6.3 - Sample spaces and systematic outcomes

6.3 - Sample spaces and systematic outcomes

Learn to record every possible outcome without omissions or double counting, then select the outcomes belonging to an event. Ordered lists and tables help you calculate probabilities and explain why some results are more likely than others. This lesson applies at Foundation and Higher.

Sample spaces and events

Two fair ordinary dice can give totals from 22 to 1212. Are those eleven totals equally likely? Keep that question in mind: the detail you record in an outcome will settle it.

You already know

For equally likely outcomes, probability is the number of favourable outcomes divided by the total. An event can contain several outcomes, such as the even faces of a die.

Start with one die. Its possible faces form the set E={1,2,3,4,5,6}\mathrm{E}=\{1,2,3,4,5,6\}. The event “prime” selects 22, 33 and 55 from that set.

Sample space

The sample space is the set of all possible outcomes of an experiment, each listed once. An event is a subset of that space. The empty event is impossible; the whole sample space is certain.

Decide what an outcome records. For a bag containing five red counters and one blue counter, “red, blue” lists the possible colours. Those two colour outcomes are not equally likely. You could instead label and list all six individual counters, which are equally likely when one is chosen uniformly at random.

Probability from a sample space

For a finite sample space of equally likely outcomes,

P(A)=n(A)n(E)P(A)=\frac{n(A)}{n(\mathrm{E})}

Here nn means the number of outcomes in the set. This rule is absent from both tier sheets. Not in the formulae sheet: learn it.

Write the event list before calculating. It lets you check whether the numerator describes the requested event.

Worked example 1: prime faces

Question. A fair six-sided die is rolled. Write the sample space and the event AA of a prime face. Find P(A)P(A).

Plan. List all faces, select the primes, then count.

E={1,2,3,4,5,6}all possible facesA={2,3,5}prime faces onlyP(A)=36=12three out of six\begin{aligned} \mathrm{E}&=\{1,2,3,4,5,6\} &&\text{all possible faces}\\ A&=\{2,3,5\} &&\text{prime faces only}\\ P(A)&=\frac36=\frac12 &&\text{three out of six} \end{aligned}

Why leave out 11? A prime has exactly two positive factors; 11 has only one.

Answer. E={1,2,3,4,5,6}\mathrm{E}=\{1,2,3,4,5,6\}, A={2,3,5}A=\{2,3,5\}, P(A)=12P(A)=\frac12.

Check. Three prime faces and three nonprime faces account for all six faces.

Change only the event from “prime” to “square”. The experiment and denominator stay fixed.

Worked example 2: square faces

Question. A fair six-sided die is rolled. Write the sample space and the event BB of a square face. Find P(B)P(B).

Plan. Use the same sample space and select its square numbers.

E={1,2,3,4,5,6}unchanged experimentB={1,4}squares in the spaceP(B)=26=13two favourable faces\begin{aligned} \mathrm{E}&=\{1,2,3,4,5,6\} &&\text{unchanged experiment}\\ B&=\{1,4\} &&\text{squares in the space}\\ P(B)&=\frac26=\frac13 &&\text{two favourable faces} \end{aligned}

Why include 11? It is 121^2, even though it was not prime.

Answer. B={1,4}B=\{1,4\}, with probability 13\frac13.

Check. The other four faces have probability 46\frac46, and 26+46=1\frac26+\frac46=1.

The event is a set of outcomes; its probability is a number. Keep those two kinds of answer distinct.

Watch out

P(red)=12P(\text{red})=\frac12 from the colour list “red, blue” is wrong for five red counters and one blue. Equal likelihood applies to the six individual counters, giving 56\frac56 for red.

Use the same event as the second example, but change the available number range.

Systematic ordered outcomes

For successive events, record the first result and then the second. The pair (a,b)(a,b) means first aa, then bb; reversing the order can describe a different result.

Ordered outcome

An ordered outcome keeps the positions of successive results. For two coin tosses, (H,T)(H,T) and (T,H)(T,H) are different outcomes, although both contain one head.

To list systematically, fix the first result and attach every possible second result. Then move to the next first result. If each of mm first results can be followed by all nn second results, there are mnmn pairs. If the experiment forbids a combination, leave it out rather than applying this count blindly.

Worked example 3: two coin tosses

Question. A fair coin is tossed twice, with neither toss affecting the other. List all ordered outcomes and find the probability of exactly one head.

Plan. List the head-first row, then the tail-first row, and select the required pairs.

E={(H,H),(H,T),(T,H),(T,T)}fix each first resultA={(H,T),(T,H)}exactly one headP(A)=24=12equally likely pairs\begin{aligned} \mathrm{E}&=\{(H,H),(H,T),(T,H),(T,T)\} &&\text{fix each first result}\\ A&=\{(H,T),(T,H)\} &&\text{exactly one head}\\ P(A)&=\frac24=\frac12 &&\text{equally likely pairs} \end{aligned}

Why keep both mixed pairs? They place the head in different toss positions.

Answer. The four pairs are listed above; exactly one head has probability 12\frac12.

Check. 2×2=42\times2=4 pairs. “At least one head” would also include (H,H)(H,H), giving 34\frac34.

Change only the target event from “exactly one head” to “at least one head”. The experiment and sample space stay fixed.

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