Probability course
Probability course
Module 1 of 8

Counting & sample spaces

Turn a story into an equally likely sample space.

14 min2 worked examples3 concept checks

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Learning objectives
  • Specify the elementary outcomes before dividing counts.
  • Distinguish selections from arrangements.
  • Use a complement when counting the unwanted event is easier.

Concepts

Start underneath the story

A probability model begins with possible outcomes and their weights. Counting favorable outcomes and dividing by all outcomes works only when the elementary outcomes are equally likely. Two fair dice have 36 ordered face pairs. Their eleven possible sums are not equally likely: a sum of 7 has six supporting pairs, while a sum of 2 has one.

Does order change the outcome?

Fill ordered positions one at a time. Choosing a captain and a deputy from n people gives n(n−1) outcomes when one person cannot hold both roles. A two-person committee has no roles; each committee was counted twice, so divide by 2. In general, ordered selections of k distinct items number n!/(n−k)!, and unordered selections number (nk)=n!/[k!(n−k)!]\binom{n}{k}=n!/[k!(n-k)!]. State whether replacement is allowed before using either.

Count the simpler side

Every event and its complement divide the sample space into two disjoint parts. Therefore P(A)=1−P(Ac)P(A)=1-P(A^c). For an 'at least one' question, 'none' is often a single product. This shortcut still needs the right dependence assumptions: multiply fixed probabilities across trials only when that is justified.

Examples

Worked example 1

Two different jobs

Seven distinct applicants are equally likely to receive two different roles: lead and reviewer. Nobody can fill both. How many assignments are possible, and what is the probability that Mira is selected for either role?

  1. Define an assignment

    An outcome is an ordered pair (lead, reviewer), so there are 7×6=427\times6=42 assignments.

  2. Count the two disjoint cases

    Mira can lead with any of six reviewers, or review for any of six leads. These cases cannot overlap. Favorable assignments: 6+6=12.

  3. Form the ratio

    The probability is 12/42=2/712/42=2/7. As a check, exactly two of seven symmetric applicants are chosen.
Worked example 2

Exactly one six

Two independent fair six-sided dice are rolled. What is the probability that exactly one die shows six?

  1. Keep the dice distinguishable

    There are 36 ordered pairs. A six on the first die and a six on the second die are different placements.

  2. Exclude the double six

    First die six, second not six gives five pairs. Second die six, first not six gives five more. The pair (6,6) belongs to neither case.

  3. Add disjoint counts

    There are ten favorable pairs, so the probability is 10/36=5/1810/36=5/18. The word 'exactly' prevents counting (6,6).

Concept checks

Choose an answer, then check it. You can retry; your latest checked answer determines completion.

0 of 3 concept checks passed
Concept check 1Three distinct roles are assigned to different people from seven applicants. How many assignments?
Concept check 2Three independent fair coins are tossed. What is the probability of at least one head?
Concept check 3Why can you not assign probability 1/11 to every sum of two fair dice?

Key results

  • Choose the elementary outcome first.
  • Multiply along a sequence of choices; add mutually exclusive cases.
  • Divide out rearrangements only when order does not matter.
  • Try the complement for 'at least one'.

Practice problems

Solve related problems or try a timed probability drill.

Practice problemsHints and worked solutions included.Timed probability drillShort questions on this module’s topic.
Next in the course

Update on the evidence

Choose the right denominator, then use Bayes naturally.

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