Probability · Basic probability

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Sample space & systematic listing

How to list every outcome once and only once — systematic listing, two-way sample-space diagrams, the product rule for counting, and why swapping two things over sometimes makes a new outcome and sometimes doesn't.

Probability · Basic probability

Sample space & systematic listing

Why it works

When every outcome of an experiment is equally likely,

P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}.

Both numbers in that fraction are counts. So this whole topic is really one skill: counting the outcomes without missing any and without counting any twice. The complete list of everything that could happen is the sample space.

Why "systematic" is the whole game. Ask someone to list the outcomes of flipping three coins and they will usually write HHH, HHT, THH, TTT, and then stare at the page wondering what's left. Scribbling down outcomes as they occur to you is exactly how you lose marks — you have no way of knowing when you've finished. Instead, fix one thing and run through all the others, then change the fixed thing:

\underbrace{HHH,\ HHT,\ HTH,\ HTT}_{\text{first coin }H},\qquad \underbrace{THH,\ THT,\ TTH,\ TTT}_{\text{first coin }T}.

Eight outcomes, in a pattern you can see is complete: within each half the last two coins run HH, HT, TH, TT — an order that never wavers. A list built this way can't skip an outcome (the pattern would have a hole in it) and can't repeat one (every entry differs from the one before it in a controlled way).

Two events at once: the sample-space diagram. When there are exactly two stages, a table is better than a list. Put one event down the side and the other across the top; every cell is one outcome. For two fair dice, put the total in each cell:

$+$	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

There are

6 \times 6 = 36

cells, and — this is the point — the 36 cells are equally likely. Reading a probability off the table is then just counting cells: a total of

5

appears

4

times, so

P(\text{total} = 5) = \tfrac{4}{36} = \tfrac{1}{9}

.

The trap, made concrete. Look at that table and a tempting shortcut appears: "the total can be anything from

2

12

, that's

11

possible outcomes, so

P(\text{total} = 5) = \tfrac{1}{11}

." It is wrong, and it's worth being precise about why. The formula at the top only applies to outcomes that are equally likely, and the eleven totals are not: a total of

2

occupies one cell, a total of

7

occupies six. Counting totals silently glues six equally likely cells together and calls the result one outcome. The

36

cells are the honest sample space; the totals are a summary of it.

The same trap wearing different clothes: list the two-dice outcomes as unordered sets —

\{1,1\}, \{1,2\}, \{1,3\}, \dots, \{6,6\}

— and you get

21

of them. But

\{1,2\}

happens two ways (red 1 blue 2, or red 2 blue 1) while

\{1,1\}

happens only one way, so those

21

are not equally likely either, and

P(\text{total} = 3) \neq \tfrac{1}{21}

. Merging outcomes that aren't identical, or splitting one outcome into two, both break the fraction.

The product rule for counting. You don't always need to see the whole list — often you only need its size. If stage 1 can happen in

m

ways and stage 2 in

n

ways, then the two stages together can happen in

m \times n \text{ ways.}

Why multiply? Because the sample-space table has

m

rows and

n

columns, and every cell is a different outcome: each of the

m

first-stage outcomes pairs up with every one of the

n

second-stage outcomes. Adding would be counting the stages as alternatives ("either a die roll or a coin flip") rather than as things that both happen.

It keeps going for more stages. A menu with

3

starters,

4

mains and

2

desserts gives

3 \times 4 \times 2 = 24

meals: each of the

12

starter-and-main pairs branches into

2

meals. A four-digit PIN using digits

0

–

9

with repeats allowed gives

10 \times 10 \times 10 \times 10 = 10\,000

. If a restriction applies to one stage, apply it to that stage's count and leave the others alone: a PIN that must start with an odd digit has

5 \times 10 \times 10 \times 10 = 5000

possibilities.

Ordered or unordered — the question to ask. Two dice, one red and one blue, are distinguishable: red 2 with blue 5 is a different cell from red 5 with blue 2, and both belong in the sample space. But choosing two people from a group of six to attend a conference is not like that: picking Ana then Ben gives the same pair as picking Ben then Ana. Here

6 \times 5 = 30

counts every pair exactly twice, so the number of pairs is

\frac{6 \times 5}{2} = 15.

The test is one sentence: swap the two things over — is the result genuinely different? For a chairperson and a secretary chosen from the six, swapping gives a different arrangement, so it stays

6 \times 5 = 30

. For two delegates with the same job, swapping changes nothing, so you halve it. Getting this backwards is the single most common way a correct method produces a wrong number — always exactly a factor of

2

out for pairs.