As I am covering this topic now with many of my students, let’s start a series on statistics.

The first concept taught when introducing statistics to students is that of probability. Let’s start with the *experiment* of rolling a die. I italicise *experiment* because it is a formal term in statistics. I will italicise other terms in this post.

If we are interested in the *outcome* or the *event* of rolling a “3”, what is the probability of that occurring? As there are six possible outcomes, all equally likely, and “3” is just one of them, then the probability is 1/6 or 1 out of 6. As maths likes to use shorthand notation to represent concepts, lets notationise (my word) this.

Let *A* represent the event of rolling a “3”. The probability of this is represented by *P*(*A*). The probability of rolling a “3” is the number of ways a “3” can occur (one way) divided by the total number of things that can occur (six). So to generalise this, for experiments where all outcomes are equally likely, the probability of an event *A* is

Now I’ve introduced some new notation here. *n*(*A*) is notation that means “number of ways *A* can occur”. The Greek letter xi, ????, is the set of all the things that can happen. In this case, ???? ={1, 2, 3, 4, 5, 6}. This is also called the *sample space* of the experiment. So *n*(????) = 6. In our experiment and the event of interest (rolling a “3”), *n*(*A*) = 1 and *n*(????) = 6 so *P*(*A*) = 1/6.

So what is the probability of rolling an even number? Here, *A* = “rolling an even number”. As there are three even numbers, or three ways, that this can occur, then *P*(*A*) = 3/6 = 1/2.

Let me say a few general things about probability. The probability of an event is always a number between 0 and 1 including 0 and 1. At the extreme ends, if *P*(*A*) = 0, then event *A* has no chance of occurring. So in our experiment, if *A* is the event of rolling a “7”, then *P*(*A*) = 0. If *P*(*A*) = 1, then the event is a certainty to happen. If *A* is the event of rolling an odd or even number, then *P*(*A*) = 1.

Now let’s look at a slightly more complex experiment: picking a card from a standard deck of cards. A standard deck of 52 cards has four suits (hearts, clubs, diamonds, spades) of 13 cards each. Each suit consists of an Ace, Jack, Queen, King, and numbered cards 2 to 10. The Jack, Queen, and King cards are called face cards. So this experiment is choosing one card out of a shuffled deck of cards.

If *A* = choosing a Heart, then *P*(*A*) = 13/52 = 1/4 because there are 13 ways to pick a Heart out of 52 ways any card can be picked. Now let *A* = choosing a Jack. Here *P*(*A*) = 4/52 = 1/13 as there are 4 Jacks in a deck of cards. Now let *A* = picking a face card. Then *P*(*A*) = 12/52 =3/13 as there are 12 face cards in a deck.

In my next post, we’ll explore how to handle more complex events like choosing a Queen or a Heart.