Skill: Represent sample spaces using tables

Explanation and Free Practice Resources

A table represents a sample space by organizing all possible outcomes of a chance experiment, with rows and columns corresponding to stages or categories and each cell showing one outcome combination. The representation must be exhaustive and non-overlapping, enabling event probabilities to be found by counting appropriate outcomes or combining their assigned probabilities, while distinguishing a single outcome from an event containing several outcomes; highly abstract or extensive multistage sample spaces are not included.

Detailed Explanation: Represent sample spaces using tables

A sample space is the list of every possible outcome of a chance experiment. A table is useful when the experiment has two stages, because:

  • the rows can represent the outcomes of the first stage;
  • the columns can represent the outcomes of the second stage;
  • each cell represents one combined outcome.

Example

A fair coin is flipped, and a fair spinner numbered (1,2,3)(1,2,3) is spun. Find the probability of getting heads and an odd number.

Step 1: List the outcomes for each stage

The coin can land:

H or TH \text{ or } T

The spinner can land:

1, 2, or 31,\ 2,\ \text{or } 3

Step 2: Make a table

Put the coin outcomes in the rows and the spinner outcomes in the columns.

Coin \ Spinner112233
HH(H1)(H1)(H2)(H2)(H3)(H3)
TT(T1)(T1)(T2)(T2)(T3)(T3)

Each cell is one outcome, such as (H2)(H2), meaning “heads and 2.”

The table is:

  • exhaustive because it includes every possible outcome;
  • non-overlapping because each result belongs to exactly one cell.

Thus, the sample space has 66 outcomes:

{H1,H2,H3,T1,T2,T3}\{H1,H2,H3,T1,T2,T3\}

Step 3: Identify the event

The event is “heads and an odd number.” The odd spinner numbers are 11 and 33, so the successful outcomes are:

H1 and H3H1 \text{ and } H3

These are two outcomes, not one outcome.

Step 4: Find the probability

All 66 outcomes are equally likely. Therefore,

P(heads and odd)=number of successful outcomestotal number of outcomes=26=13P(\text{heads and odd})=\frac{\text{number of successful outcomes}}{\text{total number of outcomes}} =\frac{2}{6} =\frac{1}{3}

So, the probability of getting heads and an odd number is

13\boxed{\frac{1}{3}}

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Probability Sample Space - Definition to Sample Space List


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