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Let us now move from the
abstract to the concrete.

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Recall the example that we
discussed earlier where we

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have two rolls of a
tetrahedral die.

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So there are 16 possible
outcomes

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illustrated in this diagram.

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To continue, now we need to
specify a probability law,

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some kind of probability
assignment.

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To keep things simple, we're
going to make the assumption

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that the 16 possible outcomes
are all equally likely.

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And each outcome has a
probability of 1 over 16.

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Given this assumption, we will
now proceed to calculate

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certain probabilities.

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Let us look first at the
probability that X, which

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stands the result of the first
roll, is equal to 1.

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The way to calculate this
probability is to identify

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what exactly that event is in
our picture of the sample

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space, and then calculate.

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The event that X is equal to 1
can happen in four different

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ways that correspond to these
four particular outcomes.

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Each one of these outcomes has
a probability of 1 over 16.

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The probability of this event
is the sum of the

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probabilities of the outcomes
that it contains.

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So it is 4 times 1 over 16,
equal to one fourth.

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Let now Z stand for the smaller
of the two numbers

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that came up in our two rolls.

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So for example, if X is 2 and
Y is equal to 3, then Z is

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equal to 2, which is the
smaller of the two.

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Let us try to calculate the
probability that the smaller

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of the two outcomes
is equal to 4.

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Now for the smaller of the two
outcomes to be equal to 4, we

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must have that both X and
Y are equal to 4.

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So this outcome here is the only
way that this particular

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event can happen.

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Since there's only one outcome
that makes the event happen,

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the probability of this event
is the probability of that

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outcome and is equal
to 1 over 16.

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For another example, let's
calculate the probability that

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the minimum is equal to 2.

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What does it mean that the
minimum is equal to 2?

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It means that one of the dice
resulted in a 2, and the other

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die resulted in a number
that's 2 or larger.

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So we could have both
equal to 2.

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We could have X equal
to 2, but Y larger.

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Or we could have Y equal to
2 and X something larger.

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This green event, this green
set, is the set of all

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outcomes for which the
minimum of the two

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rolls is equal to 2.

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There's a total of five
such outcomes.

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Each one of them has
probably 1 over 16.

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And we have discussed that for
finite sets, the probability

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of a finite set is the sum of
the probabilities of the

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elements of that set.

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So we have five elements
here, each one with

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probability 1 over 16.

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And this is the answer
to this problem.

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This particular example that we
saw here is a special case

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of what is called a discrete
uniform law.

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In a discrete uniform law,
we have a sample

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space which is finite.

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And it has n elements.

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And we assume that these n
elements are equally likely.

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Now since the probability of
omega, the probability of the

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entire sample space, is equal to
1, this means that each one

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of these elements must have
probability 1 over n.

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That's the only way that the
sum of the probabilities of

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the different outcomes would be
equal to 1 as required by

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the normalization axiom.

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Consider now some subset of the
sample space, an event A

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that, exactly k elements.

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What is the probability
of the set A?

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It's the sum of the
probabilities of its elements.

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There are k elements.

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And each one of them has a
probability of 1 over n.

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And this way we can find the
probability of the set A.

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So when we have a discrete
uniform probability law, we

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can calculate probabilities by
simply counting the number of

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elements of omega, which is n,
finding the number n, and

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counting the number of elements
of the set A. That's

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the reason why counting
will turn out to be

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an important skill.

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And there will be a whole
lecture devoted to this

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particular topic.