Probability Zero and Impossible Events

Introduction

In a probability model, “impossible” can mean two different things. An event may be impossible because it contains no outcomes at all, or it may contain outcomes but receive probability zero under the model. These cases are not the same: an event is a set, while its probability is a number assigned to that set.

The distinction matters even when the outcome space is countable. A countable probability model can include outcomes with zero probability, provided the remaining assigned probabilities still sum to one. On the other hand, if every individual outcome in a countable model has positive probability, then every nonempty event has positive probability. The model’s point probabilities determine which situation applies.

Empty Events and Null Events

Let \(\Omega\) be an outcome space, and let an event be a subset of \(\Omega\) to which the probability model assigns a probability. The event \(\varnothing\) contains no outcomes. A nonempty event contains at least one outcome, whether or not the probability model assigns it positive probability.

Definition: The empty event is \(\varnothing\), the set containing no outcomes. A null event is an event \(A\) for which \(P(A)=0\). Every empty event is a null event, but a null event need not be empty. An outcome \(\omega\) with \(P(\{\omega\})=0\) is called a zero-probability outcome (or a zero-mass outcome).

The word “impossible” is often used informally for any event with probability zero. For precision, reserve empty for an event with no outcomes, and use probability zero or null for an event assigned probability zero. Whether an outcome belongs to the stated outcome space is a set-theoretic question; what probability it receives is a property of the probability law.

One fact follows directly from the probability axioms: the empty event must have probability zero. For any event \(A\), probabilities are also monotone: if \(B\subseteq A\), then \(P(B)\leq P(A)\). To see the latter, write \(A\) as the disjoint union of \(B\) and \(A\setminus B\). Additivity gives \(P(A)=P(B)+P(A\setminus B)\), and the second term is nonnegative.

Result: \(P(\varnothing)=0\). If \(B\subseteq A\) and \(P(A)=0\), then \(P(B)=0\).

For the first claim, \(\Omega\) and \(\varnothing\) are disjoint, and their union is \(\Omega\). Additivity and the fact that \(P(\Omega)=1\) give

1=P(\Omega)=P(\Omega\cup\varnothing)=P(\Omega)+P(\varnothing)=1+P(\varnothing).

Therefore \(P(\varnothing)=0\). For the second claim, monotonicity gives \(0\leq P(B)\leq P(A)=0\), so \(P(B)=0\). This establishes that every subset of a null event is null, including the empty subset. It does not establish that a null event itself is empty.

Worked Example: A Zero-Probability Outcome in a Finite Model

Let \(\Omega=\{a,b,c,d\}\), with point probabilities \(P(\{a\})=1/2\), \(P(\{b\})=1/3\), \(P(\{c\})=1/6\), and \(P(\{d\})=0\). Check whether these assignments define a probability law and compare \(\{d\}\) with \(\varnothing\).

Check the total. The probabilities are nonnegative and sum to

\frac12+\frac13+\frac16+0 =\frac36+\frac26+\frac16 =1.

Thus they define a probability law on this finite outcome space. The event \(\{d\}\) is nonempty because it contains \(d\), but its probability is zero. By contrast, \(\varnothing\) contains no outcomes and also has probability zero. Equal probabilities do not make the events equal.

What Countability Tells Us

For a countable outcome space, suppose every subset is an event and write \(p(\omega)=P(\{\omega\})\) for the probability assigned to a single outcome. Countable additivity then gives the probability of any event \(A\) by summing the probabilities of its individual outcomes:

P(A)=\sum_{\omega\in A}p(\omega), \qquad p(\omega)\geq 0, \qquad \sum_{\omega\in\Omega}p(\omega)=1.

This formula makes the possibilities precise. If even one outcome in \(A\) has positive probability, then \(P(A)>0\). If every outcome in \(A\) has probability zero, the sum is zero, even when \(A\) contains many outcomes. In particular, an event that is a countable union of zero-probability singletons is still a null event.

Theorem: In a countable outcome space where every subset is an event, an event \(A\) has probability zero if and only if every outcome it contains has probability zero: \[ P(A)=0 \quad\Longleftrightarrow\quad p(\omega)=0\text{ for every }\omega\in A. \] Consequently, such a model has a nonempty null event if and only if it has at least one zero-probability outcome.

Proof. If \(P(A)=0\) and \(\omega\in A\), then \(\{\omega\}\subseteq A\). By monotonicity, \(0\leq p(\omega)\leq P(A)=0\), so \(p(\omega)=0\). This proves that every outcome in \(A\) has probability zero.

Conversely, suppose every \(\omega\in A\) has \(p(\omega)=0\). Since \(A\) is countable, countable additivity applied to its disjoint singleton events gives

P(A)=\sum_{\omega\in A}P(\{\omega\}) =\sum_{\omega\in A}p(\omega) =\sum_{\omega\in A}0 =0.

If \(A\) is nonempty, it contains a zero-probability outcome. And if the model has a zero-probability outcome \(\omega\), then its singleton \(\{\omega\}\) is a nonempty null event. This proves the consequence as well. \(\square\)

Worked Example: A Countable Model With One Zero-Mass Outcome

Consider \(\Omega=\{1,2,3,\ldots\}\cup\{z\}\), and assign

p(n)=2^{-n}\quad(n=1,2,3,\ldots), \qquad p(z)=0.

The probabilities on the positive integers sum to one, since

\sum_{n=1}^{\infty}2^{-n} =\frac{1/2}{1-1/2} =1.

Including \(p(z)=0\) therefore leaves the total probability equal to one. The event \(\{z\}\) is nonempty but has probability zero. For a different event, \(A=\{1,z\}\), countable additivity gives

P(A)=p(1)+p(z)=\frac12+0=\frac12.

The zero-mass outcome is part of the outcome space, but it contributes nothing to the probability of an event containing it.

Worked Example: Countably Many Zero-Probability Outcomes

Let \(\Omega\) contain the positive integers and additional outcomes \(z_1,z_2,\ldots\). Assign \(p(n)=2^{-n}\) to each positive integer and \(p(z_j)=0\) to every \(j\geq 1\). The total probability is still one because

\sum_{n=1}^{\infty}2^{-n}+\sum_{j=1}^{\infty}0=1+0=1.

Let \(Z=\{z_1,z_2,\ldots\}\). This is a nonempty, countably infinite event. Its probability is

P(Z)=\sum_{j=1}^{\infty}P(\{z_j\}) =\sum_{j=1}^{\infty}0 =0.

This calculation uses countable additivity, not a rule that an infinite event must have positive probability. The number of outcomes in an event, by itself, does not determine its probability.

Worked Example: When Every Nonempty Event Has Positive Probability

Now take \(\Omega=\{1,2,3,\ldots\}\) with \(p(n)=2^{-n}\) for each positive integer \(n\). Each outcome has positive probability. The event \(E=\{2,4,6,\ldots\}\) has probability

P(E)=\sum_{k=1}^{\infty}2^{-2k} =\sum_{k=1}^{\infty}\left(\frac14\right)^k =\frac{1/4}{1-1/4} =\frac13.

More generally, any nonempty event \(A\) contains some integer \(m\). By monotonicity, \(P(A)\geq P(\{m\})=2^{-m}>0\). Thus this countable model has no nonempty null events. Countability alone does not guarantee that zero-probability outcomes exist; the point probabilities decide.

Why the Distinction Matters

The distinction prevents a common error: concluding that an outcome cannot belong to the outcome space merely because its probability is zero. A probability law assigns numerical weights to events; the value zero does not remove an outcome from the set on which the model was defined. If a model includes a zero-mass outcome, it may be mathematically valid, as long as the probabilities remain nonnegative and sum to one.

It is equally important not to infer too much in the other direction. A countable model in which every individual outcome has positive probability cannot contain a nonempty zero-probability event. Any nonempty event contains at least one outcome, and its probability is at least the positive probability of that outcome. The precise conclusion depends on the point probabilities, not simply on whether the outcome space is finite, countable, or large.

Common mistake: “The event has probability zero, so it must be the empty event.” The reasoning confuses a numerical assignment with set membership. The correction is to check both questions separately: does the event contain any outcomes, and what probability does the law assign to it? In a countable model, inspect the point probabilities: a nonempty event has probability zero exactly when every outcome in it has probability zero.

Another mistake is to suppose that a countably infinite collection of zero-probability outcomes must accumulate positive probability simply because it contains infinitely many outcomes. Countable additivity says the probability is the sum of the singleton probabilities. A countable sum of zeros is zero; an infinite number of terms does not change that conclusion.

Key takeaway: The empty event is always a null event, but a null event can be nonempty. In a countable model, a nonempty null event exists precisely when the model assigns probability zero to at least one individual outcome.

Check Your Understanding

Use the definitions and countable-model results to answer each question.

Explain why \(\varnothing\) must have probability zero, using the probability axioms.
A countable model assigns positive probability to every individual outcome. Can it have a nonempty event with probability zero? Explain.
Suppose \(P(\{a\})=0\) in a model whose outcome space contains \(a\). Is \(\{a\}\) empty? What is its probability?
In a countable model, every outcome in \(A\) has probability zero. State the condition under which countable additivity implies \(P(A)=0\).
A nonempty countable event \(B\) has probability zero. What must be true of every individual outcome in \(B\)?