Sample spaces and counting

Idea: List the outcomes first

The sample space is the list of everything that can happen. Write it before any formula.

• Each item is an outcome
• Equally likely outcomes turn a chance into one fraction
• Most early mistakes come from a list that only looks fair

Formula: The counting rule

$P(A)=\frac{k}{n}$

With n equally likely outcomes, k of them in the event, the chance is k over n.

Idea: Size the list stage by stage

Multiply the choices at each stage to count the whole list.

• $3$ dice: $6^3=216$ outcomes
• $4$ coins: $2^4=16$ sequences
• All $3$ dice even: $3\times 3\times 3=27$ outcomes

Example: A total of 10 with 2 dice

Roll a pair of fair dice. What is the chance that the total is $10$?

1. Call the dice red and blue: $6\times 6=36$ ordered pairs
2. Total $10$: $(4,6)$, $(5,5)$, $(6,4)$
3. Chance: $\frac{3}{36}=\frac{1}{12}$

Common mistake: Listing totals, not outcomes

A pair of dice has $11$ totals, but they are not equally likely: $2$ has $\frac{1}{36}$, $7$ has $\frac{6}{36}=\frac{1}{6}$.

Instead Count ordered pairs. A mixed pair like $(4,6)$ happens in $2$ ways, a double only in $1$.

Example: Exactly 2 heads in 4 tosses

Toss a fair coin $4$ times. What is the chance of exactly $2$ heads?

1. The list: $2^4=16$ sequences
2. Choose which tosses are heads: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$
3. Chance: $\frac{6}{16}=\frac{3}{8}$

Quick check: At least one 6

Roll $2$ fair dice. What is the chance that at least one shows a $6$?

Idea: Averages come from the same list

An expected value is the average of a quantity over equally likely outcomes.

• One die: $\frac{1+2+3+4+5+6}{6}=\frac{7}{2}$
• Totals of all $36$ pairs add to $252$, so $\frac{252}{36}=7$
• Same as $2\times \frac{7}{2}=7$: averages add die by die

Idea: Say it out loud

Name the sample space and its size first, and why its outcomes are equally likely.

• Say "ordered pairs" when you mean them
• List small counts, describe large ones by what you choose
• Check that the counts fill the list