Outcomes, Events, and Probability

Read an outcome and its sample space

Two fair dice can produce seven in several different ways. The gold diagonal collects those ways. Before calculating its probability, we will move into one cell, read its two coordinates, and then return to the whole experiment. The number inside a cell is the sum, not the complete description of the roll.

Here is the cell in row two and column five. The first die shows two; the second shows five. Its displayed sum is seven. Keep the order of those faces: the first coordinate tells us which row, and the second tells us which column.

Now travel to row five and column two. The sum is still seven, but the dice have exchanged their roles. These are different complete outcomes. Recording only the sum would lose that distinction, even though our event will eventually include both cells.

Return to the grid. A complete outcome identifies one cell. An event can collect several cells. This distinction is the starting point for probability: specify what one result records, and only then decide which results answer the question you are asking.

Hold the first die at one and follow the top row. Its partner can be one, two, three, four, five, or six. That gives six possible complete outcomes for this one first face. We are counting possible results, not claiming that six rolls will visit all six.

The next row repeats those six possible second faces with a first face of two. Continue down to the sixth row. There are six rows, each containing six cells, so the whole sample space contains thirty-six ordered pairs.

Equal likelihood needs an assumption. Each die is fair, and the dice act independently. The chance of a particular first face is one sixth. Given that face, each second face still has chance one sixth. Multiplying gives one thirty-sixth for each cell.

Look again at the whole square. Thirty-six cells, each carrying one thirty-sixth of the probability, account for a total of one. A complete roll lands somewhere in this square. The experiment has no extra outcome between cells, and no missing seventh face.

The equal-sized cells help us count, but their appearance does not make the dice fair. With loaded dice, those same face pairs could have unequal weights. This display currently uses the fair independent model. Later we will see unequal weights directly on a spinner.

Build an event from individual outcomes

An event is a collection of complete outcomes satisfying a stated rule. Our rule is: the sum equals seven. Start at one with six. That cell qualifies because one plus six is seven; neighboring one with five does not.

Move along the highlighted diagonal toward six with one. Each step increases the first face by one and decreases the second by one, preserving the sum. There are six qualifying cells, with no repeated cell and no missing first face.

A single roll cannot land in two different cells at once. These six possibilities are disjoint, so their probabilities add. Six copies of one thirty-sixth give six thirty-sixths, which simplifies to one sixth.

One sixth is a statement about the event before a roll. After a roll, its complete outcome either belongs to the gold diagonal or does not. It is not one sixth inside the diagonal. Probability and membership answer different questions.

We counted favorable outcomes and divided by all possible outcomes because these outcomes are equally likely. Keep that qualification attached to the rule. For unequal probabilities, add the weights of the favorable outcomes instead of simply counting their names.

Same experiment, different events

Change the event to a sum of two. Only one with one qualifies. The whole experiment still has thirty-six equally likely outcomes; the event now contains just one of them. Its probability is one thirty-sixth.

Select a sum of three. There are two ways: one with two, and two with one. Each contributes one thirty-sixth, so together they contribute two thirty-sixths. Changing the event changes the numerator; it has not changed the denominator.

For a sum of four, the highlighted cells are one with three, two with two, and three with one. The middle cell appears only once. Exchanging equal faces gives the same ordered pair, so we must not count two with two twice.

Return to a sum of seven and compare its longer diagonal with those earlier events. Every individual cell still has the same probability. The sums have different probabilities because their events contain different numbers of cells.

Combine several sums without double counting

Now change the rule from exactly four to at most four. At most includes equality. The gold corner contains sums two, three, and four. It is a collection of three separate diagonals, not just the diagonal labelled four.

Start with the corner cell, then move toward the three-cell outer diagonal. The counts are one, two, and three. Adding them gives six favorable cells. No cell is counted twice because each complete roll has exactly one sum.

Six out of thirty-six simplifies to one sixth. This is the same probability as exactly seven, although it is a different event. Equal probabilities do not mean equal sets of outcomes. Compare the corner region now with the earlier central diagonal.

Raise the boundary to five. Four more cells join the event: one with four, two with three, three with two, and four with one. The cumulative count is now ten, giving ten thirty-sixths. The old six cells remain included.

This is why at most and exactly must be read carefully. The boundary value alone does not specify the event. Exactly five selects four cells; at most five selects ten. The comparison word determines which whole collection we count.

Account for everything outside the event

Return to at most four, with its six gold cells. Every other cell is outside that event. Since the whole square has thirty-six cells, the outside contains thirty. Subtracting the inside from the whole is called taking the complement.

The complement is now highlighted: a sum of at least five. The sum is a whole number, so being greater than four means being at least five. Notice that a sum of four belongs to the previous event, not to this one.

Its probability is thirty thirty-sixths, or five sixths. One sixth plus five sixths gives one. Every roll belongs to exactly one of these two complementary events: there is no overlap, and together they fill the entire sample space.

Select at least two. Now every cell qualifies, because the smallest possible sum is two. Its probability is one. An impossible sum, such as one, would select no cell and have probability zero; that example is outside this control’s supported threshold range.

When counting outcome names is not enough

Move to a spinner with four equally weighted sectors. Their names are A, B, C, and D. Each sector occupies one quarter of the circle. Under this spinner model, each has probability one quarter, so counting sectors works in this equal-weight case.

Focus on sector A. Its displayed percentage is twenty-five percent. That is another way to write one quarter. The sector is one outcome of a spin; selecting its name defines the event whose probability we are currently reading.

Now use weights one, two, three, and four. Their total is ten. The sectors have visibly different angles, representing different probabilities. A has one tenth, B two tenths, C three tenths, and D four tenths of the whole.

Approach sector D. Its weight is four out of the total ten, so its probability is forty percent. There are still four outcome names, but dividing one by four would now give the wrong probability for this sector.

Compare sector A, whose probability is ten percent. The model gives D four times A’s probability. The names do not carry equal weight. The general rule is to add probability weights belonging to the event; equal-outcome counting is one useful special case.

Compare a model with observed results

Keep those unequal weights and select D again. Collect two hundred spins. The gold line shows the model probability of forty percent. The mint trace shows the fraction of observed spins that landed in D as this reproducible experiment grows.

Continue the same sequence to one thousand spins. The earlier results remain part of the total. The observed frequency can move above or below forty percent. A model probability does not promise an exact quota of successes in every finite collection.

Return to the exact spinner without experimental results. The forty percent comes from its assigned weights, not from measuring the last mint trace. Experiments can test how a model describes observations, but an observed fraction and a model probability remain different quantities.

Apply the method to a different sample space

For a final check, use a four-sided first die with a six-sided second die. The rectangular grid now has four rows and six columns, making twenty-four complete outcomes. Both dice are still fair and independent, so each cell carries one twenty-fourth.

Ask for a sum of five. Follow the four highlighted cells from one with four to four with one. Four favorable outcomes out of twenty-four gives one sixth. That equals the earlier probability of seven with two six-sided dice, but the sample space has changed.

Return to the whole rectangle and check the method in order: describe a complete outcome, identify the full sample space, state the event, and establish how the outcomes are weighted. Then add its probability weights. Only use a simple count ratio when those weights are equal.

Use Interact to choose another sum on this same grid. Predict the number of qualifying cells before checking them. Then compare exactly that sum with at most that sum. Explain which cells were added and why the denominator stays twenty-four while the dice remain unchanged.