WEBVTT
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We will now use counting
to solve a simple
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probabilistic problem.
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We have in our hands an ordinary
six-sided die which
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we are going to roll
six times.
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So this is our probabilistic
experiment.
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And we're interested in the
probability of a certain
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event, the event that
the six rolls result
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in different numbers.
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So let us give a name to that
event and call it event A. So
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we wish to calculate the
probability of this event.
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But before we can even get
started answering this
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question, we need a
probabilistic model.
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We need to make some
assumptions, and the
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assumption that we're going to
make is that all outcomes of
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this experiment are
equally likely.
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This is going to place us within
a discrete uniform
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probabilistic model so
that we can calculate
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probabilities by counting.
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In particular, as we discussed
earlier in this lecture, the
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probability of an event A is
going to be the number of
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elements of the set A, the
number of outcomes that make
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event A occur, divided by the
total number of possible
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outcomes, which is the
number of elements
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in our sample space.
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So let us start with the
denominator, and let us look
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at the typical outcomes
of this experiment.
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A typical outcome is something
like this sequence,
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2, 3, 4, 3, 6, 2.
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That's one possible outcome.
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How many outcomes of this
kind are there?
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Well, we have 6 choices for the
result of the first roll,
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6 choices for the result of the
second roll, and so on.
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And since we have a total of 6
rolls, this means that there
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is a total of 6 to the 6th
power possible outcomes,
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according to the Counting
Principle.
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And by the way, since we have so
many possible outcomes and
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we assume that they are
equally likely, the
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probability of each one
of them would be 1
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over 6 to the 6th.
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Incidentally, that's the same
number, the same answer, you
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would get if you were to assume,
instead of assuming
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directly that all outcomes are
equally likely, to just assume
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that the different rolls are
rolls of a fair six-sided die,
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so the probability of getting a
2 is 1/6, and also that the
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different rolls are independent
of each other.
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So in that case, the
probability, let's say, of
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this particular sequence would
be the probability of
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obtaining a 2, which is 1/6,
times the probability that we
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get a 3 at the next roll, which
is 1/6, times 1/6 times
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1/6 and so on, and we get the
same answer, 1 over 6 to 6th.
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So we see that this assumption
of all outcomes being equally
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likely has an alternative
interpretation in terms of
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having a fair die which is
rolled independently 6 times.
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Now, let us look at the event
of interest, A. What is a
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typical element of A?
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A typical element of A is a
sequence of 6 rolls in which
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no number gets repeated.
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So, for example, it could be a
sequence of results of this
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kind, where each number
appears just once.
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So all the numbers appear
exactly once in this sequence.
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So what we need here is
basically to have a
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permutation of the numbers
1 up to 6.
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So these 6 numbers have to
appear in an arbitrary order.
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In how many ways can we
order 6 elements?
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This is the number of
permutations of a set of 6
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elements and, as we discussed
earlier, this
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is equal to 6 factorial.
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So we have now counted the
number of outcomes that make
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event A happen, which
is 6 factorial.
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And by calculating this ratio,
we have obtained the
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probability of the
desired event.
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You can now pause and try
to solve a problem
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of a similar kind.