --> 00:24:43,370
If we just see, if we just take out
this part and this part, you will see that,
288
00:24:43,370 --> 00:24:45,710
this part has been whatever the extra, that
part
289
00:24:45,710 --> 00:24:47,200
has been deducted again.
290
00:24:47,200 --> 00:24:51,220
Again, when you are taking the third pair
this is, one negative
291
00:24:51,220 --> 00:24:52,630
area is coming here.
292
00:24:52,630 --> 00:24:55,760
So, that have to add here, to get this two
are equal.
293
00:24:55,760 --> 00:24:59,750
So, this is the
extension of this last property which is discussed
294
00:24:59,750 --> 00:25:07,340
in terms of two events, here it is in
terms of any three events.
295
00:25:07,340 --> 00:25:08,340
.
296
00:25:08,340 --> 00:25:09,340
..
297
00:25:09,340 --> 00:25:14,820
Now, it is just an opposite sense of what
we discuss in this fifth property.
298
00:25:14,820 --> 00:25:23,360
This is, this
says that, if an event A must result in the
299
00:25:23,360 --> 00:25:33,100
occurrence of one of the mutually exclusive
event in A1 , A2 and up to An , then the probability
300
00:25:33,100 --> 00:25:38,820
of A can be expressed as their
individual intersection.
301
00:25:38,820 --> 00:25:49,780
If you just see it here, if this is your sample
space and this sample
302
00:25:49,780 --> 00:25:56,049
space is having the partition like that mutually
exclusive and collectively exhaustive
303
00:25:56,049 --> 00:26:02,610
partitions are there and there is another
event like this.
304
00:26:02,610 --> 00:26:13,160
And, if I say that, these are all, these partitions
are, this is total S, this is partition is
305
00:26:13,160 --> 00:26:17,299
A 1 ,
A2 and up to this, it is going and it is coming
306
00:26:17,299 --> 00:26:27,950
up to An and if I say that this event is your
A, then this probability of A can be expressed
307
00:26:27,950 --> 00:26:30,510
as a summation of this particular area.
308
00:26:30,510 --> 00:26:31,950
First is this area.
309
00:26:31,950 --> 00:26:33,680
So, what is this area?
310
00:26:33,680 --> 00:26:37,900
This area is nothing, but the intersection
between
311
00:26:37,900 --> 00:26:47,580
the event A and event A1 . So, we just write
it event A intersection event A1 plus....,
312
00:26:47,580 --> 00:26:51,610
I
am writing now, this area, this corresponding
313
00:26:51,610 --> 00:26:57,250
area in this Venn diagram, which is
nothing, but probability of intersection of
314
00:26:57,250 --> 00:27:02,440
A2 and this event A. So, A intersection A2
. In
315
00:27:02,440 --> 00:27:04,980
this way, I will go on adding up.
316
00:27:04,980 --> 00:27:08,690
Then this area will come, then this area will
come, then
317
00:27:08,690 --> 00:27:12,500
this area will come and finally, the last
event will come.
318
00:27:12,500 --> 00:27:13,500
..
319
00:27:13,500 --> 00:27:14,500
.
320
00:27:14,500 --> 00:27:23,750
So, in this way, I will just add up to probability
of A intersection An . So, this is stated
321
00:27:23,750 --> 00:27:32,110
here in this slide, that if an event A must
result in the occurrence of one of the mutually
322
00:27:32,110 --> 00:27:33,620
exclusive events A1, A2
323
00:27:33,620 --> 00:27:39,429
upto An , then probability of A is a summation
of the
324
00:27:39,429 --> 00:27:51,029
intersection of that event with the individual
events A1 up to An . So, the next property,
325
00:27:51,029 --> 00:27:57,520
the, before we go to that one, that is on
the conditional probability.
326
00:27:57,520 --> 00:28:00,330
That we will see in a
minute.
327
00:28:00,330 --> 00:28:07,909
.Before that, we will just see, we will just
quickly go through one small example problem
328
00:28:07,909 --> 00:28:09,770
which states like this.
329
00:28:09,770 --> 00:28:17,390
That if a steel section manufacturer produces
a particular section,
330
00:28:17,390 --> 00:28:23,340
and the initial quality check reveals that
the probability of producing a defective unit
331
00:28:23,340 --> 00:28:24,650
is
0.022.
332
00:28:24,650 --> 00:28:31,270
Further investigation reveals that the probability
of producing a defective unit in
333
00:28:31,270 --> 00:28:40,010
terms of the measurement is 0.01 and the defective
unit in terms of the material quality is
334
00:28:40,010 --> 00:28:47,570
0.017, then what is the probability of producing
a unit that is defective in measurement
335
00:28:47,570 --> 00:28:50,080
as well as in material quality?
336
00:28:50,080 --> 00:28:54,320
So, here what the information is given is
that, what is the
337
00:28:54,320 --> 00:29:00,110
total probability of producing a defective
unit is given.
338
00:29:00,110 --> 00:29:05,830
And these two events, that is probability
of the…, it is defective, it can be defective
339
00:29:05,830 --> 00:29:08,190
in
two different ways.
340
00:29:08,190 --> 00:29:11,900
One way of defective unit is that this measurement
is the…, the
341
00:29:11,900 --> 00:29:18,010
measurement was wrong and other one is that,
the material quality was not satisfied.
342
00:29:18,010 --> 00:29:21,740
So,
in two different ways, the particular section
343
00:29:21,740 --> 00:29:22,950
can be defective.
344
00:29:22,950 --> 00:29:30,010
So, their individual
probability that we got is 0.01 and another
345
00:29:30,010 --> 00:29:34,620
one is 0.017 and the total is given here as
0.22.
346
00:29:34,620 --> 00:29:40,000
The question is that, what is the probability
of producing a unit which is defective in
347
00:29:40,000 --> 00:29:41,000
terms of both.
348
00:29:41,000 --> 00:29:46,190
That is, it is defective in terms of the measurement
and also in the material
349
00:29:46,190 --> 00:29:47,289
quality.
350
00:29:47,289 --> 00:29:57,480
So, here, we can, we can assign two different
events.
351
00:29:57,480 --> 00:30:03,289
The first event is that, it
says that it is event A, that event says that
352
00:30:03,289 --> 00:30:07,360
the event of production of a defective unit
in
353
00:30:07,360 --> 00:30:08,559
terms of the measurement.
354
00:30:08,559 --> 00:30:09,559
.
355
00:30:09,559 --> 00:30:14,700
.So, this is your first thing and the second
is that B, that is event of production of
356
00:30:14,700 --> 00:30:17,720
the
defective unit in terms of the material quality.
357
00:30:17,720 --> 00:30:22,730
So, from the problem, the information that
is supplied is, what is the probability of
358
00:30:22,730 --> 00:30:28,669
A and what is the probability of B. So,
probability of A is your 0.01 and probability
359
00:30:28,669 --> 00:30:37,570
B is 0.017 and it is also that supplied, what
is the probability A union B, that is, whatever
360
00:30:37,570 --> 00:30:39,850
way it is the defective unit, probability
of
361
00:30:39,850 --> 00:30:42,860
defective unit is 0.022.
362
00:30:42,860 --> 00:30:50,580
So, here the idea is that, when we are saying
that, that probability
363
00:30:50,580 --> 00:31:01,880
of A union B, that means, if you, if we again
refer to that Venn diagram, that A and this
364
00:31:01,880 --> 00:31:05,090
B.
So, this A union B that is when and saying
365
00:31:05,090 --> 00:31:11,309
that, the unit, a particular section is defective
it can be anywhere.
366
00:31:11,309 --> 00:31:16,210
It can be either, it can be defective in case
of only for the reason,
367
00:31:16,210 --> 00:31:22,730
only for the event A, it can be defective
by both or it can be defective by the second
368
00:31:22,730 --> 00:31:29,470
event, that is event B. So, that is why, whatever
the defect, the probability of a defective
369
00:31:29,470 --> 00:31:32,360
unit, that value is given 0.022.
370
00:31:32,360 --> 00:31:37,410
This should refer to this probability of A
union B. So,
371
00:31:37,410 --> 00:31:42,820
there is the probability of A union B is given,
so 0.022.
372
00:31:42,820 --> 00:31:46,771
Also, we have seen that
probability of individual event, that is the
373
00:31:46,771 --> 00:31:58,800
probability of A is given as 0.01 and
probability of B is given as 0.017.
374
00:31:58,800 --> 00:32:08,899
So, now, it is asked that, what is the probability
that a particular unit is defective in terms
375
00:32:08,899 --> 00:32:13,890
of both, that is in terms of its measurement
as well as the material quality?
376
00:32:13,890 --> 00:32:18,260
Now, so, here
according to this Venn diagram, we are basically
377
00:32:18,260 --> 00:32:25,270
referring to this intersection of this two
two event, that is, we want to know what is
378
00:32:25,270 --> 00:32:33,970
the probability A intersection B. Now, we
know, to do this one, one simple property
379
00:32:33,970 --> 00:32:41,640
that we have to use, we should use this one,
that is probability of A union B is equal
380
00:32:41,640 --> 00:32:47,840
to probability of A plus probability of B
minus
381
00:32:47,840 --> 00:32:54,750
probability of A intersection B. Now, this
is known , this is known and this is known.
382
00:32:54,750 --> 00:33:01,279
Obviously, this should be known, which is
probability A intersection B equals to
383
00:33:01,279 --> 00:33:16,080
probability of A plus probability of B minus
probability of A union B, which is 0.01,
384
00:33:16,080 --> 00:33:25,520
0.017, minus 0.022 equals to 0.005.
385
00:33:25,520 --> 00:33:33,660
So, this refers to this particular area, where
I can say that, this unit is defective, in
386
00:33:33,660 --> 00:33:36,710
terms
of both measurement as well as your material
387
00:33:36,710 --> 00:33:38,480
quality.
388
00:33:38,480 --> 00:33:44,640
This is a simple application of that
property that we discussed, that if there
389
00:33:44,640 --> 00:33:49,870
are two events which are are any two events,
it
390
00:33:49,870 --> 00:33:54,529
does not mean that these two are mutually
exclusive, then we can use this property to
391
00:33:54,529 --> 00:33:59,549
get
that simple answer.
392
00:33:59,549 --> 00:34:00,549
..
393
00:34:00,549 --> 00:34:08,500
The next thing which which we are going to
cover is known as the conditional
394
00:34:08,500 --> 00:34:09,500
probability.
395
00:34:09,500 --> 00:34:15,169
Here this conditional probability means in
the sample space what happens,
396
00:34:15,169 --> 00:34:24,460
these two events are given and we are interested
to know the probability of a particular
397
00:34:24,460 --> 00:34:30,619
event giving some condition, that other events
have already occurred.
398
00:34:30,619 --> 00:34:34,909
Now, if these two
events are shown here, as you can see in these
399
00:34:34,909 --> 00:34:42,829
Venn diagram, that is, this is the area that
corresponds to these event A and these red
400
00:34:42,829 --> 00:34:47,409
corresponds to this B and this overlap
between A and B is nothing but, the intersection
401
00:34:47,409 --> 00:34:48,989
of this two.
402
00:34:48,989 --> 00:34:56,849
So, now, if we want to know, what is the probability
of B, this can be obtained from this,
403
00:34:56,849 --> 00:34:59,920
that technique, that we discussed in the previous
class.
404
00:34:59,920 --> 00:35:08,450
Obviously, we can also know
what is the probability of the individual
405
00:35:08,450 --> 00:35:11,410
event, probability of A as well as probability
of
406
00:35:11,410 --> 00:35:17,989
B. Now, here, the question that we are putting
is, the conditional, conditional means that
407
00:35:17,989 --> 00:35:22,979
I want to know the probability of one event,
any one of these two events that, probability
408
00:35:22,979 --> 00:35:25,729
of A or probability of B, with certain condition.
409
00:35:25,729 --> 00:35:33,230
Here the condition is that, I am ensuring
that occurrence of the other event, the occurrence
410
00:35:33,230 --> 00:35:43,690
of the other event here means that,
here it means that, if A and B are two events,
411
00:35:43,690 --> 00:35:53,259
to be specific, any two events in the sample
space S and the probability of A is not equal
412
00:35:53,259 --> 00:35:57,960
to 0, then the probability of B given that
A
413
00:35:57,960 --> 00:36:00,070
has already occurred.
414
00:36:00,070 --> 00:36:01,070
..
415
00:36:01,070 --> 00:36:07,249
Now, the probability of B means, there is
no condition is given here, which we generally,
416
00:36:07,249 --> 00:36:17,229
which we simply write in terms of the probability
of B. Now, what, here, what we are
417
00:36:17,229 --> 00:36:37,279
seeing is that, this is your probability of
B. Now, when we are saying that, that
418
00:36:37,279 --> 00:36:52,209
probability of B, given A has already occurred.
419
00:36:52,209 --> 00:36:53,209
.
420
00:36:53,209 --> 00:36:54,209
..
421
00:36:54,209 --> 00:37:03,930
So, this we generally denote as probability
of B given A. So, this is the notation for
422
00:37:03,930 --> 00:37:04,930
this
one.
423
00:37:04,930 --> 00:37:11,430
So, the probability of B, given that A has
occurred and this just simply the
424
00:37:11,430 --> 00:37:17,619
probability of B. Now, if you come back to
this one, then we see that, which is denoted
425
00:37:17,619 --> 00:37:25,009
as probability of B, given A, and this is
expressed as probability of B given A is equal
426
00:37:25,009 --> 00:37:30,180
to,
this should be equals to, not 3 one, sorry
427
00:37:30,180 --> 00:37:31,180
for this mistake.
428
00:37:31,180 --> 00:37:35,869
So, this is equals to probability
of A intersection B divided by probability
429
00:37:35,869 --> 00:37:45,459
of A. Now, if you just refer to this Venn
diagram, that is shown here, is that, this
430
00:37:45,459 --> 00:37:49,640
is your probability A and this is your probability
B.
431
00:37:49,640 --> 00:37:54,670
Now, from the traditional definition of this
probability, we will just see, so, we are
432
00:37:54,670 --> 00:38:01,079
interested to know that, probability of B
on condition A. Now, there must be something
433
00:38:01,079 --> 00:38:03,700
in the numerator and something in the denominator.
434
00:38:03,700 --> 00:38:10,309
So, in this denominator what we
should get, it is the total possible space
435
00:38:10,309 --> 00:38:14,510
that we know from this, in the previous classes
we discussed that, these denominator should
436
00:38:14,510 --> 00:38:16,180
have the total possible case.
437
00:38:16,180 --> 00:38:18,249
Now, here,
what is the possible case?
438
00:38:18,249 --> 00:38:23,420
The possible case is that, given A has already
occurred.
439
00:38:23,420 --> 00:38:29,719
So, now, if A has already occurred, so, I
should ensure that, whatever the outcome of
440
00:38:29,719 --> 00:38:32,450
the
experiment, that outcome of this experiment
441
00:38:32,450 --> 00:38:38,319
is certainly within this region, should
correspond to this particular area of this
442
00:38:38,319 --> 00:38:39,319
Venn diagram.
443
00:38:39,319 --> 00:38:40,709
So, this is already in this one.
444
00:38:40,709 --> 00:38:48,420
When we are talking about this probability
of B, the total feasible space or the total
445
00:38:48,420 --> 00:38:50,759
.sample space is the full sample space.
446
00:38:50,759 --> 00:38:56,259
Now, when we are giving some condition, the
condition that A has already occurred, so,
447
00:38:56,259 --> 00:39:03,569
my feasible space is this total, this A.
So, obviously in the denominator, this probability
448
00:39:03,569 --> 00:39:08,619
of A should occur and in the
numerator we know that, we should, we should
449
00:39:08,619 --> 00:39:12,200
put the quantity which is the favorable
case.
450
00:39:12,200 --> 00:39:17,319
Now, we want to know what is the probability
of B, on the condition that A has
451
00:39:17,319 --> 00:39:18,319
already occurred.
452
00:39:18,319 --> 00:39:20,380
A has already occurred, that is why I put
in the denominator
453
00:39:20,380 --> 00:39:23,609
probability of A. Now, what is the probability
of B?
454
00:39:23,609 --> 00:39:30,809
Now, the success area being, within
this area, the success area is nothing but,
455
00:39:30,809 --> 00:39:31,950
this area.
456
00:39:31,950 --> 00:39:36,669
If it is in this area, then we can say
that B has also occurred.
457
00:39:36,669 --> 00:39:37,669
.
458
00:39:37,669 --> 00:39:39,739
So, what does this area correspond to?
459
00:39:39,739 --> 00:39:50,609
This area is nothing but probability of A
intersection B. So, this is your expression
460
00:39:50,609 --> 00:39:54,359
for this conditional probability, which says
that
461
00:39:54,359 --> 00:40:02,150
probability of B given A has already occurred,
which is equals to probability of A
462
00:40:02,150 --> 00:40:11,119
intersection B divided by probability of A.
So, here, so, the same thing has been shown
463
00:40:11,119 --> 00:40:17,089
here, which is the probability of A intersection
B is this shaded area, which is orange
464
00:40:17,089 --> 00:40:23,339
type area and probability of A means this
green area plus this orange, this intersection
465
00:40:23,339 --> 00:40:24,339
area.
466
00:40:24,339 --> 00:40:27,999
So, this is the conditional probability.
467
00:40:27,999 --> 00:40:28,999
..
468
00:40:28,999 --> 00:40:37,049
Now, we will see the application of this simple
conditional probability equation to some
469
00:40:37,049 --> 00:40:40,170
civil engineering related problem.
470
00:40:40,170 --> 00:40:46,410
Here, on a national highway, a stretch of
10 kilo meter
471
00:40:46,410 --> 00:40:50,400
is declared to be the accident prone zone.
472
00:40:50,400 --> 00:40:58,819
Now, over this stretch, the probability of
accident, probability of accident at any place,
473
00:40:58,819 --> 00:41:02,430
sorry for this spelling mistake, at any place
is equally likely.
474
00:41:02,430 --> 00:41:06,809
Now, here when, if you just recall, some of
our previous class, here the
475
00:41:06,809 --> 00:41:12,459
equally likely means that I am just giving
an attribute, how to assign the probability
476
00:41:12,459 --> 00:41:13,920
for a
particular event.
477
00:41:13,920 --> 00:41:14,920
.
478
00:41:14,920 --> 00:41:22,449
.So, here the event means that one accident
is taking place at any, over any particular
479
00:41:22,449 --> 00:41:27,460
substretch of this 10 kilo meter area and
it is equally likely.
480
00:41:27,460 --> 00:41:34,499
So, in the middle of this stretch,
this 10 kilometers stretch, there is a bridge
481
00:41:34,499 --> 00:41:38,249
of 755 meter length.
482
00:41:38,249 --> 00:41:44,440
Now, given, an accident
has occurred within the first 6 kilo meter
483
00:41:44,440 --> 00:41:50,670
stretch, what is the probability that it has
occurred on the bridge?
484
00:41:50,670 --> 00:41:57,000
Now, this is interesting, in the sense that...
485
00:41:57,000 --> 00:42:04,880
So, my total stretch is
this, starting from 0 to 10 kilometer.
486
00:42:04,880 --> 00:42:08,910
Now, I say that, this is equally likely accident
prone
487
00:42:08,910 --> 00:42:10,769
area and it is equally likely.
488
00:42:10,769 --> 00:42:16,410
So, if given that any stretch, if I just give
from 0 to or from
489
00:42:16,410 --> 00:42:23,650
the any kilometers range from h1 to say h2,
the probability of the accident between this
490
00:42:23,650 --> 00:42:31,779
two range, the probability, if this is some
event say E, then, probability of E should
491
00:42:31,779 --> 00:42:38,200
be
equals to h2 minus h1 divided by 10.
492
00:42:38,200 --> 00:42:42,309
So, this we can get from for the any event,
that we
493
00:42:42,309 --> 00:42:45,279
can assign as these are equally likely.
494
00:42:45,279 --> 00:42:50,829
Now, for this, for this problem, if we just
assign
495
00:42:50,829 --> 00:43:06,729
that, this is say, A is one event, that event,
that, event of accident, within the first
496
00:43:06,729 --> 00:43:14,749
6
kilometer, so, what is the probability of
497
00:43:14,749 --> 00:43:18,359
A that we will get here is nothing, but so,
6
498
00:43:18,359 --> 00:43:22,319
minus 0, divided by 10 minus 0, which is 0.6.
499
00:43:22,319 --> 00:43:30,140
So, from, starting from here up to 6
kilometer range, the probability of the accident
500
00:43:30,140 --> 00:43:33,269
in between these two is your 0.6.
501
00:43:33,269 --> 00:43:38,731
Now, if
I assign, if I just denote another event,
502
00:43:38,731 --> 00:43:53,079
the event of accident on the bridge, here
the
503
00:43:53,079 --> 00:43:57,390
bridge is 755 meter length.
504
00:43:57,390 --> 00:43:58,390
.
505
00:43:58,390 --> 00:44:04,289
So, the probability of B will be, it is equally
likely all over the stretch.
506
00:44:04,289 --> 00:44:06,970
So, this is also
equally likely.
507
00:44:06,970 --> 00:44:09,499
So, this is 0.755.
508
00:44:09,499 --> 00:44:12,859
I am just converting them into kilometer which
is
509
00:44:12,859 --> 00:44:14,190
0.0755.
510
00:44:14,190 --> 00:44:25,729
So, our condition, so, here, the, what is
the probability that we want to know is
511
00:44:25,729 --> 00:44:36,640
.that, the probability of B, that is, it should
occur on the bridge.
512
00:44:36,640 --> 00:44:39,430
What is the probability
that it has occurred on the bridge?
513
00:44:39,430 --> 00:44:45,959
That is the event B, on condition that, it
has occurred
514
00:44:45,959 --> 00:44:48,420
within the first 6 kilometer of the stretch.
515
00:44:48,420 --> 00:44:52,559
So, this is B given A, just by putting it
here, the
516
00:44:52,559 --> 00:45:04,020
probability of, then A intersection B, divided
by probability of A. Now, this probability
517
00:45:04,020 --> 00:45:13,029
of A intersection B can be…, so, you have
to see at the middle of this bridge.
518
00:45:13,029 --> 00:45:18,759
So, if this
is the location of this bridge, then this
519
00:45:18,759 --> 00:45:20,910
is your 5 kilometer.
520
00:45:20,910 --> 00:45:26,259
So, bridge has some length
and this is within this first 6 kilometer,
521
00:45:26,259 --> 00:45:29,759
most probably comes here.
522
00:45:29,759 --> 00:45:32,799
So, this is your 755
meter.
523
00:45:32,799 --> 00:45:38,209
So, the full bridge is coming within this
first 6 kilometer range.
524
00:45:38,209 --> 00:45:45,210
So, that means, thus A
intersection B is nothing but…, so, B is
525
00:45:45,210 --> 00:45:46,930
a subset of A. So, the intersection obviously
will
526
00:45:46,930 --> 00:45:55,339
be within this region which is nothing but,
equal to the probability of B. So, which is
527
00:45:55,339 --> 00:46:02,699
the
probability of B, that you got there is a
528
00:46:02,699 --> 00:46:06,760
0.0755 divided by probability of A is 0.6,
which
529
00:46:06,760 --> 00:46:12,779
is 0.126, as we got in this calculation.
530
00:46:12,779 --> 00:46:20,420
So, here if you just compare that, what is
the probability of B is your 0.0755.
531
00:46:20,420 --> 00:46:24,729
So, this is
the probability that we got.
532
00:46:24,729 --> 00:46:29,699
Now, when we are giving some condition, that
probability
533
00:46:29,699 --> 00:46:30,949
information changes.
534
00:46:30,949 --> 00:46:38,069
This is the useful thing that, we should know
that if we give some
535
00:46:38,069 --> 00:46:43,329
condition, then the probability of the same
event may change, as we have seen in this
536
00:46:43,329 --> 00:46:46,649
particular problem as well.
537
00:46:46,649 --> 00:46:47,649
.
538
00:46:47,649 --> 00:46:48,649
..
539
00:46:48,649 --> 00:46:57,640
So, here you can see that, this probability
of B given that A, it comes to be 0.126.
540
00:46:57,640 --> 00:47:03,479
So, in
this class, we have seen that, there are three
541
00:47:03,479 --> 00:47:11,069
axioms that defines the basic properties of
the probability of events in a sample space
542
00:47:11,069 --> 00:47:18,279
and the probability theorems formulate the
probabilities of an event, when other events
543
00:47:18,279 --> 00:47:23,289
exist in the same space derived from this
axioms.
544
00:47:23,289 --> 00:47:28,880
The special probability theorems based on
the theorems explained here will be
545
00:47:28,880 --> 00:47:31,130
discussed in the next lecture.
546
00:47:31,130 --> 00:47:35,119
But, before I conclude, I just want to add
to one point here.
547
00:47:35,119 --> 00:47:39,150
When you are talking about this axioms of
this probability, one thing should be kept
548
00:47:39,150 --> 00:47:42,819
in
mind, that this axioms of probability never
549
00:47:42,819 --> 00:47:48,430
tells how should we, how should we assign
the probability.
550
00:47:48,430 --> 00:47:53,450
It only gives some guidelines, that how the
probability should be
551
00:47:53,450 --> 00:47:54,450
assigned.
552
00:47:54,450 --> 00:47:57,970
But, what should be the actual probability
of a particular event, it has nothing
553
00:47:57,970 --> 00:47:59,700
to do with the axioms of probability.
554
00:47:59,700 --> 00:48:00,700
..
555
00:48:00,700 --> 00:48:01,700
.
556
00:48:01,700 --> 00:48:07,599
So, if we just see that the last example that,
we have done it is that when we are, when
557
00:48:07,599 --> 00:48:13,049
we are talking about this probability of B
and the probability of B and probability of
558
00:48:13,049 --> 00:48:15,449
A or
whatever, we have just now seen, but these
559
00:48:15,449 --> 00:48:21,949
are coming from this, from the equally likely
probable events, all these things that we
560
00:48:21,949 --> 00:48:23,019
discussed in the last class.
561
00:48:23,019 --> 00:48:25,949
Only thing the
axioms states that there are some certain
562
00:48:25,949 --> 00:48:29,759
guidelines that must be followed to get to
assign
563
00:48:29,759 --> 00:48:30,759
this probabilities.
564
00:48:30,759 --> 00:48:33,619
But, what should be the actual probability
for the particular event, it
565
00:48:33,619 --> 00:48:38,039
has nothing to do with the axioms of probability.
566
00:48:38,039 --> 00:48:41,259
So, with this I conclude this class.
567
00:48:41,259 --> 00:48:42,259
In
568
00:48:42,259 --> 00:48:46,329
.the next class, some more properties and
the special probability theorem will be
569
00:48:46,329 --> 00:48:47,839
discussed in the next lecture.
570
00:48:47,839 --> 00:48:48,839
Thank you.
571
00:48:48,839 --> 00:48:48,839
.