Translating Event Language, introduction to probability lesson

A machine needs attention if its warning count is even or greater than four. Does a count of six qualify, and what about a count of three? Questions like these are settled by testing each possible outcome against the whole condition, not by spotting one familiar word. This lesson turns the common phrases of probability problems into sets, using a constructed inspection that records a whole-number count from 1 to 8, so that $\Omega = \{1, 2, \dots, 8\}$ . No probabilities are assigned yet, and a list of eight counts says nothing about whether they are equally likely.

Key terms

Inclusive or

Definition In mathematics the phrase ” $A$ or $B$ ” means the inclusive or. It is the event $A \cup B$ , which contains the outcomes in $A$ , in $B$ or in both. The phrase ” $A$ and $B$ ”, also read “both $A$ and $B$ ”, is the intersection $A \cap B$ . A problem that means to leave out the overlap has to say so, for example with the words “exactly one of $A$ and $B$ ”.

In plain words An outcome qualifies for “or” as soon as it passes at least one of the two tests, and passing both is no reason to leave it out. A list of the union names each qualifying outcome once.

Example Let $A$ be the event that the count is even and $B$ the event that it is greater than 4,

\begin{gathered} A = \{2, 4, 6, 8\}, \\[0.4em] B = \{5, 6, 7, 8\}. \end{gathered}

The count 6 passes both tests, so it lies in $A \cap B = \{6, 8\}$ and also in the union. Counts 2 and 4 qualify through $A$ , counts 5 and 7 through $B$ , and 6 and 8 through both, so

A \cup B = \{2, 4, 5, 6, 7, 8\}.

The count 3 passes neither test, so it does not qualify under this rule.

Neither and not both

Definition The phrase “neither $A$ nor $B$ ” means that both conditions fail. It is the event

(A \cup B)^{c} = A^{c} \cap B^{c}.

The phrase “not both $A$ and $B$ ” means that at least one condition fails. It is the event

(A \cap B)^{c} = A^{c} \cup B^{c}.

Both complements are taken in the stated sample space. Every outcome in “neither” is also in “not both”, and the two events are equal exactly when $A = B$ .

In plain words Neither rejects an outcome that passes even one test. Not both rejects only the outcomes that pass both tests, so it keeps the outcomes that pass one test as well as those that pass none.

Example With $A$ the even counts and $B$ the counts greater than 4, the event “not $A$ ” is $A^{c} = \{1, 3, 5, 7\}$ , the odd counts. Both conditions fail only for 1 and 3, so “neither” is $\{1, 3\}$ , which is also what remains after removing $A \cup B$ from $\Omega$ . “Not both” removes only 6 and 8, which leaves $\{1, 2, 3, 4, 5, 7\}$ , six outcomes rather than two.

Exactly one

Definition The event that exactly one of $A$ and $B$ occurs is the symmetric difference

\begin{aligned} A \mathbin{\triangle} B &= (A \setminus B) \\[0.4em] &\quad \cup (B \setminus A). \end{aligned}

Here $A \setminus B$ , read ” $A$ but not $B$ ” or “only $A$ ”, contains the outcomes in $A$ that are not in $B$ . Any two events split $\Omega$ into four pairwise disjoint pieces, only $A$ , only $B$ , both and neither, and every outcome lies in exactly one of these pieces. In those terms “or” is only $A$ , only $B$ and both together, “exactly one” is only $A$ and only $B$ , and “not both” is everything except both.

In plain words Exactly one keeps the outcomes that pass one test and fail the other. It leaves out the overlap, which the inclusive or keeps, and it leaves out the outcomes that pass neither test, which not both keeps.

Example For the warning counts, only $A$ is $\{2, 4\}$ and only $B$ is $\{5, 7\}$ , so exactly one is $\{2, 4, 5, 7\}$ . The four pieces are $\{2, 4\}$ , $\{5, 7\}$ , $\{6, 8\}$ and $\{1, 3\}$ . They share no count and together contain all eight, which gives a quick check on any translation that involves two events.

At least and at most

Definition For a recorded number $n$ and a fixed value $k$ , the phrase “at least $k$ ” means $n \ge k$ and “at most $k$ ” means $n \le k$ , and both include the boundary value $k$ . The phrase “more than $k$ ” means $n > k$ and “fewer than $k$ ” means $n < k$ , and both exclude it. The phrase “exactly $k$ ” means $n = k$ . Within a sample space of numbers, the complement of “at least $k$ ” is “fewer than $k$ ”, and the complement of “at most $k$ ” is “more than $k$ ”. For two events, “at least one of $A$ and $B$ ” is $A \cup B$ , and “at most one of $A$ and $B$ ” is $(A \cap B)^{c}$ , the event not both.

In plain words At least and at most keep the boundary, while more than and fewer than drop it. A single boundary outcome is easy to lose, and later it can change a probability.

Example For the warning counts and the boundary 4,

Phrase	Condition	Event
at least 4	$n \ge 4$	$\{4, 5, 6, 7, 8\}$
more than 4	$n > 4$	$\{5, 6, 7, 8\}$
at most 4	$n \le 4$	$\{1, 2, 3, 4\}$
fewer than 4	$n < 4$	$\{1, 2, 3\}$

Because the counts are whole numbers, “more than 3” is the same event as “at least 4”. That shortcut depends on the whole-number recording rule and fails for a measurement such as 3.2. A phrase like “between 3 and 6” is incomplete until it says whether 3 and 6 are included, so write the inequalities that are meant, such as $3 \le n \le 6$ , $3 < n < 6$ or $3 \le n < 6$ .

Only if

Definition For events $A$ and $B$ , the statement ” $A$ happens only if $B$ happens” means that every outcome in $A$ also belongs to $B$ , that is $A \subseteq B$ . Equivalently, $A \setminus B = \varnothing$ , and the statement can also be read “if $A$ , then $B$ ”. The statement ” $A$ happens if $B$ happens” reverses the direction and means $B \subseteq A$ , and ” $A$ if and only if $B$ ” means $A = B$ . These are claims about two events, each true or false as a whole, unlike the event $A \cap B$ , which is a set of outcomes.

In plain words Only if names a requirement. Whenever $A$ happens, $B$ happens too. It does not say that $B$ guarantees $A$ . If $A$ is empty, the claim is true automatically, since there is no outcome to check.

Example For the warning counts, “the count is even only if it is greater than 4” claims $A \subseteq B$ . It is false, because 2 is even but not greater than 4. The reversed claim “the count is even if it is greater than 4” claims $B \subseteq A$ , and it is false too, because 5 is greater than 4 but odd. By contrast, “the count is greater than 6 only if it is greater than 4” is true, since $\{7, 8\} \subseteq \{5, 6, 7, 8\}$ .

Counterexample

Definition A counterexample to a claim of the form “every outcome in $A$ belongs to $B$ ” is an outcome that belongs to $A$ but not to $B$ , that is, an element of $A \setminus B$ . One counterexample shows that the claim is false. Outcomes that agree with the claim do not prove it, unless every outcome in $A$ has been checked, which is possible when $A$ is finite.

In plain words To refute a general claim, one failure is enough. To confirm it, every case has to pass.

Example The count 2 is a counterexample to ” $A$ only if $B$ ” above, and 5 is a counterexample to ” $A$ if $B$ ”. The count 6 lies in both events, so it agrees with each claim and proves neither. Counterexamples and complements also depend on the sample space. If the inspection may also record 0 warnings, the sample space becomes $\{0, 1, \dots, 8\}$ and the even event becomes $\{0, 2, 4, 6, 8\}$ , since 0 is even. Then 0 is a new counterexample to ” $A$ only if $B$ ”, and the complement of $B$ grows from $\{1, 2, 3, 4\}$ to $\{0, 1, 2, 3, 4\}$ , although the meanings of even and greater than 4 have not changed.

Check yourself

On $\Omega = \{1, 2, \dots, 8\}$ , let $C$ be the event that the count is at most 5 and $D$ the event that the count is odd. Which outcomes are in both events, and which are in $C$ but not $D$ ?

Show the answer

The events are $C = \{1, 2, 3, 4, 5\}$ and $D = \{1, 3, 5, 7\}$ . Both conditions hold for 1, 3 and 5, so $C \cap D = \{1, 3, 5\}$ . The difference $C \setminus D$ is $\{2, 4\}$ , the even counts from 1 to 5.

For the same events, find neither $C$ nor $D$ , exactly one of $C$ and $D$ , and not both. Does not both mean the same as neither here?

Show the answer

Not $C$ leaves 6, 7 and 8, and requiring not $D$ as well removes 7, so neither is $\{6, 8\}$ . Only $C$ is $\{2, 4\}$ and only $D$ is $\{7\}$ , so exactly one is $\{2, 4, 7\}$ . Not both removes the intersection $\{1, 3, 5\}$ from $\Omega$ , which leaves $\{2, 4, 6, 7, 8\}$ . The two events differ. Not both contains the neither outcomes 6 and 8 and also the exactly one outcomes 2, 4 and 7.

Decide whether each claim is true on $\Omega = \{1, 2, \dots, 8\}$ , and give a counterexample when it is false. The first claim is “the count is odd only if it is at most 5”. The second is “the count is at least 7 only if it is more than 5”.

Show the answer

The first claim says $D \subseteq C$ . The count 7 is odd but not at most 5, so it is a counterexample and the claim is false. The odd counts 1, 3 and 5 agree with the claim, but agreement from some outcomes cannot rescue it.

The second claim says $\{7, 8\} \subseteq \{6, 7, 8\}$ . Both 7 and 8 are more than 5, so every outcome of the first event has been checked and the claim is true. No counterexample exists.

Translating Event Language

Key terms

Inclusive or

Neither and not both

Exactly one

At least and at most

Only if

Counterexample

Check yourself

How this lesson connects

Builds on

Used later in