1
00:00:00,000 --> 00:00:00,600
2
00:00:00,600 --> 00:00:01,480
Hi.
3
00:00:01,480 --> 00:00:03,990
In this problem, we'll be
talking about communication
4
00:00:03,990 --> 00:00:05,790
across a noisy channel.
5
00:00:05,790 --> 00:00:08,520
But before we dive into the
problem itself, I wanted to
6
00:00:08,520 --> 00:00:12,340
first motivate the context a
little bit and talk more about
7
00:00:12,340 --> 00:00:14,900
what exactly a communication
channel is and
8
00:00:14,900 --> 00:00:16,430
what "noise" means.
9
00:00:16,430 --> 00:00:19,690
So in our everyday life,
we deal with a lot of
10
00:00:19,690 --> 00:00:22,790
communication channels, for
example, the internet, where
11
00:00:22,790 --> 00:00:28,610
we download data and we watch
videos online, or even just
12
00:00:28,610 --> 00:00:29,980
talking to a friend.
13
00:00:29,980 --> 00:00:32,210
And the air could be
your communication
14
00:00:32,210 --> 00:00:33,680
channel for our voice.
15
00:00:33,680 --> 00:00:36,970
But as you probably have
experienced, sometimes these
16
00:00:36,970 --> 00:00:39,900
channels have noise, which
just means that what the
17
00:00:39,900 --> 00:00:42,950
sender was trying to send isn't
necessarily exactly what
18
00:00:42,950 --> 00:00:44,630
the receiver receives.
19
00:00:44,630 --> 00:00:50,070
And so in probability, we try
to model these communication
20
00:00:50,070 --> 00:00:53,350
channels and noise and
try to understand the
21
00:00:53,350 --> 00:00:55,740
probability behind it.
22
00:00:55,740 --> 00:00:58,585
And so now, let's go into
the problem itself.
23
00:00:58,585 --> 00:01:00,860
In this problem, we're dealing
with a pretty simple
24
00:01:00,860 --> 00:01:02,090
communication channel.
25
00:01:02,090 --> 00:01:05,110
It's just a binary channel,
which means that what we're
26
00:01:05,110 --> 00:01:07,400
sending is just one
bit at a time.
27
00:01:07,400 --> 00:01:09,980
And here, a bit just means
either 0 or 1--
28
00:01:09,980 --> 00:01:13,300
so essentially, the simplest
case of information that you
29
00:01:13,300 --> 00:01:14,830
could send.
30
00:01:14,830 --> 00:01:18,710
But sometimes when you send
a 0, the receiver actually
31
00:01:18,710 --> 00:01:21,490
receives a 1 instead,
or vice versa.
32
00:01:21,490 --> 00:01:24,500
And that is where
noise comes in.
33
00:01:24,500 --> 00:01:27,640
So here in this problem, we
actually have a probabilistic
34
00:01:27,640 --> 00:01:31,920
model of this channel and the
noise that hits the channel.
35
00:01:31,920 --> 00:01:35,250
36
00:01:35,250 --> 00:01:40,260
What we're trying to send
is either 0 or a 1.
37
00:01:40,260 --> 00:01:42,810
38
00:01:42,810 --> 00:01:53,760
And what we know is that on
the receiving end, a 0 can
39
00:01:53,760 --> 00:01:58,020
either be received when a 0 is
sent, or a 1 can be received
40
00:01:58,020 --> 00:02:00,490
when a 0 is sent.
41
00:02:00,490 --> 00:02:04,970
And when a 1 is sent, we
could also have noise
42
00:02:04,970 --> 00:02:05,960
that corrupts it.
43
00:02:05,960 --> 00:02:08,190
And you get a 0 instead.
44
00:02:08,190 --> 00:02:14,370
Or you can have a 1 being
successfully transmitted.
45
00:02:14,370 --> 00:02:16,470
And the problem actually
tells us what the
46
00:02:16,470 --> 00:02:18,180
probabilities here are.
47
00:02:18,180 --> 00:02:22,940
So we know that if a 0 is sent,
then with probability 1
48
00:02:22,940 --> 00:02:27,190
minus epsilon naught,
a 0 is received.
49
00:02:27,190 --> 00:02:30,450
And with the remaining
probability, it actually gets
50
00:02:30,450 --> 00:02:32,460
corrupted and turned into a 1.
51
00:02:32,460 --> 00:02:35,760
And similarly, if a 1 is sent,
then with probability 1 minus
52
00:02:35,760 --> 00:02:39,050
epsilon 1, the 1 is correctly
transmitted.
53
00:02:39,050 --> 00:02:42,300
And with the remaining
probability epsilon 1, it's
54
00:02:42,300 --> 00:02:44,460
turned into a 0 instead.
55
00:02:44,460 --> 00:02:47,800
And the last bit of information
is that we know
56
00:02:47,800 --> 00:02:52,530
that with the probability p, any
random bit is actually is
57
00:02:52,530 --> 00:02:54,040
0 that is being sent.
58
00:02:54,040 --> 00:02:56,890
And with probability 1 minus
p, we're actually
59
00:02:56,890 --> 00:02:59,840
trying to send a 1.
60
00:02:59,840 --> 00:03:03,290
So that is the basic setup
for the problem.
61
00:03:03,290 --> 00:03:08,430
And the first part that the
problem asks us to find, what
62
00:03:08,430 --> 00:03:12,310
is the probability of a
successful transmission when
63
00:03:12,310 --> 00:03:18,360
you have just any arbitrary
bit that's being sent.
64
00:03:18,360 --> 00:03:23,660
So what we can do here is, use
this tree that we've already
65
00:03:23,660 --> 00:03:29,240
drawn and identify what are the
cases, the outcomes where
66
00:03:29,240 --> 00:03:32,140
a bit is actually successfully
transmitted.
67
00:03:32,140 --> 00:03:37,850
So if a 0 is sent and a 0
is received, then that
68
00:03:37,850 --> 00:03:40,770
corresponds to a successful
transmission.
69
00:03:40,770 --> 00:03:45,290
Similarly, if a 1 is sent and
a 1 is received, that also
70
00:03:45,290 --> 00:03:48,250
corresponds to a successful
transmission.
71
00:03:48,250 --> 00:03:52,170
And then we can calculate what
these probabilities are,
72
00:03:52,170 --> 00:03:53,540
because we just calculate the
73
00:03:53,540 --> 00:03:55,190
probabilities along the branches.
74
00:03:55,190 --> 00:03:58,360
And so here implicitly, what
we're doing is invoking the
75
00:03:58,360 --> 00:04:00,490
multiplication rule.
76
00:04:00,490 --> 00:04:02,810
So we can calculate the
probabilities of these two
77
00:04:02,810 --> 00:04:05,880
individual outcomes and their
disjoint outcomes.
78
00:04:05,880 --> 00:04:08,990
So we can actually just sum the
two probabilities to find
79
00:04:08,990 --> 00:04:10,140
the answer.
80
00:04:10,140 --> 00:04:16,170
So the probability here is p
times 1 minus epsilon naught--
81
00:04:16,170 --> 00:04:18,339
that's the probability of
a 0 being successfully
82
00:04:18,339 --> 00:04:19,200
transmitted--
83
00:04:19,200 --> 00:04:26,120
plus 1 minus p times 1 minus
epsilon, 1, which is the
84
00:04:26,120 --> 00:04:28,650
probability that a 1 is
successfully transmitted.
85
00:04:28,650 --> 00:04:32,500
And so what we've done here is
actually just looked at this
86
00:04:32,500 --> 00:04:35,340
kind of diagram, this tree
to find the answer.
87
00:04:35,340 --> 00:04:37,980
What we also could have done
is been a little bit more
88
00:04:37,980 --> 00:04:41,020
methodical maybe and actually
apply the law of total
89
00:04:41,020 --> 00:04:44,000
probability, which is really
what we're trying to do here.
90
00:04:44,000 --> 00:04:46,745
So you can see that this
actually corresponds to--
91
00:04:46,745 --> 00:04:52,560
the p corresponds to the
probability of 0 being sent.
92
00:04:52,560 --> 00:04:59,250
And 1 minus epsilon naught is
the probability of success,
93
00:04:59,250 --> 00:05:01,690
given that a 0 is sent.
94
00:05:01,690 --> 00:05:06,830
And this second term
is analogous.
95
00:05:06,830 --> 00:05:11,100
It's the probability that a 1
was sent times the probability
96
00:05:11,100 --> 00:05:16,970
that you have a success, given
that a 1 was sent.
97
00:05:16,970 --> 00:05:25,190
And this is just an example of
applying the law of total
98
00:05:25,190 --> 00:05:29,020
probability, where we
partitioned into the two cases
99
00:05:29,020 --> 00:05:32,270
of a 0 being sent and a 1 being
sent and calculated the
100
00:05:32,270 --> 00:05:33,820
probabilities for each
of those two cases
101
00:05:33,820 --> 00:05:36,210
and added those up.
102
00:05:36,210 --> 00:05:39,570
So that's kind of a review of
the multiplication rule and
103
00:05:39,570 --> 00:05:40,820
law of total probability.
104
00:05:40,820 --> 00:05:43,500
105
00:05:43,500 --> 00:05:48,800
So now, let's move on to part
B. Part B is asking, what is
106
00:05:48,800 --> 00:05:51,950
the probability that a
particular sequence of bits,
107
00:05:51,950 --> 00:05:55,090
not just a single one, but a
sequence of four bits in a row
108
00:05:55,090 --> 00:05:57,240
is successfully transmitted?
109
00:05:57,240 --> 00:05:59,830
And the sequence that we're
looking for is, 1, 0, 1, 1.
110
00:05:59,830 --> 00:06:02,710
111
00:06:02,710 --> 00:06:05,820
So this is how I'll
denote this event.
112
00:06:05,820 --> 00:06:09,450
1, 0, 1, 1 gets successfully
transmitted into 1, 0, 1, 1.
113
00:06:09,450 --> 00:06:12,690
114
00:06:12,690 --> 00:06:16,180
Now, instead of dealing with
single bits in isolation, we
115
00:06:16,180 --> 00:06:17,420
have a sequence of four bits.
116
00:06:17,420 --> 00:06:20,700
But we can really just break
this out into the four
117
00:06:20,700 --> 00:06:26,730
individual bits and look
at those one by one.
118
00:06:26,730 --> 00:06:30,050
So in order to transmit
successfully 1, 0, 1, 1, that
119
00:06:30,050 --> 00:06:34,210
whole sequence, we first need to
transmit a 1 successfully,
120
00:06:34,210 --> 00:06:38,640
then a 0 successfully, then
another 1 successfully, and
121
00:06:38,640 --> 00:06:40,860
then finally, the last
1 successfully.
122
00:06:40,860 --> 00:06:49,230
So really, this is the same as
the intersection of four
123
00:06:49,230 --> 00:06:57,150
different smaller events, a 1
being successfully transmitted
124
00:06:57,150 --> 00:07:03,310
and a 0 being successfully
transmitted and two more 1's
125
00:07:03,310 --> 00:07:04,560
being successfully
transmitted.
126
00:07:04,560 --> 00:07:07,310
127
00:07:07,310 --> 00:07:12,210
So why are we able to do
this, first of all?
128
00:07:12,210 --> 00:07:15,250
We are using an important
assumption that we make in the
129
00:07:15,250 --> 00:07:21,170
problem that each transmission
of an individual bit has the
130
00:07:21,170 --> 00:07:25,560
same probabilistic structure
so that no matter which bit
131
00:07:25,560 --> 00:07:29,050
you're talking about, they all
have the same [? error ?]
132
00:07:29,050 --> 00:07:31,770
probability, the same
probabilities of being either
133
00:07:31,770 --> 00:07:37,380
successfully transmitted or
having noise corrupt it.
134
00:07:37,380 --> 00:07:40,230
So because of that, it doesn't
really matter which particular
135
00:07:40,230 --> 00:07:42,400
1 or 0 we're talking about.
136
00:07:42,400 --> 00:07:46,280
And now, we'll make one more
step, and we'll invoke
137
00:07:46,280 --> 00:07:50,050
independence, which is
the third topic here.
138
00:07:50,050 --> 00:07:52,680
And the other important
assumption here we're making
139
00:07:52,680 --> 00:07:56,430
is that every single
bit is independent
140
00:07:56,430 --> 00:07:57,770
from any other bit.
141
00:07:57,770 --> 00:08:02,130
So the fact that this one was
successfully transmitted has
142
00:08:02,130 --> 00:08:06,180
no impact on the probability
of the 0 being successfully
143
00:08:06,180 --> 00:08:07,440
transmitted or not.
144
00:08:07,440 --> 00:08:10,260
And so because of that, we can
now break this down into a
145
00:08:10,260 --> 00:08:12,990
product of four probabilities.
146
00:08:12,990 --> 00:08:16,940
So this becomes the probability
of 1 transmitted
147
00:08:16,940 --> 00:08:22,255
into a 1 times the probability
of 0 transmitted into a 0, 1
148
00:08:22,255 --> 00:08:26,030
to a 1, and 1 to 1.
149
00:08:26,030 --> 00:08:28,610
150
00:08:28,610 --> 00:08:30,990
And that simplifies things,
because we know what each one
151
00:08:30,990 --> 00:08:32,340
of these are.
152
00:08:32,340 --> 00:08:35,170
The probability of 1 being
successful transmitted into a
153
00:08:35,170 --> 00:08:39,280
1, we know that's just
1 minus epsilon 1.
154
00:08:39,280 --> 00:08:42,539
And similarly, probability of
0 being transmitted into a 0
155
00:08:42,539 --> 00:08:44,470
is 1 minus epsilon naught.
156
00:08:44,470 --> 00:08:46,960
So our final answer
then is just--
157
00:08:46,960 --> 00:08:50,500
well, we have three of these
and one of these.
158
00:08:50,500 --> 00:08:55,660
So the answer is going to be 1
minus epsilon naught times 1
159
00:08:55,660 --> 00:08:59,220
minus epsilon 1 to
the third power.
160
00:08:59,220 --> 00:09:05,390
161
00:09:05,390 --> 00:09:10,690
Now, let's move on go on to
part C, which adds another
162
00:09:10,690 --> 00:09:11,930
wrinkle to the problem.
163
00:09:11,930 --> 00:09:16,110
So now, maybe we're not
satisfied with the success
164
00:09:16,110 --> 00:09:17,630
rate of our current channel.
165
00:09:17,630 --> 00:09:19,110
And we want to improve
it somehow.
166
00:09:19,110 --> 00:09:22,520
And one way of doing this is
to add some redundancy.
167
00:09:22,520 --> 00:09:27,010
So instead of just sending a
single 0 and hoping that it
168
00:09:27,010 --> 00:09:30,140
gets successfully transmitted,
instead what we can do is,
169
00:09:30,140 --> 00:09:34,920
send three 0's in a row to
represent a single 0 and hope
170
00:09:34,920 --> 00:09:38,950
that because we've added some
redundancy, we can somehow
171
00:09:38,950 --> 00:09:43,780
improve our error rates.
172
00:09:43,780 --> 00:09:47,590
So in particular what we're
going to do is, for a 0, when
173
00:09:47,590 --> 00:09:52,780
we want to send a 0, which I'll
put in quotes here, what
174
00:09:52,780 --> 00:09:59,240
we're actually going to send
is a sequence of three 0s.
175
00:09:59,240 --> 00:10:06,500
And what's going to happen is,
this sequence of three 0s,
176
00:10:06,500 --> 00:10:07,910
each one of these bits
is going to go
177
00:10:07,910 --> 00:10:09,320
through the same channel.
178
00:10:09,320 --> 00:10:14,490
So the 0, 0, 0 can stay and get
transmitted successfully
179
00:10:14,490 --> 00:10:15,990
as a 0, 0, 0.
180
00:10:15,990 --> 00:10:21,040
Or maybe the last 0 gets turned
into a 1, or the second
181
00:10:21,040 --> 00:10:25,400
0 gets turned into a 1, or we
can have any one of these
182
00:10:25,400 --> 00:10:30,950
eight possible outcomes
on the receiving end.
183
00:10:30,950 --> 00:10:36,580
184
00:10:36,580 --> 00:10:41,360
And then similarly, for a 1,
when we want to send a 1, what
185
00:10:41,360 --> 00:10:43,050
we'll actually send
is a sequence of
186
00:10:43,050 --> 00:10:46,410
three 1's, three bits.
187
00:10:46,410 --> 00:10:54,230
And just like above, this 1, 1,
1, due to the noise in the
188
00:10:54,230 --> 00:11:01,630
channel, it can get turned into
any one of these eight
189
00:11:01,630 --> 00:11:03,960
sequences on the
receiving end.
190
00:11:03,960 --> 00:11:09,490
191
00:11:09,490 --> 00:11:14,250
So what we're going to do now
is, instead of sending just a
192
00:11:14,250 --> 00:11:16,880
single 0, we'll send three 0s,
and instead of sending a 1,
193
00:11:16,880 --> 00:11:18,130
we'll send three 1s.
194
00:11:18,130 --> 00:11:20,910
But now, the question is, this
is what you'll get on the
195
00:11:20,910 --> 00:11:21,860
receiving end.
196
00:11:21,860 --> 00:11:24,610
How do you interpret--
197
00:11:24,610 --> 00:11:26,790
0, 0, 0, maybe intuitively
you'll say
198
00:11:26,790 --> 00:11:27,960
that's obviously a 0.
199
00:11:27,960 --> 00:11:33,930
But what if you get something
like 0, 1, 0 or 1, 0, 1, when
200
00:11:33,930 --> 00:11:38,250
there's both 0s and 1s in
the received message?
201
00:11:38,250 --> 00:11:38,870
What are you going to do?
202
00:11:38,870 --> 00:11:44,230
So one obvious thing to do is
to take a majority rule.
203
00:11:44,230 --> 00:11:47,900
So because there's three of
them, if there's two or more
204
00:11:47,900 --> 00:11:50,400
0s, we'll say that what
was meant to be sent
205
00:11:50,400 --> 00:11:51,770
was actually a 0.
206
00:11:51,770 --> 00:11:55,840
And if there's two or more 1s,
then we'll interpret that as a
207
00:11:55,840 --> 00:11:58,030
1 being sent.
208
00:11:58,030 --> 00:12:02,130
So in this case, let's look
at the case of 0.
209
00:12:02,130 --> 00:12:04,990
The majority rule here would say
that, well, if 0, 0, 0 was
210
00:12:04,990 --> 00:12:08,540
sent, then the majority is 0s.
211
00:12:08,540 --> 00:12:12,870
And similarly, in these two
cases, 0, 0, 1 or 0, 1, 0, the
212
00:12:12,870 --> 00:12:14,580
majority is also 0s.
213
00:12:14,580 --> 00:12:19,300
And then finally, in this last
case, 1, 0, 0, you get a
214
00:12:19,300 --> 00:12:20,300
majority of 0s.
215
00:12:20,300 --> 00:12:24,030
So in these four received
messages, we'll interpret that
216
00:12:24,030 --> 00:12:27,990
as a 0 have being set.
217
00:12:27,990 --> 00:12:31,760
So part C is asking, given this
majority rule and this
218
00:12:31,760 --> 00:12:35,630
redundancy, what is the
probability that a 0 is
219
00:12:35,630 --> 00:12:38,170
correctly transmitted?
220
00:12:38,170 --> 00:12:41,030
Well, to answer that, we've
already identified these are
221
00:12:41,030 --> 00:12:44,210
the four outcomes, where
a 0 would be correctly
222
00:12:44,210 --> 00:12:45,690
transmitted.
223
00:12:45,690 --> 00:12:49,520
So to find the answer to this
question, all we have to do is
224
00:12:49,520 --> 00:12:52,210
find the probability that a
sequence of 0, 0, 0 gets
225
00:12:52,210 --> 00:12:56,860
turned into one of these
four sequences.
226
00:12:56,860 --> 00:12:58,540
So let's do that.
227
00:12:58,540 --> 00:13:00,890
What is the probability that
a 0, 0, 0 gets turned
228
00:13:00,890 --> 00:13:03,240
into a 0, 0, 0?
229
00:13:03,240 --> 00:13:04,865
Well, that means that
all three of
230
00:13:04,865 --> 00:13:07,290
these 0s had no errors.
231
00:13:07,290 --> 00:13:15,480
So we would have the answer
being 1 minus epsilon 0 cubed,
232
00:13:15,480 --> 00:13:18,250
because all three of these
bits had to have been
233
00:13:18,250 --> 00:13:20,130
successfully transmitted.
234
00:13:20,130 --> 00:13:22,520
Now, let's consider
the other ones.
235
00:13:22,520 --> 00:13:24,500
For example, what's the
probability that a 0, 0, 0
236
00:13:24,500 --> 00:13:26,440
gets turned into a 0, 0, 1?
237
00:13:26,440 --> 00:13:28,560
Well, in this case, we need two
successful transmissions
238
00:13:28,560 --> 00:13:34,370
of 0s, plus one transmission
of 0 that had an error.
239
00:13:34,370 --> 00:13:40,140
So that is going to be 1 minus
epsilon naught squared for the
240
00:13:40,140 --> 00:13:43,720
two successful transmissions of
0, times epsilon naught for
241
00:13:43,720 --> 00:13:46,270
the single one that was wrong.
242
00:13:46,270 --> 00:13:49,470
And if you think about it, that
was only for this case--
243
00:13:49,470 --> 00:13:50,420
0, 0, 1.
244
00:13:50,420 --> 00:13:54,630
But the case where 0, 1, 0 and
1, 0, 0 are the same, because
245
00:13:54,630 --> 00:13:56,980
for all three of these, you
have two successful
246
00:13:56,980 --> 00:14:02,380
transmissions of 0, plus one
that was corrupted with noise.
247
00:14:02,380 --> 00:14:05,730
And so it turns out that all
three of those probabilities
248
00:14:05,730 --> 00:14:06,540
are going to be the same.
249
00:14:06,540 --> 00:14:09,080
So this is our final answer
for this part.
250
00:14:09,080 --> 00:14:14,780
251
00:14:14,780 --> 00:14:22,190
Now, let's move on to part D.
Part D is asking now a type of
252
00:14:22,190 --> 00:14:23,340
inference problem.
253
00:14:23,340 --> 00:14:24,780
And we'll talk more
about inference
254
00:14:24,780 --> 00:14:27,310
later on in this course.
255
00:14:27,310 --> 00:14:28,870
The purpose of this problem--
256
00:14:28,870 --> 00:14:37,310
what it's asking is, suppose
you received a 1, 0, 1.
257
00:14:37,310 --> 00:14:40,940
That's the sequence of
three messages, three
258
00:14:40,940 --> 00:14:42,990
bits that you received.
259
00:14:42,990 --> 00:14:48,640
Given that you received a 1, 0,
1, what's the probability
260
00:14:48,640 --> 00:14:51,800
that 0 was actually the thing
that was being sent.
261
00:14:51,800 --> 00:14:54,330
262
00:14:54,330 --> 00:15:00,680
So if you look at this, you'll
look at it and say, this looks
263
00:15:00,680 --> 00:15:03,510
like something where we
can apply Bayes' rule.
264
00:15:03,510 --> 00:15:05,030
So that's the fourth
thing that we're
265
00:15:05,030 --> 00:15:06,610
covering in this problem.
266
00:15:06,610 --> 00:15:10,740
And if you apply Bayes' rule,
what you'll get is, this is
267
00:15:10,740 --> 00:15:15,832
equal to the probability of 0
times the probability of 1, 0,
268
00:15:15,832 --> 00:15:21,250
1 being received, given that 0
was what was sent, divided by
269
00:15:21,250 --> 00:15:25,930
the probability that 1,
0, 1 is received.
270
00:15:25,930 --> 00:15:29,590
So we have this basic
structure.
271
00:15:29,590 --> 00:15:32,860
And we also know that we can
use the law of total
272
00:15:32,860 --> 00:15:35,310
probability again on
this denominator.
273
00:15:35,310 --> 00:15:38,970
So we know that the probability
that 1, 0, 1 is
274
00:15:38,970 --> 00:15:44,570
received is equal to the
probability of 0 being sent
275
00:15:44,570 --> 00:15:48,570
times probability of 1, 0, 1
being received, given that 0
276
00:15:48,570 --> 00:15:53,840
was sent, plus the probability
that 1 was sent times the
277
00:15:53,840 --> 00:15:56,150
probability that 1,
0, 1 is received,
278
00:15:56,150 --> 00:15:58,780
given that 1 is sent.
279
00:15:58,780 --> 00:16:02,240
And as you'll notice in
applications of Bayes' rule,
280
00:16:02,240 --> 00:16:05,610
usually what you'll have is a
numerator is then repeated as
281
00:16:05,610 --> 00:16:08,615
one of the terms in the
denominator, because it's just
282
00:16:08,615 --> 00:16:12,010
an application of total
probability.
283
00:16:12,010 --> 00:16:15,010
So if you put these pieces
together, really, what we need
284
00:16:15,010 --> 00:16:20,700
is just these four terms.
285
00:16:20,700 --> 00:16:23,660
Once we have those four terms,
we can just plug them into
286
00:16:23,660 --> 00:16:26,530
this equation, and we'll
have our answer.
287
00:16:26,530 --> 00:16:29,520
So let's figure out what
those four terms are.
288
00:16:29,520 --> 00:16:31,450
The probability of 0
being sent-- well,
289
00:16:31,450 --> 00:16:32,560
we said that earlier.
290
00:16:32,560 --> 00:16:37,420
Probability of 0 being
sent is just p.
291
00:16:37,420 --> 00:16:45,440
And the probability of 1 being
sent is 1 minus p.
292
00:16:45,440 --> 00:16:47,080
That's just from the
model that we're
293
00:16:47,080 --> 00:16:48,890
given in the problem.
294
00:16:48,890 --> 00:16:50,970
Now, let's figure
out this part.
295
00:16:50,970 --> 00:16:56,460
What is the probability of a 1,
0, 1 being received, given
296
00:16:56,460 --> 00:17:00,690
that 0 was sent?
297
00:17:00,690 --> 00:17:04,420
So if 0 was sent, then we know
that what really was sent was
298
00:17:04,420 --> 00:17:07,490
0, 0, 0, that sequence
of three bits.
299
00:17:07,490 --> 00:17:09,990
And now, what's the probability
that 0, 0, 0 got
300
00:17:09,990 --> 00:17:14,440
turned into 1, 0, 1?
301
00:17:14,440 --> 00:17:16,839
Wall, in this case, what we
have is one successful
302
00:17:16,839 --> 00:17:22,230
transmission of a 0, plus two
failed transmission of a 0.
303
00:17:22,230 --> 00:17:26,000
So that one successful
transmission of a 0, that
304
00:17:26,000 --> 00:17:30,000
probability is 1 minus
epsilon naught.
305
00:17:30,000 --> 00:17:32,840
And now, we have two failed
transmissions of a 0.
306
00:17:32,840 --> 00:17:37,930
So we have to multiply that
by epsilon naught squared.
307
00:17:37,930 --> 00:17:41,870
And now, for the last piece,
what's the probability of
308
00:17:41,870 --> 00:17:47,040
receiving the 1, 0, 1, given
that 1 was actually sent?
309
00:17:47,040 --> 00:17:49,680
Well, in that case, if a 1 was
sent, what was really sent was
310
00:17:49,680 --> 00:17:51,470
a sequence of three 1s.
311
00:17:51,470 --> 00:17:54,810
And now, we want the probability
that a 1, 1, 1 got
312
00:17:54,810 --> 00:17:56,480
turned into a 1, 0, 1.
313
00:17:56,480 --> 00:17:58,620
In this case, we have two
successful transmissions of
314
00:17:58,620 --> 00:18:01,620
the 1 with one failed
transmission.
315
00:18:01,620 --> 00:18:04,240
So the two successful
transmissions will have 1
316
00:18:04,240 --> 00:18:06,250
minus epsilon 1 squared.
317
00:18:06,250 --> 00:18:08,080
And then the one failed
transmission will give us an
318
00:18:08,080 --> 00:18:11,380
extra term of epsilon 1.
319
00:18:11,380 --> 00:18:16,820
So just for completeness, let's
actually write out what
320
00:18:16,820 --> 00:18:18,340
this final answer would be.
321
00:18:18,340 --> 00:18:20,930
So probability of 0 is p.
322
00:18:20,930 --> 00:18:25,950
Probability of 1, 0, 1, given 0
is, we calculated that as 1
323
00:18:25,950 --> 00:18:30,090
minus epsilon naught times
epsilon naught squared.
324
00:18:30,090 --> 00:18:31,860
The same term appears again
in the denominator.
325
00:18:31,860 --> 00:18:36,610
326
00:18:36,610 --> 00:18:43,410
Plus the other term is,
probability of 1 times the
327
00:18:43,410 --> 00:18:48,540
probability of 1,
0, 1, given 1.
328
00:18:48,540 --> 00:18:53,280
So that is 1 minus epsilon
squared times epsilon 1.
329
00:18:53,280 --> 00:18:55,010
So that is our final answer.
330
00:18:55,010 --> 00:18:59,980
And it's really just a
application of Bayes' rule.
331
00:18:59,980 --> 00:19:05,030
So this was a nice problem,
because it represents a real
332
00:19:05,030 --> 00:19:07,440
world phenomenon that happens.
333
00:19:07,440 --> 00:19:10,570
And we can see that you can
apply a pretty simple
334
00:19:10,570 --> 00:19:13,380
probabilistic model to it and
still be able to answer some
335
00:19:13,380 --> 00:19:14,530
interesting questions.
336
00:19:14,530 --> 00:19:18,700
And there are other extensions
that you can ask also.
337
00:19:18,700 --> 00:19:22,320
For example, we've talked about
adding redundancy by
338
00:19:22,320 --> 00:19:25,140
tripling the number of bits,
but tripling the number of
339
00:19:25,140 --> 00:19:29,650
bits also reduces the
throughputs, because instead
340
00:19:29,650 --> 00:19:31,710
of sending one, you have
to send three bits
341
00:19:31,710 --> 00:19:33,160
just to send one.
342
00:19:33,160 --> 00:19:37,840
So if there's a cost of that, at
what point does the benefit
343
00:19:37,840 --> 00:19:42,770
of having lower ever outweigh
the cost of having to send
344
00:19:42,770 --> 00:19:43,800
more things?
345
00:19:43,800 --> 00:19:47,220
And so that's a question that
you can answer with some more
346
00:19:47,220 --> 00:19:48,960
tools in probability.
347
00:19:48,960 --> 00:19:51,030
So we hope you enjoyed
this problem.
348
00:19:51,030 --> 00:19:52,670
And we'll see you
again next time.
349
00:19:52,670 --> 00:19:54,834