[MUSIC]

Our goal in this recording is to develop some simple probability distributions.

So recall that the sample space was the set of all

possible outcomes in our experiment.

And hence, really, by definition,

the occurrence of the sample space would equate to a certain event.

Namely, the probability of S, the sample space would be equal to 1.

So our goal now is to consider a probability distribution whereby we take

this one unit of probability, the certain event that the sample space would occur.

And we would like to break this up and to distribute it across all possible outcomes

within that experiment, i.e., across every possible value within that sample space,

to reflect how likely those different outcomes are.

Now, of course, there are many different ways we could distribute this unit of

probability, yet in a given application, we would want to do it in a way such that

it fairly reflects the likelihood, the occurrence, of each of those events.

So let's start with some baby steps, and

consider perhaps the simplest possible example.

So let's imagine your sample space has N possible outcomes within it.

And we'll consider the very special case where each of these outcomes is equally

likely.

So the simplest example we could consider would be that of a fair coin.

Remember, there are two outcomes when you toss a coin, head and tails, and

let's assume that it is a fair coin, not a biased one, such that heads and

tails are equally likely.

So here, N, the size of our sample space, is equal to 2.

And we said some event or set of interest would be some subset of that sample space.

So let's suppose we are interested in the outcome when we obtain heads.

So, in a classical world of probability, we can determine theoretically

the probability of some particular event to be the number of all of the elementary

outcomes within the sample space which agree with that event of interest.

So let's break it down and consider the simple coin toss example.

So for a fair coin, two equally likely outcomes of heads and tails.

And if we define the event A to be tossing, let's say heads, then only one

of those two outcomes, the H, out of the H and T, equate to getting heads.

And so we would assign the probability of A, i.e.,

the probability of getting heads, to be 1 over 2.

So more generally, we would say that when our sample space has equally

likely outcomes and then if little n of them reflect the number of those

which correspond to the event of interest A, then the probability of

A would be n over N, and this would reflect its probability of occurrence.

Well, we talk about baby steps, so we did a little bit of crawling.

Let's try and sort of stand up a little bit now and

consider maybe a larger example where the sample space has more possible outcomes.

We've considered tossing a coin.

Let's now consider rolling a fair die.

So we saw this in the previous session.

So in this case, our sample space has six possible outcomes,

those positive integers 1, 2, 3, 4, 5, and 6.

And provided it is a fair die,

then each of those scores would be considered to equally likely.

So what would our probability distribution look like?

Well, in this setting,

we want to construct our probability distribution in tabular form.

In that sense, we would have two rows.

Our top row would reflect every possible outcome of this experiment,

and then in the second row,

assigned to each of these values would be its probability of occurrence.

So, when rolling a fair die, we know that there were six possible values for

this variable.

And indeed let's now introduce this notation of X, a capital

X to denote the random variable which here would equate to the score on a die.

So we know for

this random experiment that there are six possible values that this score can take.

So this random variable X is either going to be a 1, 2, 3, 4, 5, or 6.

So assuming this being a fair die, rather than a loaded or biased die,

then each of those six outcomes would be equally likely.

So in our probability distribution, if we consider the probability of rolling a 1.

Well, if we consider our little n over a big N notation for

the probability of event A when the outcomes are equally likely,

then our capital N in this case is equal to 6.

Only one of these six outcomes corresponds to rolling a 1, and

hence the probability of a 1 would be 1 over 6.