Counting Outcomes and Identifying Events

An organized list writes the outcomes in a steady pattern so that you skip none and repeat none. Hold the first result still and run through every second result. Then move to the next first result and do the same.

For the roll and the flip, each number from 1 to 6 pairs with heads (H) and with tails (T):

• $(1, \text{H})$, $(1, \text{T})$
• $(2, \text{H})$, $(2, \text{T})$
• $(3, \text{H})$, $(3, \text{T})$
• $(4, \text{H})$, $(4, \text{T})$
• $(5, \text{H})$, $(5, \text{T})$
• $(6, \text{H})$, $(6, \text{T})$

Count the entries: the sample space has 12 ordered outcomes. You can check the count by multiplying, because the list has 6 groups with 2 outcomes in each: $6 \times 2 = 12$. This check works only when the list shows each ordered outcome exactly once and every first result pairs with the same number of second results.

An event is a part of the sample space: the outcomes that match its description. To find them, turn the description into a yes-or-no test. Test every outcome in the sample space, one at a time, and keep the ones that pass. The outcomes you keep are the favourable outcomes. Here, favourable simply means that they match the description.

You also need to say what you assume about chance. A cube or a coin is fair when each of its results has the same chance of coming up. With a fair cube and a fair coin, all 12 ordered outcomes are equally likely, which means each one has the same chance of happening. This matters because a count of favourable outcomes out of total outcomes describes chance only when the listed outcomes are equally likely.

A two-way table is another way to hold a sample space for two actions. Write the results of the first action down the side and the results of the second action across the top. When every first result can be followed by every second result, each cell pairs one row with one column and holds one ordered outcome, so the table skips no pair and repeats none.

Here is the table for two spinners, A and B. Each spinner has 4 equal sections numbered 1, 2, 3 and 4. You spin A first, then B, and write each outcome as (number on A, number on B).

With the spinners, any second result could follow any first result. Drawing cards can be different. With replacement means the first card goes back before the second draw, so the second draw has the same choices as the first. Without replacement means the first card stays out.

Picture 4 cards numbered 1, 2, 3 and 4, drawn without replacement. The first draw has 4 possible cards, but the second draw has only 3. The first draw changes what can happen on the second draw, so the two draws are dependent events. Outcomes such as $(3, 3)$ cannot happen, so you must not list them.

This shrinks every total that involves the second draw: 3 choices instead of 4, and $4 \times 3 = 12$ ordered outcomes instead of $4 \times 4 = 16$. When you later write favourable over total as a fraction, these smaller totals are the denominators.

Every count for a compound event starts with the whole sample space. Build an organized list or a two-way table in which every ordered outcome appears exactly once, and keep different orders as different outcomes. Count all the entries to get the total. Then test each entry against the description to find the outcomes that make up the event.

Say when you are assuming that the outcomes are equally likely, because favourable over total describes chance only then. Remember that drawing without replacement leaves fewer choices for the later draw, so the totals you use as denominators change. Next, you will use these counts to find the probability of a compound event.