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Hi.

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In this problem, we'll be
talking about communication

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across a noisy channel.

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But before we dive into the
problem itself, I wanted to

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first motivate the context a
little bit and talk more about

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what exactly a communication
channel is and

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what "noise" means.

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So in our everyday life,
we deal with a lot of

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communication channels, for
example, the internet, where

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we download data and we watch
videos online, or even just

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talking to a friend.

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And the air could be
your communication

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channel for our voice.

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But as you probably have
experienced, sometimes these

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channels have noise, which
just means that what the

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sender was trying to send isn't
necessarily exactly what

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the receiver receives.

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And so in probability, we try
to model these communication

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channels and noise and
try to understand the

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probability behind it.

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And so now, let's go into
the problem itself.

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In this problem, we're dealing
with a pretty simple

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communication channel.

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It's just a binary channel,
which means that what we're

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sending is just one
bit at a time.

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And here, a bit just means
either 0 or 1--

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so essentially, the simplest
case of information that you

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could send.

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But sometimes when you send
a 0, the receiver actually

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receives a 1 instead,
or vice versa.

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And that is where
noise comes in.

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So here in this problem, we
actually have a probabilistic

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model of this channel and the
noise that hits the channel.

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What we're trying to send
is either 0 or a 1.

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And what we know is that on
the receiving end, a 0 can

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either be received when a 0 is
sent, or a 1 can be received

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when a 0 is sent.

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And when a 1 is sent, we
could also have noise

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that corrupts it.

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And you get a 0 instead.

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Or you can have a 1 being
successfully transmitted.

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And the problem actually
tells us what the

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probabilities here are.

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So we know that if a 0 is sent,
then with probability 1

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minus epsilon naught,
a 0 is received.

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And with the remaining
probability, it actually gets

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corrupted and turned into a 1.

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And similarly, if a 1 is sent,
then with probability 1 minus

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epsilon 1, the 1 is correctly
transmitted.

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And with the remaining
probability epsilon 1, it's

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turned into a 0 instead.

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And the last bit of information
is that we know

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that with the probability p, any
random bit is actually is

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0 that is being sent.

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And with probability 1 minus
p, we're actually

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trying to send a 1.

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So that is the basic setup
for the problem.

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And the first part that the
problem asks us to find, what

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is the probability of a
successful transmission when

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you have just any arbitrary
bit that's being sent.

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So what we can do here is, use
this tree that we've already

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drawn and identify what are the
cases, the outcomes where

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a bit is actually successfully
transmitted.

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So if a 0 is sent and a 0
is received, then that

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corresponds to a successful
transmission.

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Similarly, if a 1 is sent and
a 1 is received, that also

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corresponds to a successful
transmission.

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And then we can calculate what
these probabilities are,

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because we just calculate the

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probabilities along the branches.

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And so here implicitly, what
we're doing is invoking the

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multiplication rule.

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So we can calculate the
probabilities of these two

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individual outcomes and their
disjoint outcomes.

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So we can actually just sum the
two probabilities to find

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the answer.

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So the probability here is p
times 1 minus epsilon naught--

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that's the probability of
a 0 being successfully

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transmitted--

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plus 1 minus p times 1 minus
epsilon, 1, which is the

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probability that a 1 is
successfully transmitted.

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And so what we've done here is
actually just looked at this

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kind of diagram, this tree
to find the answer.

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What we also could have done
is been a little bit more

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methodical maybe and actually
apply the law of total

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probability, which is really
what we're trying to do here.

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So you can see that this
actually corresponds to--

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the p corresponds to the
probability of 0 being sent.

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And 1 minus epsilon naught is
the probability of success,

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given that a 0 is sent.

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And this second term
is analogous.

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It's the probability that a 1
was sent times the probability

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that you have a success, given
that a 1 was sent.

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And this is just an example of
applying the law of total

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probability, where we
partitioned into the two cases

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of a 0 being sent and a 1 being
sent and calculated the

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probabilities for each
of those two cases

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and added those up.

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So that's kind of a review of
the multiplication rule and

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law of total probability.

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So now, let's move on to part
B. Part B is asking, what is

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the probability that a
particular sequence of bits,

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not just a single one, but a
sequence of four bits in a row

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is successfully transmitted?

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And the sequence that we're
looking for is, 1, 0, 1, 1.

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So this is how I'll
denote this event.

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1, 0, 1, 1 gets successfully
transmitted into 1, 0, 1, 1.

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Now, instead of dealing with
single bits in isolation, we

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have a sequence of four bits.

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But we can really just break
this out into the four

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individual bits and look
at those one by one.

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So in order to transmit
successfully 1, 0, 1, 1, that

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whole sequence, we first need to
transmit a 1 successfully,

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then a 0 successfully, then
another 1 successfully, and

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then finally, the last
1 successfully.

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So really, this is the same as
the intersection of four

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different smaller events, a 1
being successfully transmitted

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and a 0 being successfully
transmitted and two more 1's

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being successfully
transmitted.

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So why are we able to do
this, first of all?

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We are using an important
assumption that we make in the

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problem that each transmission
of an individual bit has the

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same probabilistic structure
so that no matter which bit

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you're talking about, they all
have the same [? error ?]

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probability, the same
probabilities of being either

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successfully transmitted or
having noise corrupt it.

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So because of that, it doesn't
really matter which particular

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1 or 0 we're talking about.

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And now, we'll make one more
step, and we'll invoke

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independence, which is
the third topic here.

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And the other important
assumption here we're making

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is that every single
bit is independent

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from any other bit.

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So the fact that this one was
successfully transmitted has

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no impact on the probability
of the 0 being successfully

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transmitted or not.

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And so because of that, we can
now break this down into a

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product of four probabilities.

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So this becomes the probability
of 1 transmitted

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into a 1 times the probability
of 0 transmitted into a 0, 1

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to a 1, and 1 to 1.

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And that simplifies things,
because we know what each one

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of these are.

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The probability of 1 being
successful transmitted into a

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1, we know that's just
1 minus epsilon 1.

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And similarly, probability of
0 being transmitted into a 0

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is 1 minus epsilon naught.

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So our final answer
then is just--

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well, we have three of these
and one of these.

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So the answer is going to be 1
minus epsilon naught times 1

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minus epsilon 1 to
the third power.

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Now, let's move on go on to
part C, which adds another

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wrinkle to the problem.

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So now, maybe we're not
satisfied with the success

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rate of our current channel.

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And we want to improve
it somehow.

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And one way of doing this is
to add some redundancy.

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So instead of just sending a
single 0 and hoping that it

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gets successfully transmitted,
instead what we can do is,

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send three 0's in a row to
represent a single 0 and hope

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that because we've added some
redundancy, we can somehow

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improve our error rates.

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So in particular what we're
going to do is, for a 0, when

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we want to send a 0, which I'll
put in quotes here, what

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we're actually going to send
is a sequence of three 0s.

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And what's going to happen is,
this sequence of three 0s,

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each one of these bits
is going to go

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through the same channel.

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So the 0, 0, 0 can stay and get
transmitted successfully

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as a 0, 0, 0.

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Or maybe the last 0 gets turned
into a 1, or the second

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0 gets turned into a 1, or we
can have any one of these

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eight possible outcomes
on the receiving end.

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And then similarly, for a 1,
when we want to send a 1, what

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we'll actually send
is a sequence of

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three 1's, three bits.

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And just like above, this 1, 1,
1, due to the noise in the

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channel, it can get turned into
any one of these eight

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sequences on the
receiving end.

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So what we're going to do now
is, instead of sending just a

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single 0, we'll send three 0s,
and instead of sending a 1,

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we'll send three 1s.

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But now, the question is, this
is what you'll get on the

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receiving end.

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How do you interpret--

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0, 0, 0, maybe intuitively
you'll say

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that's obviously a 0.

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But what if you get something
like 0, 1, 0 or 1, 0, 1, when

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there's both 0s and 1s in
the received message?

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What are you going to do?

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So one obvious thing to do is
to take a majority rule.

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So because there's three of
them, if there's two or more

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0s, we'll say that what
was meant to be sent

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was actually a 0.

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And if there's two or more 1s,
then we'll interpret that as a

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1 being sent.

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So in this case, let's look
at the case of 0.

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The majority rule here would say
that, well, if 0, 0, 0 was

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sent, then the majority is 0s.

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And similarly, in these two
cases, 0, 0, 1 or 0, 1, 0, the

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majority is also 0s.

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And then finally, in this last
case, 1, 0, 0, you get a

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majority of 0s.

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So in these four received
messages, we'll interpret that

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as a 0 have being set.

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So part C is asking, given this
majority rule and this

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redundancy, what is the
probability that a 0 is

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correctly transmitted?

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Well, to answer that, we've
already identified these are

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the four outcomes, where
a 0 would be correctly

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transmitted.

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So to find the answer to this
question, all we have to do is

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find the probability that a
sequence of 0, 0, 0 gets

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turned into one of these
four sequences.

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So let's do that.

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What is the probability that
a 0, 0, 0 gets turned

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into a 0, 0, 0?

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Well, that means that
all three of

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these 0s had no errors.

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So we would have the answer
being 1 minus epsilon 0 cubed,

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because all three of these
bits had to have been

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successfully transmitted.

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Now, let's consider
the other ones.

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For example, what's the
probability that a 0, 0, 0

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gets turned into a 0, 0, 1?

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Well, in this case, we need two
successful transmissions

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of 0s, plus one transmission
of 0 that had an error.

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So that is going to be 1 minus
epsilon naught squared for the

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two successful transmissions of
0, times epsilon naught for

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the single one that was wrong.

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And if you think about it, that
was only for this case--

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0, 0, 1.

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But the case where 0, 1, 0 and
1, 0, 0 are the same, because

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for all three of these, you
have two successful

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transmissions of 0, plus one
that was corrupted with noise.

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And so it turns out that all
three of those probabilities

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are going to be the same.

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So this is our final answer
for this part.

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00:14:09,080 --> 00:14:14,780

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Now, let's move on to part D.
Part D is asking now a type of

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inference problem.

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And we'll talk more
about inference

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later on in this course.

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The purpose of this problem--

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what it's asking is, suppose
you received a 1, 0, 1.

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That's the sequence of
three messages, three

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bits that you received.

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Given that you received a 1, 0,
1, what's the probability

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that 0 was actually the thing
that was being sent.

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00:14:51,800 --> 00:14:54,330

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So if you look at this, you'll
look at it and say, this looks

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like something where we
can apply Bayes' rule.

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So that's the fourth
thing that we're

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covering in this problem.

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And if you apply Bayes' rule,
what you'll get is, this is

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equal to the probability of 0
times the probability of 1, 0,

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1 being received, given that 0
was what was sent, divided by

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the probability that 1,
0, 1 is received.

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So we have this basic
structure.

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And we also know that we can
use the law of total

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probability again on
this denominator.

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So we know that the probability
that 1, 0, 1 is

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received is equal to the
probability of 0 being sent

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times probability of 1, 0, 1
being received, given that 0

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was sent, plus the probability
that 1 was sent times the

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00:15:53,840 --> 00:15:56,150
probability that 1,
0, 1 is received,

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given that 1 is sent.

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And as you'll notice in
applications of Bayes' rule,

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usually what you'll have is a
numerator is then repeated as

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one of the terms in the
denominator, because it's just

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an application of total
probability.

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So if you put these pieces
together, really, what we need

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is just these four terms.

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00:16:20,700 --> 00:16:23,660
Once we have those four terms,
we can just plug them into

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00:16:23,660 --> 00:16:26,530
this equation, and we'll
have our answer.

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00:16:26,530 --> 00:16:29,520
So let's figure out what
those four terms are.

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The probability of 0
being sent-- well,

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00:16:31,450 --> 00:16:32,560
we said that earlier.

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Probability of 0 being
sent is just p.

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00:16:37,420 --> 00:16:45,440
And the probability of 1 being
sent is 1 minus p.

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That's just from the
model that we're

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given in the problem.

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Now, let's figure
out this part.

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What is the probability of a 1,
0, 1 being received, given

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00:16:56,460 --> 00:17:00,690
that 0 was sent?

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00:17:00,690 --> 00:17:04,420
So if 0 was sent, then we know
that what really was sent was

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0, 0, 0, that sequence
of three bits.

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And now, what's the probability
that 0, 0, 0 got

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turned into 1, 0, 1?

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00:17:14,440 --> 00:17:16,839
Wall, in this case, what we
have is one successful

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00:17:16,839 --> 00:17:22,230
transmission of a 0, plus two
failed transmission of a 0.

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00:17:22,230 --> 00:17:26,000
So that one successful
transmission of a 0, that

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00:17:26,000 --> 00:17:30,000
probability is 1 minus
epsilon naught.

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00:17:30,000 --> 00:17:32,840
And now, we have two failed
transmissions of a 0.

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00:17:32,840 --> 00:17:37,930
So we have to multiply that
by epsilon naught squared.

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00:17:37,930 --> 00:17:41,870
And now, for the last piece,
what's the probability of

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00:17:41,870 --> 00:17:47,040
receiving the 1, 0, 1, given
that 1 was actually sent?

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00:17:47,040 --> 00:17:49,680
Well, in that case, if a 1 was
sent, what was really sent was

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00:17:49,680 --> 00:17:51,470
a sequence of three 1s.

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00:17:51,470 --> 00:17:54,810
And now, we want the probability
that a 1, 1, 1 got

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00:17:54,810 --> 00:17:56,480
turned into a 1, 0, 1.

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00:17:56,480 --> 00:17:58,620
In this case, we have two
successful transmissions of

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00:17:58,620 --> 00:18:01,620
the 1 with one failed
transmission.

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00:18:01,620 --> 00:18:04,240
So the two successful
transmissions will have 1

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00:18:04,240 --> 00:18:06,250
minus epsilon 1 squared.

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00:18:06,250 --> 00:18:08,080
And then the one failed
transmission will give us an

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00:18:08,080 --> 00:18:11,380
extra term of epsilon 1.

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00:18:11,380 --> 00:18:16,820
So just for completeness, let's
actually write out what

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00:18:16,820 --> 00:18:18,340
this final answer would be.

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00:18:18,340 --> 00:18:20,930
So probability of 0 is p.

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00:18:20,930 --> 00:18:25,950
Probability of 1, 0, 1, given 0
is, we calculated that as 1

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00:18:25,950 --> 00:18:30,090
minus epsilon naught times
epsilon naught squared.

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00:18:30,090 --> 00:18:31,860
The same term appears again
in the denominator.

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00:18:31,860 --> 00:18:36,610

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00:18:36,610 --> 00:18:43,410
Plus the other term is,
probability of 1 times the

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00:18:43,410 --> 00:18:48,540
probability of 1,
0, 1, given 1.

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00:18:48,540 --> 00:18:53,280
So that is 1 minus epsilon
squared times epsilon 1.

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00:18:53,280 --> 00:18:55,010
So that is our final answer.

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00:18:55,010 --> 00:18:59,980
And it's really just a
application of Bayes' rule.

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00:18:59,980 --> 00:19:05,030
So this was a nice problem,
because it represents a real

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00:19:05,030 --> 00:19:07,440
world phenomenon that happens.

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00:19:07,440 --> 00:19:10,570
And we can see that you can
apply a pretty simple

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00:19:10,570 --> 00:19:13,380
probabilistic model to it and
still be able to answer some

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00:19:13,380 --> 00:19:14,530
interesting questions.

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00:19:14,530 --> 00:19:18,700
And there are other extensions
that you can ask also.

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00:19:18,700 --> 00:19:22,320
For example, we've talked about
adding redundancy by

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00:19:22,320 --> 00:19:25,140
tripling the number of bits,
but tripling the number of

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00:19:25,140 --> 00:19:29,650
bits also reduces the
throughputs, because instead

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00:19:29,650 --> 00:19:31,710
of sending one, you have
to send three bits

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00:19:31,710 --> 00:19:33,160
just to send one.

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00:19:33,160 --> 00:19:37,840
So if there's a cost of that, at
what point does the benefit

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00:19:37,840 --> 00:19:42,770
of having lower ever outweigh
the cost of having to send

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00:19:42,770 --> 00:19:43,800
more things?

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00:19:43,800 --> 00:19:47,220
And so that's a question that
you can answer with some more

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00:19:47,220 --> 00:19:48,960
tools in probability.

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00:19:48,960 --> 00:19:51,030
So we hope you enjoyed
this problem.

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00:19:51,030 --> 00:19:52,670
And we'll see you
again next time.

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