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The coefficients n-choose-k
that we calculated in the
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previous segment are known as
the binomial coefficients.
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They are intimately related
to certain probabilities
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associated with coin tossing
models, the so-called binomial
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probabilities.
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This is going to
be our subject.
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We consider a coin which we
toss n times in a row,
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independently.
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For each one of the tosses of
this coin, we assume that
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there is a certain probability,
p, that the
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result is heads, which of
course, implies that the
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probability of obtaining tails
in any particular toss is
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going to be 1 minus p.
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The question we want to address
is the following.
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We want to calculate the
probability that in those n
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independent coin tosses,
we're going to
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observe exactly k heads.
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Let us start working our way
towards the solution to this
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problem by looking first
at a simple setting
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and then move on.
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So let us answer this
simple question.
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What is the probability
that we observe
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this particular sequence?
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Of course here we take n equal
to six, and we wish to
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calculate this probability.
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Now, because we have assumed
that the coin tosses are
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independent, we can multiply
probabilities.
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So the probability of this
sequence is equal to the
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probability that the first
toss is heads times the
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probability that the second
toss is tails, which is 1
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minus p, times the probability
that the third toss is tails,
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which is 1 minus p, times the
probability of heads, times
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the probability of
heads, times the
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probability of heads.
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And by collecting terms, this is
p to the 4th times 1 minus
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p to the second power.
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More generally, if I give you a
particular sequence of heads
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and tails, as in this example,
and I ask you, what is the
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probability that this particular
sequence is
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observed, then by generalizing
from this answer or from the
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derivation of this answer, you
see that you're going to get p
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to the power number of heads.
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And the reason is that each
time that there's a head
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showing up in this sequence,
there's a corresponding factor
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of p in this numerical answer.
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And then there are factors
associated with tails.
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Each tail contributes a
factor of 1 minus p.
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And so we're going to have
here 1 minus p to a power
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equal to the number of tails.
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Now, if I ask you about the
probability of a particular
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sequence and that particular
sequence has happened to have
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exactly k heads, what is the
probability of that sequence?
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Well, we already calculated
what it is.
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It is the previous answer,
except we use the symbol k
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instead of just writing out
explicitly "number of heads."
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And the number of tails is the
number of tosses minus how
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many tosses resulted in heads.
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Now, we're ready to consider
the actual problem that we
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want to solve, which
is calculate the
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probability of k heads.
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The event of obtaining
k heads can happen in
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many different ways.
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Any particular k-head sequence
makes that event to occur.
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Any particular k-head sequence
has a probability equal to
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this expression.
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The overall probability of k
heads is going to be the
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probability of any particular
k-head sequence, times the
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number of k-head sequences
that we have.
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Now, the reason why we can carry
out this argument is the
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fact that any k-head sequence
has the same probability.
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Otherwise, we wouldn't be able
to write down an answer which
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is just the product
of two terms.
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But because every k-head
sequence has the same
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probability, to find the overall
probability, we take
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the probability of each one of
them and multiply it with the
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number of how many
of these we have.
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So to make further progress, now
we need to calculate the
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number of possible
k-head sequences.
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How many are there?
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Well, specifying a k-head
sequence is
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the same as the following.
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You think of having
n time slots.
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These time slots corresponds
to the different
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tosses of your coin.
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And to specify a k-head
sequence, you need to tell me
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which ones of these slots happen
to contain a head.
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You need to tell me
k of those slots.
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So in other words, what you're
doing is you're specifying a
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subset of the set of
these n slots, a
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subset that has k elements.
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You need to choose k of the
slots out of the n and tell me
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that those k slots have heads.
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That's the way of specifying a
particular k-head sequence.
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So what's the number of
k-head sequences?
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Well, it's the same as the
number of ways that you can
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choose k slots out of the n
slots, which is our binomial
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coefficient, n-choose-k.
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Therefore, the answer to our
problem is this expression
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here, times n-choose-k, which
is shown up here.
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At this point, we can pause and
consider a simple question
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to check your understanding of
the binomial probabilities.