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Hi.

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In this problem, we'll be going
over practice with the

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calculation of conditional
probabilities.

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We'll start with a game where
our friend Alice will be

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tossing a coin with certain
bias of having a head, and

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tosses this coin twice.

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And we're interested in
knowing, what's the

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probability that both
coin tosses will

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end up being a head?

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The first step we're going to
do is to convert the problem

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into a mathematical form
by defining two

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events as the following.

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Event A is where the first
coin toss is a head.

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And similarly, event B will be
having the second coin toss

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also being a head.

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Having these two events will
allow us to say, well, the

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event that A intersection B will
be the event that both

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coin tosses are a head.

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And we'd like to know the
probability of such an event.

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In particular, the probability
of A and B will be calculated

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under two types of
information.

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In the first case, we'll be
conditioning on that we know

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the first coin toss is a head.

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I'd like to know what the
probability of A and B is.

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In the second case, we know that
at least one of the two

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coin tosses is a head expressed
in the form A union

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B.

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And under this conditioning,
what is the probability of A

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and B, A intersection B?

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So Alice, in this problem,
says-- well, her guess will be

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that the first quantity
is no smaller

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than the second quantity.

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Namely, knowing that the first
coin toss is a head somehow

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more strongly implies that both
coin tosses will be a

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head, compared to the case that
we only know at least one

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of the two coin tosses
is a head.

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And we'd like to verify if this

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inequality is indeed true.

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To do so, let's just use the
basic calculation of

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conditional probability.

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Now, from the lectures, you've
already learned that to

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calculate this quantity, we'll
write out a fraction where the

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numerator is the probability of
the intersection of these

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two events.

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So we have A intersect B
intersection A divided by the

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probability of the event that
we're conditioning on, which

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is A.

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Now, the top quantity, since
we know that A and B is a

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subset of event A, then taking
the intersection of these two

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quantities will just give
us the first event.

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So we have A and B. And the
bottom is still probability of

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A.

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Let's do the same thing for
the second quantity here.

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We have the top probability of A
and B intersection the event

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A union B, and on the bottom,
probability of the event A and

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B. Again, we see the event A and
B is a subset of the event

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A union B. So the top will be
A and B. And the bottom--

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A union B.

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OK, now let's stop
for a little bit.

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We've computed the probability
for each expression in the

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following fractional form.

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And we observed that for
both fractions, the

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numerator is the same.

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So the numerator is a
probability of A and B. And

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the denominator in the first
case is probably of A, and the

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second case, probably
of A union B.

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Since we know that A is a subset
of the event A union B,

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and by the monotonicity of
probabilities, we know that

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the probability of A is hence no
greater than a probability

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of A union B. Substituting
this back into these

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expressions, we know that
because they lie in the

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denominators, the first
expression is indeed no

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smaller than the second
expression.

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So our friend Alice
was correct.

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So throughout this problem, we
never used the fact that the

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probability of a particular coin
toss results, let's say,

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in a head is a certain number.

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Actually, this bias for the
coin is irrelevant.

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Whether the coin is fair
or unfair, this

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fact is always true.

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So indeed, it does
not depend on the

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probability of the coin.

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But if you're really curious
what happens when the coin is

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fair, we can plug
in the numbers.

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And here, we're assuming the
coin is fair, which means

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probability of having
a head is 1/2.

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Then, we'll see after going
through the calculations that

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the first probability is 1/2,
whereas the second probability

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is 1/3, which means, in this
case, the [? dominance ?]

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actually is strict.

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So the first one is strictly
greater than

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the second one, OK?

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So this completes the first
part of the problem.

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How do we generalize this into
more general settings?

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There are multiple ways,
but we'll go over

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one particular form.

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And to do so, we'll be defining
three events somewhat

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more abstractly.

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Let's say we have
three events--

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C, D, and E. Imagine any event,
but all three events

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have to satisfy the following
condition.

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First, event D will be a subset
of E. And second, the

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intersection of C and
D is equal to the

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intersection of C and E, OK?

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So this will be our
choice events.

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And let's see a particular
example.

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Let's say you have a sample
space here and some event E.

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Now, by the first condition, D
will have to lie somewhere in

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E. For the second condition,
we'll pick some event C such

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that this is true.

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And one way to do so is simply
picking C that lies within

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both D and E. And you can see
C intersection D will be C.

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And C intersection E will still
be C. Hence, the second

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equality is true.

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So if both equalities are true,
we have the following

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relationship, that the
probability of C conditional

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on D will be no smaller than
the probability of C

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conditional on event E. And this
will be the more general

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form of the inequality
that we saw before.

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So first of all, the
way to prove this

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is in fact the same.

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We simply write out the
value of this using

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the fractional form.

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And based on these two facts,
we can arrive at this

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equation, which I shall
now go over.

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But just to see why this form
is more general, in fact, if

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we-- say we let C be the event
A intersection B, D be the

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event A, and E be the event A
and B where A and B are the

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events that we defined
earlier.

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We can verify that, indeed,
these conditions are true,

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namely D is a subset of E.
Because A is a subset of A

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union B, and C is a subset of
both D and E. And hence,

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condition two is also true.

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And if that's the case, we
will actually recover the

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result we got earlier for events
A and B. And hence,

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this equation here is
a more general form.

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So that's the end
of the problem.

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See you next time.

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