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Let us now put to use our
understanding of the

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coin-tossing model and the
associated binomial

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probabilities.

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We will solve the following
problem.

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We have a coin, which
is tossed 10 times.

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And we're told that exactly
three out of the 10 tosses

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resulted in heads.

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Given this information, we would
like to calculate the

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probability that the first
two tosses were heads.

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This is a question of
calculating a conditional

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probability of one event
given another.

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The conditional probability of
event A, namely that the first

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two tosses were heads, given
that another event B has

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occurred, namely that we
had exactly three heads

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out of the 10 tosses.

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However, before we can start
working towards the solution

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to this problem, we need to
specify a probability model

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that we will be working with.

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We need to be explicit about
our assumptions.

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To this effect, let us introduce
the following

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assumptions.

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We will assume that the
different coin tosses are

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independent.

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In addition, we will assume
that each coin toss has a

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fixed probability, p, the same
for each toss, that the

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particular toss results
in heads.

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These are the exact same
assumptions that we made

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earlier when we derived the
binomial probabilities.

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And in particular, we have the
following formula that if we

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have n tosses, the probability
that we obtain exactly k heads

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is given by this expression.

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So now, we have a model in place
and also the tools that

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we can use to analyze this
particular model.

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Let us start working
towards a solution.

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Actually, we will develop two
different solutions and

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compare them at the end.

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The first approach,
which is the

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safest one, is the following.

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Since we want to calculate a
conditional probability, let

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us just start with the
definition of conditional

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probabilities.

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The conditional probability of
an event given another event

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is the probability that both
events happen, divided by the

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probability of the conditioning
event.

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Now, let us specialize to the
particular example that we're

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trying to solve.

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So in the numerator, we're
talking about the probability

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that event A happens and
event B happens.

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What does that mean?

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This means that event
A happens--

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that is, the first two tosses
resulted in heads, which I'm

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going to denote symbolically
this way.

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But in addition to that,
event B happens.

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And event B requires that there
is a total of three

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heads, which means that we
had one more head in

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the remaining tosses.

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So we have one head in tosses
3 all the way to 10.

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As for the denominator, let's
keep it the way it is for now.

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So let's continue with
the numerator.

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We're talking about the
probability of two events

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happening, that the first two
tosses were heads and that in

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tosses 3 up to 10, we had
exactly one head.

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Here comes the independence
assumption.

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Because the different tosses
are independent, whatever

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happens in the first two tosses
is independent from

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whatever happened in
tosses 3 up to 10.

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So the probability of these two
events happening is the

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product of their individual
probabilities.

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So we first have the probability
that the first two

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tosses were heads, which
is p squared.

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And we need to multiply it
with the probability that

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there was exactly one head
in the tosses numbered

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from 3 up to 10.

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These are eight tosses.

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The probability of one head in
eight tosses is given by the

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binomial formula, with k equal
to 1 and n equal to 8.

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So this expression, this part,
becomes 8 choose 1, p to the

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first power times 1 minus
p to the seventh power.

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So this is the numerator.

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The denominator is
easier to find.

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This is the probability
that we had

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three heads in 10 tosses.

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So we just use this formula.

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The probability of three heads
is given by: 10 tosses choose

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three, p to the third, times 1
minus p to the seventh power.

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And here we notice that terms
in the numerator and

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denominator cancel out, and we
obtain 8 choose 1 divided by

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10 choose 3.

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And to simplify things
just a little more,

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what is 8 choose 1?

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This is the number of ways that
we can choose one item

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out of eight items.

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And this is just 8.

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And let's leave the denominator
the way it is.

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So this is the answer to the
question that we had.

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And now let us work towards
developing a second approach

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towards this particular
answer.

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In our second approach, we start
first by looking at the

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sample space and understanding
what

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conditioning is all about.

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In our model, we have
a sample space.

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As usual we can denote
it by omega.

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And the sample space contains a
bunch of possible outcomes.

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A typical outcome is going to
be a sequence of length 10.

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It's a sequence of
heads or tails.

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And it's a sequence that
has length 10.

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We want to calculate conditional
probabilities.

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And this places us in a
conditional universe.

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We have the conditioning event
B, which is some set.

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And conditional probabilities
are probabilities defined

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inside this set B and define
the probabilities, the

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conditional probabilities of
the different outcomes.

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What are the elements
of the set B?

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A typical element of the set B
is a sequence, which is, again

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of length 10, but has
exactly three heads.

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So these are the three-head
sequences.

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Now, since we're conditioning
on event B, we can just work

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with conditional
probabilities.

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So let us find the conditional
probability law.

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Recall that any three-head
sequence has the same

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probability of occurring in
the original unconditional

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probability model, namely as
we discussed earlier, any

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particular three-head sequence
has a probability equal to

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this expression.

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So three-head sequences are
all equally likely.

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This means that the
unconditional probabilities of

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all the elements of
B are the same.

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When we construct conditional
probabilities given an event

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B, what happens is that the
ratio or the relative

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proportions of the probabilities
remain the same.

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So conditional probabilities
are proportional to

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unconditional probabilities.

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These elements of B were
equally likely in

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the original model.

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Therefore, they remain equally
likely in the conditional

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model as well.

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What this means is that the
conditional probability law on

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the set B is uniform.

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Given that B occurred, all the
possible outcomes now have the

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same probability.

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Since we have a uniform
probability law, this means

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that we can now answer
probability

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questions by just counting.

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We're interested in the
probability of a certain

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event, A, given that
B occurred.

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Now, given that B occurred, this
part of A cannot happen.

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So we're interested in the
probability of outcomes that

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belong in this shaded region,
those outcomes that belong

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within the set B. To find the
probability of this shaded

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region occurring, we just need
to count how many outcomes

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belong to the shaded region and
divide them by the number

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of outcomes that belong
to the set B.

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That is, we work inside this
conditional universe.

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All of the elements in this
conditional universe are

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equally likely.

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And therefore, we calculate
probabilities by counting.

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So the desired probability is
going to be the number of

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elements in the shaded region,
which is the intersection of A

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with B, divided by the number of
elements that belong to the

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set B.

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How many elements are there in
the intersection of A and B?

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These are the outcomes or
sequences of length 10, in

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which the first two tosses
were heads--

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no choice here.

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And there is one more head.

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That additional head can appear
in one out of eight

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possible places.

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So there's eight possible
sequences that have the

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desired property.

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How many elements are
there in the set B?

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How many three-head sequences
are there?

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Well, the number of three-head
sequences is the same as the

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number of ways that we can
choose three elements out of a

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set of cardinality 10.

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And this is 10 choose 3, as
we also discussed earlier.

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So this is the same answer as
we derived before with our

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first approach.

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So both approaches, of course,
give the same solution.

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This second approach is a little
easier, because we

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never had to involve any
p's in our calculation.

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We go to the answer directly.

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The reason that this approach
worked was that the

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conditional universe, the
event B, had a uniform

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probability law on it.