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In this problem, we'll be
working with a object called

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random walk, where we have a
person on the line-- or a

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tight rope, according
to the problem.

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Let's start from the origin, and
each time step, it would

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randomly either go forward
or backward with certain

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probability.

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In our case, with probability
P, the person would go

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forward, and 1 minus
P going backwards.

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Now, the walk is random in the
following sense-- that the

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choice going forward or backward
in each step is

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random, and it's completely
independent

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from all past history.

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So let's look at the problem.

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It has three parts.

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In the first part, we'd like to
know what's the probability

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that after two steps the person
returns to the starting

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point, which in this
case is 0?

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Now, throughout this problem,
I'm going to be using the

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following notation.

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F indicates the action of going
forward and B indicates

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the action of going backwards.

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A sequence says F and B implies
the sample that the

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person first goes forward,
and then backwards.

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If I add another F, it will
mean, forward, backward,

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forward again.

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OK?

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So in order for the person to
go somewhere after two steps

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and return to the origin, the
following must happen.

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Either the person went forward
followed by backward, or

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backward followed by forward.

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And indeed, this event--

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namely, the union of these
two possibilities--

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defines the event of interest
in our case.

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And we'd like to know what's
the probability of A, which

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we'll break down into the
probability of forward,

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backward, backward, forward.

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Now, forward, backward and
backward, forward-- they are

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two completely different
outcomes.

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And we know that because they're
disjoint, this would

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just be the sum of the
two probabilities--

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plus probability of
backward/forward.

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Here's where the independence
will come in.

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When we try to compute the
probability of going forward

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and backward, because
the action--

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each step is completely
independent from the past, we

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know this is the same as saying,
in the first step, we

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have probability P of going
forward, in the next step,

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probability 1 minus P
of going backwards.

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We can do so-- namely, writing
the probability of forward,

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backward as a product of going
forward times the probability

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of going backwards, because
these actions are independent.

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And similarly, for the second
one, we have going backwards

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first, times going forward
the second time.

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Adding these two up, we have 2
times P times 1 minus P. And

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that will be the answer to the
first part of the problem.

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In the second part of the
problem, we're interested in

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the probability that after three
steps, the person ends

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up in position 1, or
one step forward

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compared to where he started.

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Now, the only possibilities here
are that among the three

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steps, exactly two steps are
forward, and one step is

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backwards, because otherwise
there's no way the person will

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end up in position 1.

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To do so, there, again, are
three possibilities in which

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we go forward, forward,
backward, or forward,

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backward, forward, or backward,
forward, forward.

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And that exhausts all the
possibilities that the person

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can end up in position
1 after three steps.

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And we'll define the collection
of all these

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outcomes as event C. The
probability of event C--

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same as before--

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is simply the sum of
the probability of

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each individual outcome.

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Now, based on the independence
assumption that we used

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before, each outcome here has
the same probability, which is

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equal to P squared times 1 minus
P. The P squared comes

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from the fact that two forward
steps are taken, and 1 minus

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P, the probability of that
one backwards step.

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And since there are three of
them, we multiply 3 in front,

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and that will give us
the probability.

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In the last part of the problem,
we're asked to

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compute that, conditional on
event C already took place,

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what is the probability that the
first step he took was a

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forward step?

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Without going into the details,
let's take a look at

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the C, in which we have three
elements, and only the first

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two elements correspond
to a forward step

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in the first step.

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So we can define event
D as simply

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the first two outcomes--

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forward, forward, backward, and

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forward, backward, forward.

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Now, the probability we're
interested in is simply

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probability of D conditional on
C. We'd write it out using

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the law of conditional
probability--

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D intersection C conditional
on C. Now, because D is a

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subset of C, we have probability
of D divided by

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the probability of C.

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Again, because all samples
here have the same

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probability, all we need to do
is to count the number of

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samples here, which is 2, and
divide by the number of

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samples here, which is 3.

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So we end up with 2 over 3.

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And that concludes
the problem.

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See you next time.

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