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In this problem, we're going
to look at the probability
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that when you take a stick and
break it into three pieces
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randomly that these three pieces
can actually be used to
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form a triangle.
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All right, so we start
out with a stick of
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unit length, so--
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length 1.
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And we'll choose a point along
the stick to break.
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And we'll choose that point
uniformly at random.
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So let's say that we chose it
here, that was the point where
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we'll break it.
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And then independently of this
first choice we'll again
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choose a second point
to break it.
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Again, uniformly at random
along the entire stick.
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So let's say the second point
we chose was here.
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So what we have now is, we'll
break it here, here, and so
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we'll have three pieces--
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the first one, the left
one and the right one.
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And we want to know, what's the
probability that when you
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take these three pieces you
could form a triangle?
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So the first thing we
should ask ourselves
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is, what do you need--
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what conditions must be
satisfied in order to actually
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be able to form a triangle
with three pieces?
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So you could think about, what
would stop you from being able
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to do that?
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Well, one possibility is
that you have pieces
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that look like this.
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So in that case you would
try to form something
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that looks like this.
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But you can't get a triangle
because these two pieces are
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too short and they can't
touch each other.
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So actually the condition that
must be satisfied is that when
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you take any two of the three
pieces, their combined length
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has to be greater than
the length of the
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remaining third piece.
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And that has to be true
for any two pieces.
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And really that's just so that
any two pieces, they can touch
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and still form a triangle.
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So let's try to add some
probability to this.
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So we have a unit
length stick.
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So let's actually give
a coordinate system.
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The stick goes from 0 to 1.
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And let's say that we break
it at these two points.
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So the first point where we
choose, we'll call that x.
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So that's the first point that
we choose to break it.
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And then the second point we
choose, we'll call that y.
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Now note that I've drawn it so
that x is to the left of y.
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But it could actually be the
case that the first point I
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chose is here and the
second point that I
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chose is to the left.
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But for now, let's
first assume that
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this scenario holds.
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That the first point is to the
left of the second point.
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So under this assumption,
we can see that--
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from the definition of these
random variables-- we can
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actually see that the
lengths are given
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by these three lengths.
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So the lengths are x, the left
most piece has length x.
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The second, middle piece
has length y minus x.
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And the last piece has
length 1 minus y.
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And now let's recall our
three conditions.
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So the conditions were that any
two of these, the sum of
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any two lengths, has
to be at least--
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has to be greater than the
length of the third piece.
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So let's do these together.
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So x plus y minus x has to be
greater than 1 minus y.
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So with these two pieces you
can cover this third piece.
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We also need that with the first
and third pieces, we can
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cover the middle piece.
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And we need with the second and
third pieces, we can cover
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the first piece.
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Now this looks kind of messy,
but in fact we can actually
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simplify this.
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So this actually simplifies.
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x minus x, that disappears.
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And so this actually simplifies
to 2y has
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to be at least 1.
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Or even more simply, y has
to be greater than 1/2.
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What about this one?
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This one, we can rearrange
things again.
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x we can move over.
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y we can move over here.
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And we get that 2x plus 1 has
to be greater than 2y.
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Or put in other words, y is
less than x plus 1/2.
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And for the last one, again
we can simplify.
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The y's cancel each other out.
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And we're left with
2x is less than 1.
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Or x is less than 1/2.
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So these are our three
conditions
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that need to be satisfied.
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So now we just have to figure
out what's the probability
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that this is actually
satisfied?
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Now let's go back to original
definition and see what are
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the actual distributions
for these random
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variables, x and y.
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Remember, we defined them to
be x is the location of the
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first break and y is the
location of the second break.
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And as we said in the problem,
these are chosen uniformly at
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random and they're
independent.
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And so we can actually draw
out their joint PDF.
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So x and y, you can cover
any point in the square.
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And moreover, it's actually
uniform within the square.
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Because each one is chosen
uniformly at random and
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they're independent.
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So it's anywhere in here.
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And so what do we need to do?
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We just need to identify, what
is the probability that these
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three conditions hold?
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Rewrite this, line these up.
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So these are our three
conditions that we need.
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And now remember, we're still
working under the assumption
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that the first point that we
chose is actually to the left
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of the second point.
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So what does that mean?
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That means that we are
actually in this top
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triangle, top half--
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x is less than y.
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All right, so what do we need?
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We need y to be at least
1/2, so here's 1/2.
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So we need y to be
above this line.
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We need x to be less than 1/2.
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So we need x to be to
the left of here.
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So now so far we're stuck
in this upper square.
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And the last thing we need is y
to be less than x plus 1/2.
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What is y?
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The line y equals x and a 1/2,
x plus 1/2, is this one.
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So y has to be less than that,
so it would have to be in this
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triangle here.
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So these three conditions tell
us that in order for us to
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have a triangle we need to for
x and y to fall jointly in
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this small triangle here.
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Now because the joint
distribution is uniform, we
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know that the density
is just 1, right?
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Because the area
here is just 1.
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So the height is
just 1 as well.
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And so the density, or the
probability of falling within
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this small triangle, is just
going to be also the area of
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this triangle.
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And what is the area
of this triangle?
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Well, you can fit 8 of these
triangles in here, or you
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could think of it as 1/2
times 1/2 times 1/2.
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So the area is just 1/8.
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So assuming that x is less than
y, then the probability
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of forming a triangle is 1/8.
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Now, that's only half
this story, though.
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Because it's possible that when
you chose these two break
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points that we actually had
the opposite result.
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That x, the point that you
chose first, falls to the
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right of the point that
you chose second.
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In which case everything
kind of flips.
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Now we assume that y is less
than x, which means that now
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we're in this lower triangle
in the square.
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Now we can go through this
whole exercise again.
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But really, what we can see is
that all we've really done is
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just swap the names.
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Instead of having x and y we now
call x-- we call x y and
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we call y x.
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And so if we just swap names,
we can see that--
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let's just fast forward through
all these steps and
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see that we could just swap
names here, too, as well, in
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the three conditions.
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So instead of needing y to be
greater than 1/2, we just need
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x to be greater than 1/2.
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Instead of having x less than
1/2, we need y less than 1/2.
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We also swap this.
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So we need x to be less than y
plus 1/2 or y is greater than
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x minus 1/2.
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All right, now let's figure out
what this corresponds to.
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We need x to be greater than
1/2, so it needs to be to the
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right of here.
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We need y to be less than
1/2, so we need it to
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be below this line.
185
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And we need y to be greater
than x minus 1/2.
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What is the line y equals
x minus 1/2?
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That is this line here.
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And we need y to be greater than
that, so it needs to be
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above this line.
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00:10:57,510 --> 00:11:02,930
And so we get that this is the
triangle, the small triangle
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that we need in this case.
192
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And notice that it's exactly
the same area
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as this one, right?
194
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And so we get another
contribution of 1/8 here.
195
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So the final answer is
1/8 plus 1/8 is 1/4.
196
00:11:19,490 --> 00:11:23,430
So the probability of forming
a triangle using this three
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00:11:23,430 --> 00:11:26,270
pieces is exactly 1/4.
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00:11:26,270 --> 00:11:31,250
And so notice that we've done
is, you've set things up very
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00:11:31,250 --> 00:11:33,660
methodically in the beginning
by assigning
200
00:11:33,660 --> 00:11:34,550
these random variables.
201
00:11:34,550 --> 00:11:37,540
And you consider different
cases.
202
00:11:37,540 --> 00:11:41,020
Because you don't actually know
the order in which x and
203
00:11:41,020 --> 00:11:44,630
y might fall, let's just assume
that one particular
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00:11:44,630 --> 00:11:46,190
order and work from there.
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And then do the other
case, as well.
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And it just so happened that
because of the symmetry of the
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00:11:52,770 --> 00:11:55,570
problem the second case was
actually very simple.
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00:11:55,570 --> 00:11:58,820
We could just see that it is
actually symmetric and so we
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00:11:58,820 --> 00:12:01,160
get the same answer.
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So this is kind of an
interesting problem because
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00:12:03,930 --> 00:12:06,550
it's actually a practical
application of something that
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00:12:06,550 --> 00:12:07,540
you might actually do.
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00:12:07,540 --> 00:12:11,460
And you can see that just by
applying these probability
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00:12:11,460 --> 00:12:13,540
concepts you can actually--