Why Basic Probability Rules Matter
A probability model must satisfy more than the values assigned to particular outcomes. Its rules also constrain the probabilities of every event formed from those outcomes. These basic consequences let you check calculations, compare events, and recognize when a set of probability statements cannot describe any valid model.
This tutorial derives three such consequences from the probability axioms: the empty event has probability zero, every event probability lies between zero and one, and an event contained in another event cannot have a greater probability. The arguments rely only on nonnegativity, normalization, and additivity for disjoint events.
Probability Axioms Used Here
Let \(S\) be a sample space and let events be subsets of \(S\). For any event \(A\), its probability is written \(P(A)\). The probability axioms used below are:
- Nonnegativity: \(P(A)\geq 0\) for every event \(A\).
- Normalization: \(P(S)=1\).
- Additivity for disjoint events: if \(A\) and \(B\) cannot occur together, then \(P(A\cup B)=P(A)+P(B)\).
Two events are disjoint when their intersection is empty. The set difference \(B\setminus A\) contains outcomes in \(B\) that are not in \(A\). The notation \(A\subseteq B\) means every outcome in \(A\) is also in \(B\); equality is allowed.
Deriving the Probability of the Empty Event
The empty event, written \(\varnothing\), contains no outcomes. It is disjoint from every event \(A\), and combining it with \(A\) does not change \(A\): \(A\cup\varnothing=A\). Apply additivity to this union:
But \(A\cup\varnothing=A\), so the left side is \(P(A)\). Therefore \(P(A)=P(A)+P(\varnothing)\). Subtracting \(P(A)\) from both sides gives the result.
This conclusion does not require the empty event to be assigned a probability separately. It follows from additivity and the fact that adjoining the empty event changes nothing. The argument works for any event \(A\), including \(S\).
Deriving the Bounds on Event Probabilities
Nonnegativity gives the lower bound \(P(A)\geq 0\). For the upper bound, \(A\) and its complement \(A^c=S\setminus A\) are disjoint, and together they make up all of \(S\). Normalization and additivity therefore give:
Since \(P(A^c)\geq 0\), rearranging yields \(P(A)\leq 1\). Combining the lower and upper bounds gives the full result.
The complement identity \(P(A^c)=1-P(A)\) follows from the same equation. It is a consequence of the axioms, not a separate rule that permits probabilities outside the interval from zero to one.
Deriving Monotonicity from Event Inclusion
Suppose \(A\subseteq B\). Every outcome in \(A\) is already in \(B\), so the portion of \(B\) not in \(A\) is \(B\setminus A\). The events \(A\) and \(B\setminus A\) are disjoint, and their union is \(B\). Applying additivity gives:
By nonnegativity, \(P(B\setminus A)\geq 0\). Hence \(P(B)\geq P(A)\). This proves monotonicity: including more outcomes cannot reduce an event's probability.
The proof also gives a useful exact difference: whenever \(A\subseteq B\), the probability increase from \(A\) to \(B\) is precisely the probability of the outcomes added to form \(B\). Inclusion alone guarantees that this difference is nonnegative, not that it is positive.
Worked Applications
Worked Example: Checking an Event and Its Complement
A finite experiment has outcomes \(r,s,t,u\), with probabilities \(0.10, 0.20, 0.30,\) and \(0.40\), respectively. Let \(A=\{s,u\}\). Find \(P(A)\), use the axioms to determine \(P(\varnothing)\), and verify the bounds using the complement of \(A\).
Method. Add the probabilities of the outcomes in \(A\), then apply the empty-event and complement results. The common mistake to avoid is treating the probability of an event and its complement as unrelated quantities.
First, the event \(A\) contains \(s\) and \(u\), so \(P(A)=P(\{s\})+P(\{u\})=0.20+0.40=0.60\).
Next, the empty-event result gives \(P(\varnothing)=0\). The complement is \(A^c=\{r,t\}\), so \(P(A^c)=0.10+0.30=0.40\). The complement identity also gives \(P(A^c)=1-P(A)=1-0.60=0.40\).
Finally, \(0\leq 0.60\leq 1\), and \(P(A)+P(A^c)=0.60+0.40=1\). Both the bounds and complement calculation check.
Worked Example: Comparing Nested Events
In a hypothetical review of submitted benefit claims, let \(A\) be the event that a claim is flagged for manual review and \(B\) the event that it is either flagged for manual review or referred for additional documentation. Thus \(A\subseteq B\). Suppose \(P(A)=0.34\) and \(P(B\setminus A)=0.21\). Find \(P(B)\), and check the inclusion and probability bounds.
Method. Decompose \(B\) into the disjoint events \(A\) and \(B\setminus A\). The common mistake is to compare \(P(A)\) with the probability of the additional category alone, rather than with the probability of all of \(B\).
Because \(A\) and \(B\setminus A\) are disjoint and their union is \(B\), \(P(B)=P(A)+P(B\setminus A)=0.34+0.21=0.55\).
The inclusion rule requires \(P(A)\leq P(B)\), and indeed \(0.34\leq 0.55\). The added part has probability \(0.21\), so the increase from \(A\) to \(B\) is \(0.55-0.34=0.21\), as required. Both probabilities satisfy the bounds: \(0\leq0.34\leq1\) and \(0\leq0.55\leq1\).
As a final check, the three disjoint categories \(A\), \(B\setminus A\), and \(B^c\) have probabilities \(0.34\), \(0.21\), and \(1-0.55=0.45\). Their total is \(0.34+0.21+0.45=1.00\).
Worked Example: Detecting Inconsistent Probability Claims
For two events \(C\) and \(D\) in a proposed probability model, suppose a report states that \(D\subseteq C\), \(P(D)=0.58\), and \(P(C)=0.42\). It also states that \(P(C\setminus D)=0.16\). Determine whether all three claims can be true.
Method. Apply monotonicity and then check the disjoint decomposition \(C=D\cup(C\setminus D)\). The common mistake is to accept a stated probability just because it lies between zero and one; valid individual bounds do not guarantee that several claims are mutually consistent.
Since \(D\subseteq C\), monotonicity requires \(P(D)\leq P(C)\). The stated values would require \(0.58\leq0.42\), which is false. Thus the claims are already inconsistent.
The decomposition gives a second check. Because \(D\subseteq C\), the events \(D\) and \(C\setminus D\) are disjoint and make up \(C\). The claims would imply \(P(C)=P(D)+P(C\setminus D)=0.58+0.16=0.74\), but the report states \(P(C)=0.42\). This is another contradiction.
Therefore no probability model can satisfy all the stated claims. In particular, the issue is not that \(0.58\), \(0.42\), or \(0.16\) individually falls outside the probability bounds; it is their incompatibility with inclusion and decomposition.
Common Mistakes and Exam Checks
- Assuming the empty event has positive probability because it is an event. Additivity with \(A\cup\varnothing=A\) forces \(P(\varnothing)=0\).
- Checking only nonnegativity. A proposed probability of \(1.2\) is nonnegative but violates the upper bound. Check both \(0\leq P(A)\) and \(P(A)\leq1\).
- Reversing the inclusion inequality. If \(A\subseteq B\), then \(P(A)\leq P(B)\), not the reverse.
- Assuming proper inclusion guarantees a strict inequality. A larger event can add outcomes whose total probability is zero. Strict inequality follows only when \(P(B\setminus A)>0\).
- Using bounds to validate an entire model. Each stated probability can lie between zero and one while the statements still contradict event inclusion or a disjoint decomposition.
Check Your Understanding
For each problem, use the axioms or the results derived from them, and check that your conclusion is consistent with the stated information.
- Events \(E\) and \(F\) are disjoint, \(P(E)=0.27\), and \(P(F)=0.38\). Let \(G=E\cup F\). Find \(P(G)\) and state whether it satisfies the probability bounds. (Answer: \(P(G)=0.65\), which satisfies \(0\leq P(G)\leq1\).)
- A quality-control event \(H\) is contained in event \(J\). If \(P(H)=0.19\) and \(P(J\setminus H)=0.26\), find \(P(J)\) and the probability difference between the two events. (Answer: \(P(J)=0.45\) and \(P(J)-P(H)=0.26\).)
- A proposed model assigns \(P(K)=0.73\) and \(P(K^c)=0.32\). Can these values both be correct? Use the complement relationship to explain. (Answer: No; they sum to \(1.05\), but complementary event probabilities must sum to \(1\).)
- Events \(L\subseteq M\) have \(P(L)=0.48\) and \(P(M)=0.48\). Find \(P(M\setminus L)\). Does the equality of probabilities contradict proper inclusion if \(L\) is a proper subset of \(M\)? (Answer: \(P(M\setminus L)=0\); no, proper inclusion is possible when the added event has probability zero.)
- Events \(R\subseteq T\) are reported to have \(P(R)=0.61\), \(P(T)=0.83\), and \(P(T\setminus R)=0.17\). Decide whether the values are consistent and identify the conflicting calculation. (Answer: No; the decomposition requires \(P(T)=0.61+0.17=0.78\), not \(0.83\).)