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In this video, we'll look at an
example in which we compute

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the expectation and cumulative
density function of a mixed

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random variable.

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The problem is as follows.

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Al arrives at some bus stand
or taxi stand at a given

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time-- let's say time
t equals 0.

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He finds a taxi waiting for him
with probability 2/3 in

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which he takes it.

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Otherwise, he takes the next
arriving taxi or bus.

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The time that the next taxi
arrives between 0 and 10

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minutes, and it's uniformly
distributed.

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The next bus leaves exactly
in 5 minutes.

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So the question is, if X is Al's
waiting time, what is the

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CDF and expectation of X?

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So one way to view this problem
that's convenient is

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the tree structure.

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So I've drawn it for you here
in which the events of

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interest are B1, B2, and B3,
B1 being Al catches the

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waiting taxi, B2 being Al
catches the next taxi, which

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arrives between 0 and 5 minutes,
and B3 being Al

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catches the bus at the
time t plus 5.

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Notice that these three
events are disjoint.

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So Al catching the waiting taxi
means he can't catch the

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bus or the next arriving taxi.

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And it also covers the entire
set of outcomes.

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So, in fact, B1, B2, and
B3 are a partition.

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So let's look at the relevant
probabilities.

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Whether or not B1 happens
depends on whether or not the

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taxi's waiting for Al.

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So if the taxi is waiting for
him, which happens with 2/3

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probability, B1 happens.

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Otherwise, with 1/3 probability,
we see whether or

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not a taxi is going to arrive
between 0 and 5 minutes.

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If it arrives, which is going
to happen with what

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probability?

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Well, we know that the next
taxi is going to arrive

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between 0 and 10 minutes
uniform.

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It's a uniform distribution.

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And so half the mass is going
to be between 0 and 5.

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And the other half is going
to be between 5 and 10.

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And so this is going
to be 1/2 and 1/2.

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And let's look at what
X looks like.

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If B1 happens, Al isn't waiting
at all, so x is going

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to be equal to 0.

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If B3 happens, which is the
other easy case, Al's going to

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be waiting for 5 minutes
exactly.

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And if B2 happens, well, it's
going to be some value

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between 0 and 5.

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We can actually draw the
density, so let's see if we

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can do that here.

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The original next taxi was
uniformly distributed

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between 0 and 10.

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But now, we're told two
pieces of information.

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We're told that B2 happens,
which means that there's no

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taxi waiting, and the
next taxi arrives

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between 0 and 5 minutes.

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Well, the fact that there was no
taxi waiting has no bearing

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on that density.

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But the fact that the next taxi
arrives between 0 and 5

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does make a difference, because
the density then is

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going to be definitely 0 in any
region outside 0 and 5.

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Now, the question is, how
is it going to look

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between 0 and 5?

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Well, it's not going
to look crazy.

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It's not going to look like
something different.

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It's simply going to be a scale
version of the original

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density between 0 and 5.

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You can verify this by looking
at the actual formula for when

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you condition events on
a random variable.

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Here, it's going to be 1/5
in order for this to

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integrate out to 1.

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And now we can jump right into
figuring out the expectation.

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Now, notice that X is actually
a mixed random variable?

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What does that mean?

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Well, X either takes on values
according to either a discrete

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probability law or
a continuous one.

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So if B1 happens, for example, X
is going to be exactly equal

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to 0 with probability
1, which is a

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discrete probability problem.

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On the other hand, if B2
happens, then the value of X

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depends on the density, which
is going to be continuous.

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So X is going to
be a continuous

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random variable here.

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So how do you define an
expectation in this case?

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Well, you can do it so that
it satisfies the total

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expectation theorem, which means
that the expectation of

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X is the probability of B1 times
the expectation given B1

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plus the probability of B2 times
the expectation given B2

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plus the probability
of B3 times the

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expectation given B3.

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So this will satisfy the total
expectation theorem.

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So the probability of B1 is
going to be exactly 2/3.

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It's simply the probability
of a taxi waiting for Al.

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The expected value of X-- well,
when B1 happens, X is

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going to be exactly
equal to 0.

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So the expected value is
also going to be 0.

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The probability of B2 happening
is the probability

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of a taxi not being there times
the probability of a

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taxi arriving between 0 and 5.

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It's going to be
1/3 times 1/2.

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And the expected value of X
given B2 is going to be the

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expected value of
this density.

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The expected value of this
density is the midpoint

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between 0 and 5.

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And so it's going to be 5/2.

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And the probability of B3 is
going to be 1/3 times 1/2.

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Finally, the expected
value of X given B3.

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Well, when B3 happens,
X is going to be

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exactly equal to 5.

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So the expected value is
also going to be 5.

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Now we're left with 5/12
plus 5/6, which

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is going to be 15/12.

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And we can actually fill that
in here so that we can clear

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up the board to do
the other part.

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Now we want to compute the CDF
of X. Well, what is the CDF?

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Well, the CDF of X is going to
be equal to the probability

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that the random variable
X is less than or equal

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to some little x.

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It's a constant [INAUDIBLE].

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Before we jump right in, let's
try to understand what's the

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form of the CDF.

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And let's consider some
interesting cases.

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You know that the random
variable X, the waiting time,

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is going to be somewhere
between 0 and 5, right?

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So let's consider what happens
if little x is going to be

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less than 0.

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That's basically saying, what's
the probability of the

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random variable X being
less than some number

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that's less than 0?

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Waiting time can't be negative,
so the probablility

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of this is going to be 0.

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Now, what if X is between
equaling 0 and

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strictly less than 5?

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In that case, either X can fall
between 0 and 5 according

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to this case, in the case
of B2, or X can be

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exactly equal to 0.

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It's not clear.

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So let's do that later.

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Let's fill that in later.

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What about if x is greater
than or equal to 5?

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Little x, right?

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That's the probability that the
random variable X is less

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than some number that's bigger
than or equal to 5.

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The waiting time X, the random
variable, is definitely going

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to be less than or equal to 5,
so the probability of this is

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going to be 1.

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So now this case.

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How do we do it?

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Well, let's try to use a similar
kind of approach that

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we did for the expected value
and use the total probability

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theorem in this case.

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So let's try to review this.

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First of all, let's assume that
this is true, that little

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x is between 0 and
5, including 0.

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And let's use the total
probability theorem, and use

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the partitions B1, B2, and B3.

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So what's the probability
of B1?

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It's the probability that Al
catches waiting taxi, which

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happens with probability 2/3.

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What's the probability that the
random variable X, which

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is less than or equal to little
x under this condition,

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when B1 happens?

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Well, if B1 happens, then random
variable X is going to

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be exactly equal to 0, right?

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So in that case, it's definitely
going to be less

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than or equal to any value
of x, including 0.

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So the probability will be 1.

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What's the probability
that B2 happens now?

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The probability that B2 happens
is 1/3 times 1/2, as

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we did before.

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And the probability that the
random variable X is less than

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or equal to little x
when B2 happens.

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Well, if B2 happens, this
is your density.

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And this is our condition.

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And so x is going to
be somewhere in

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between these spots.

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And we'd like to compute what's
the probably that

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random variable X is less than
or equal to little x.

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So we want this area.

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And that area is going to have
height of 1/5 and width of x.

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And so the area's going
to be 1/5 times x.

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And finally, the probability
that B3 happens is going to be

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1/3 times 1/2 again times the
probability that the random

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variable X is less than or equal
to little x given B3.

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Well, when B3 happens, X is
going to be exactly 5 as a

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random variable.

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But little x, you know-- we're
assuming in this condition--

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is going to be between 0 and 5,
but strictly less than 5.

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So there's no way that if the
random variable X is 5 and

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this is strictly less than 5,
this is going to be true.

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And so that probability
will be 0.

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So we're now left with
2/3 plus 1/30.

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And now we can fill this in.

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2/3 plus 1/30 x.

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And this is our CDF.

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So now we've finished the
problem, computed the expected

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value here and then the CDF
here, and this was a great

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illustration of how you
would do so for a

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mixed random variable.

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