Building and Using Uniform Probability Models

A probability model has two parts: a sample space, and a probability for each outcome in it. Every model in this lesson describes one trial of an experiment: one roll, one spin or one draw.

A model must meet two conditions:

• Each probability is a number from $0$ to $1$.
• The probabilities of all the outcomes add to $1$.

Why $1$? On every trial, exactly one outcome from the sample space happens. So it is sure that some outcome on the list happens, and an event that is sure to happen has probability $1$. The outcomes share that total of $1$ among them.

Some devices give every outcome the same chance. Outcomes like that are equally likely. A fair number cube is one example: fair means that no face is more likely to land on top than any other.

A uniform probability model is a probability model in which every outcome has the same probability. (Uniform means the same throughout.) Use one only when the description of the device gives you a reason to say the outcomes are equally likely.

What is that shared probability? Let $n$ be the number of outcomes in the sample space, and let $p$ be the probability of each outcome. The $n$ equal probabilities add to $1$, so $n \times p = 1$. Because $n$ counts at least one outcome, $n$ is not $0$, so you can divide both sides by $n$ to get $p = \frac{1}{n}$.

In words: $n$ equally likely outcomes share the total of $1$ equally, so each one gets $\frac{1}{n}$. A fair number cube has $n = 6$ outcomes, so each one gets $\frac{1}{6}$. The table shows the whole model.

Now put a model to work. Write $P(\text{event})$ for the probability of an event. An event happens when any one of its outcomes happens, and only one outcome happens on each trial. So the probability of an event is the sum of the probabilities of its outcomes.

Take the fair number cube and the event of rolling a number greater than 4. Its outcomes are 5 and 6, so

$P(\text{greater than } 4) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}$

Dividing the top and bottom by $2$ gives the equal fraction $\frac{1}{3}$.

You added $\frac{1}{6}$ once for each outcome in the event. The same thing happens in every uniform model. Let $k$ be the number of outcomes in the event. In a uniform model with $n$ outcomes, $P(\text{event}) = k \times \frac{1}{n} = \frac{k}{n}$.

Since $k$ is at least $0$ and at most $n$, this probability is always from $0$ to $1$. The rule depends on every outcome having the same probability $\frac{1}{n}$, so it works only when the outcomes are equally likely.

The total of $1$ gives you one more tool. For an event $A$, the event "not $A$" is made of every outcome in the sample space that is not in $A$. Each outcome is in exactly one of the two events, so together they hold every outcome once. That means their probabilities add to $1$:

$P(A) + P(\text{not } A) = 1$

Subtract $P(A)$ from both sides, which keeps the two sides equal:

$P(\text{not } A) = 1 - P(A)$

This is true in any probability model, as long as $A$ and "not $A$" are events from the same sample space.

Try it on the full bag of 10 tiles. $P(\text{not blue}) = 1 - \frac{3}{10} = \frac{10}{10} - \frac{3}{10} = \frac{7}{10}$. Counting agrees: the tiles that are not blue are the 5 red and 2 green tiles, which is 7 of the 10 equally likely tiles.

When you can explain from the device why its outcomes are equally likely, a uniform probability model fits. List the $n$ outcomes, give each one the probability $\frac{1}{n}$, and find the probability of an event by dividing the number of outcomes in it by $n$.

Keep the conditions with the rule: favourable over total is a probability only for equally likely outcomes, every probability is from $0$ to $1$, and the probabilities of all the outcomes add to $1$. If the available outcomes change, as when a tile is not put back, build a new model.

Next, you will develop a probability model from observed frequencies and use a probability model to predict outcomes.