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Represent a multi-stage finite sample space

Start a set of 8
Step 3 of 5
Stage counts
\(\displaystyle n_1=3,\quad n_2=4\)
Every stage pair is possible
\(\displaystyle S=S_1\times S_2\)

Find the total number of ordered outcomes

Year 11, VCE Units 1 & 2, Chapter 8, 4 difficulty levels, No calculatorBuild a worksheet

Product sample spaces

A multi-stage sample space contains one ordered tuple for every choice at every stage. Count all tuples first, then count the tuples that satisfy the event

\[N=n_1\times n_2\times\cdots\times n_k\]
  1. Keep stage order The position of a result in the tuple records the stage where it occurred \(\displaystyle (A,B)\ne(B,A)\)
  2. Multiply stage choices When every first-stage outcome can pair with every second-stage outcome, multiply their choice counts \(\displaystyle N=3\times4=12\)
  3. Count favourable tuples Count the ordered tuples that satisfy the event condition \(\displaystyle F=3\)
  4. Divide by total outcomes For equally likely tuples, divide the favourable count by the full sample-space count \(\displaystyle \Pr(E)=\frac{F}{N}=\frac{3}{12}=\frac{1}{4}\)

Worked example

A two-stage experiment has 2 possible first-stage outcomes and 3 possible second-stage outcomes. All ordered outcomes are equally likely. Event E contains 2 ordered outcomes. Find the probability of E

\[N=2\times3,\quad\Pr(E)=\frac{F}{N}\]
  1. \(\displaystyle N=2\times3=6\)Multiply the two stage counts to find the full ordered sample space
  2. \(\displaystyle F=2\)The event contains 2 of the ordered outcomes
  3. \(\displaystyle \Pr(E)=\frac{2}{6}=\frac{1}{3}\)Divide the favourable count by the total count and reduce the fraction

Common mistakes

Adding stage counts
✕\(\displaystyle N=2+3=5\)✓\(\displaystyle N=2\times3=6\)
Successive stages create one ordered outcome, so their choice counts are multiplied
Collapsing order
✕\(\displaystyle (A,B)=(B,A)\)✓\(\displaystyle (A,B)\ne(B,A)\)
The tuple positions record the stages, so reversing the results gives a different outcome
Using only favourable outcomes as the denominator
✕\(\displaystyle \Pr(E)=\frac{2}{2}=1\)✓\(\displaystyle \Pr(E)=\frac{2}{6}=\frac{1}{3}\)
The denominator must count every ordered outcome in the sample space, not only favourable outcomes

Difficulty levels

introUse two short equiprobable stages and a grid or table to form the product space
standardUse three binary stages or unequal stage sizes and preserve every ordered tuple
challengeHandle an order-sensitive event or a tree with unequal exact branch masses
examChoose the complete representation and calculate one event probability from it

Questions about this topic

Why does reversing the stage order change an outcome?+

The first entry records the first stage and the second entry records the second stage. Reversing the entries changes the ordered result

Why are stage counts multiplied?+

Each choice at one stage can be paired with every choice at the next stage, so the product counts all ordered tuples

How is an event probability found in an equiprobable product space?+

Count the ordered tuples that satisfy the event and divide by the total number of ordered tuples

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 8: Probability and counting

In this chapter

Finite events, probability assignments and complementsRepresent a multi-stage finite sample spaceCompound events, the addition rule and mutual exclusivityComplete and use a two-way probability tableConditional probability from exact representationsDistribution of a finite random variableDesign and interpret a probability simulationCount restricted selections and calculate exact probabilitiesCount arrangements with product, sum and restriction rulesTest exact independenceExact probability of repeated independent successesMulti-stage probability, multiplication and total probabilityProbability from a uniform area modelProbabilities of sequences of independent events