Sample spaces, events, and the complement

Equally likely outcomes counted; an event as a subset; and why "at least one" is almost always 1 − P(none).

Three words

A random experiment is any process whose outcome is not known in advance — drawing one invoice from a file, auditing one claim. The sample space $S$ is the set of every possible outcome. An event is a subset of $S$ : the outcomes that make a sentence true. When every outcome in $S$ is equally likely, probability is counting:

$P (A) = \frac{number of outcomes in A}{number of outcomes in S}$

Equally likely is what "draw one at random" means. It is the assumption that lets you count instead of model, and this whole module rests on it.

Worked example — Maple Ledger's 50 invoices

Maple Ledger, a Kitchener bookkeeping firm, reviews the 50 invoices it issued last month. Twelve were paid late, eight contain a data-entry error, and three are both late and erroneous. One invoice is drawn at random, so $S$ has 50 equally likely outcomes.

$P (late) = 12/50 = 0.24$ .
$P (not late) = 38/50 = 0.76$ .

Not late is the complement of late, written late $^{'}$ (or late $^{c}$ ): every outcome in $S$ that is not in the event. Because an outcome is in $A$ or in $A^{'}$ and never both,

$P (A^{'}) = 1 - P (A)$

Two events at once

"Late or error" means late, or error, or both — inclusive or, always. Count it once: 12 late + 8 error counts the 3 both-invoices twice, so the union has $12 + 8 - 3 = 17$ invoices and $P (late or error) = 17/50 = 0.34$ . "Late and error" is the overlap: $3/50 = 0.06$ .

The complement of "late or error" is "neither late nor an error":

$P (neither) = 1 - 0.34 = 0.66$

— thirty-three clean invoices out of fifty. The wrong route, $1 - 0.24 - 0.16 = 0.60$ , subtracts the three overlap invoices twice. Lesson 5.2 turns this into a rule.

"At least one"

Maple Ledger's partner asks: if two invoices are drawn, what is the chance at least one is late? "At least one" means one or two — two cases to add. Its complement is a single case: none is late. So

$P (at least one) = 1 - P (none)$

and the paper's "at least one" question is almost always a complement question in disguise. Lesson 5.3 supplies $P (none)$ for two draws; the binomial (Module 6) supplies it for twenty.

Reading the question

A probability question has three moving parts: what is drawn (one invoice, two claims), what is equally likely (each invoice, each pair), and which sentence is the event. Write the event as a set of outcomes before counting. Half the errors in this module are counting the wrong set beautifully.