Why Three Events Need an Extra Check
For two events, independence is checked by comparing the probability of their intersection with the product of their probabilities. With three events, checking each pair is not enough to establish mutual independence. A three-event counterexample makes the distinction precise: every pair can satisfy the independence equation even though the three events together do not.
The central calculation is therefore not complicated, but it has several parts. Find the probability of each event, check all three pairs, and then compare the probability of the three-way intersection with the product of all three marginal probabilities. The last comparison is the one that can expose a failure missed by the pairwise checks.
A Four-Outcome Construction
A useful construction uses two fair binary results. Let the sample space be \(\{00,01,10,11\}\), where each outcome has probability \(1/4\). Define \(A\) to mean that the first digit is 1, \(B\) to mean that the second digit is 1, and \(C\) to mean that the two digits are equal.
The event sets are \(A=\{10,11\}\), \(B=\{01,11\}\), and \(C=\{00,11\}\). Each event contains two of the four equally likely outcomes, so each has probability \(1/2\). Each pair intersects only at \(11\), so every pairwise intersection has probability \(1/4\). But all three events also intersect only at \(11\), giving a three-way intersection of \(1/4\), not \(1/8\).
Worked Example: Checking the Equality-of-Digits Model
A hypothetical device generates one of four labels, \(00\), \(01\), \(10\), or \(11\), each with probability \(1/4\). Let \(A\) be the event that the first digit is 1, \(B\) the event that the second digit is 1, and \(C\) the event that the digits match. Determine whether the events are pairwise independent and whether they are mutually independent.
Method. List the outcomes in each event, calculate the three event probabilities, check each pair, and then check the triple intersection. This avoids the mistake of deciding mutual independence after checking only one or two pairs.
The event sets are \(A=\{10,11\}\), \(B=\{01,11\}\), and \(C=\{00,11\}\). Hence $$ P(A)=P(B)=P(C)=\frac{2}{4}=\frac12. $$
For the first pair, \(A\cap B=\{11\}\), so \(P(A\cap B)=1/4\), while \(P(A)P(B)=(1/2)(1/2)=1/4\). For the second pair, \(A\cap C=\{11\}\), so \(P(A\cap C)=1/4=P(A)P(C)\). For the third pair, \(B\cap C=\{11\}\), so \(P(B\cap C)=1/4=P(B)P(C)\). All three pairs are independent.
The triple intersection is \(A\cap B\cap C=\{11\}\), giving \(P(A\cap B\cap C)=1/4\). The product of the three marginal probabilities is $$ P(A)P(B)P(C)=\left(\frac12\right)^3=\frac18. $$ Since \(1/4\ne 1/8\), the events are not mutually independent. The check is complete: they are pairwise independent, but not mutually independent.
Why Pairwise Checks Do Not Determine the Triple
The pairwise equations constrain the probabilities of \(A\cap B\), \(A\cap C\), and \(B\cap C\). They do not, by themselves, require the probability of \(A\cap B\cap C\) to equal the product of all three marginal probabilities. The four-outcome model shows this directly: the pairwise intersections each have probability \(1/4\), but the triple intersection is also \(1/4\), whereas the product of the three marginals is \(1/8\).
This is a finite-model justification rather than a general shortcut: the construction verifies that the pairwise conditions can all hold while the additional mutual-independence condition fails. Therefore, when the question asks whether three events are mutually independent, the triple-intersection equation must be checked in addition to all pairwise equations.
Worked Example: A Counterexample With No Triple Intersection
Suppose a sample space has four equally likely outcomes, labeled \(1,2,3,4\). Define \(A=\{1,2\}\), \(B=\{1,3\}\), and \(C=\{2,3\}\). Decide whether the events are pairwise independent and whether they are mutually independent.
Method. Use the event sets to count each event, pair intersection, and the triple intersection. This example avoids the mistaken belief that the triple intersection must be too large whenever mutual independence fails; it can instead be too small.
Each event contains two outcomes, so \(P(A)=P(B)=P(C)=2/4=1/2\). The pair intersections are \(A\cap B=\{1\}\), \(A\cap C=\{2\}\), and \(B\cap C=\{3\}\). Each therefore has probability \(1/4\), equal to the corresponding product of marginal probabilities: $$ P(A)P(B)=P(A)P(C)=P(B)P(C)=\frac12\cdot\frac12=\frac14. $$ Thus the events are pairwise independent.
No outcome belongs to all three event sets, so \(A\cap B\cap C=\varnothing\) and \(P(A\cap B\cap C)=0\). But \(P(A)P(B)P(C)=(1/2)^3=1/8\). Since \(0\ne1/8\), the events are not mutually independent. The calculation confirms pairwise independence and failure of the required triple condition.
Accounting for Unequal Outcome Probabilities
A finite sample space does not automatically have equally likely outcomes. The event probability must be found by adding the probabilities of the outcomes it contains. Assuming equal probabilities when the model assigns unequal weights can invalidate every later comparison.
One way to adapt the four-outcome construction is to split each of its four labels into two outcomes while preserving a total probability of \(1/4\) for each label. The event calculations then use the assigned weights, not the number of listed outcomes.
Worked Example: Verifying a Weighted Eight-Outcome Model
A hypothetical system produces one of eight outcomes. Outcomes \(x_1,x_2\) each correspond to label \(00\), with probabilities \(0.10,0.15\); \(y_1,y_2\) correspond to \(01\), with probabilities \(0.20,0.05\); \(z_1,z_2\) correspond to \(10\), with probabilities \(0.12,0.13\); and \(w_1,w_2\) correspond to \(11\), with probabilities \(0.08,0.17\). Define \(A\) as first digit 1, \(B\) as second digit 1, and \(C\) as matching digits. Check pairwise and mutual independence.
Method. First verify the probability model and aggregate the probabilities within each two-outcome label. Then calculate event and intersection probabilities using those totals. This avoids treating the eight listed outcomes as equally likely.
The total probability is \(0.10+0.15+0.20+0.05+0.12+0.13+0.08+0.17=1.00\). The label totals are $$ P(00)=0.10+0.15=0.25,\quad P(01)=0.20+0.05=0.25, $$ $$ P(10)=0.12+0.13=0.25,\quad P(11)=0.08+0.17=0.25. $$
The events consist of labels \(10,11\) for \(A\), labels \(01,11\) for \(B\), and labels \(00,11\) for \(C\). Thus \(P(A)=0.25+0.25=0.50\), \(P(B)=0.25+0.25=0.50\), and \(P(C)=0.25+0.25=0.50\).
Each pair intersects at label \(11\). In the listed outcomes, \(A\cap B=\{w_1,w_2\}\), \(A\cap C=\{w_1,w_2\}\), and \(B\cap C=\{w_1,w_2\}\). Each intersection has probability \(0.08+0.17=0.25\), matching each pairwise product \(0.50(0.50)=0.25\). Therefore all pairs are independent.
The triple intersection is also \(\{w_1,w_2\}\), so \(P(A\cap B\cap C)=0.25\). The product of the marginals is \(0.50^3=0.125\). Because \(0.25\ne0.125\), the events are not mutually independent. The model is valid, pairwise independent, and not mutually independent.
Common Exam Trap
A frequent error is to check only the first pair and then claim that the events are pairwise independent—or to check all pairs and then claim they must be mutually independent. Neither conclusion follows without the required calculations.
For example, in the equality-of-digits model, the statement “\(A\) and \(B\) are independent because \(P(A\cap B)=1/4=P(A)P(B)\), so the three events are mutually independent” is wrong. The first equation establishes only that this one pair is independent. The correction is to check the other two pairs and the triple: the other pairs also factor, but \(P(A\cap B\cap C)=1/4\ne1/8=P(A)P(B)P(C)\). The events are pairwise independent, not mutually independent.
A related error is to conclude that events are not pairwise independent because one pair has equal intersection and product probabilities. Equality means that pair passes the independence check. A failure of pairwise independence must be demonstrated by at least one pair for which the two quantities differ.
Confirm that the outcome probabilities are nonnegative and sum to 1; do not assume equal likelihood without support.
Find \(P(A)\), \(P(B)\), and \(P(C)\) from the stated outcomes and weights.
Compare each pair-intersection probability with the product of the corresponding marginals.
Compare \(P(A\cap B\cap C)\) with \(P(A)P(B)P(C)\); do not infer this equality from the pair checks.
When the Distinction Matters
Pairwise independence says that examining any two of the events alone gives the factorization associated with independence. Mutual independence makes the stronger assertion that the joint occurrence of all three also factors. In probability models for several indicators, confusing these conditions can produce incorrect joint probabilities even when every two-event calculation appears consistent.
For exam work, a compact but complete response should state the event probabilities, show the three pairwise comparisons, and then show the triple comparison. If the question asks for a counterexample, list the sample space and the event definitions clearly enough that every intersection can be checked. This is aligned with the SOA VEE Mathematical Statistics syllabus.
Check Your Understanding
For each finite model, calculate the requested probabilities before checking the answer.
- A uniform four-outcome space has outcomes \(00,01,10,11\). Let \(A\) mean first digit 1, \(B\) mean second digit 1, and \(C\) mean that the digits differ. Are the events pairwise independent, and are they mutually independent? (Answer: Pairwise independent but not mutually independent; each marginal is \(1/2\), each pair intersection is \(1/4\), and the triple intersection is 0 rather than \(1/8\).)
- On a uniform four-outcome space \(\{1,2,3,4\}\), let \(A=\{1,2\}\), \(B=\{1,3\}\), and \(C=\{2,3\}\). What is the triple-intersection probability, and are the events mutually independent? (Answer: The triple-intersection probability is 0; they are not mutually independent, since the product of their marginals is \(1/8\).)
- On a uniform eight-outcome space, let \(A=\{1,2,3,4\}\), \(B=\{1,2,5,6\}\), and \(C=\{1,3,5,7\}\). Are the events mutually independent? (Answer: Yes; each marginal is \(1/2\), each pair intersection has probability \(1/4\), and the triple intersection \(\{1\}\) has probability \(1/8\).)
- Four labels \(00,01,10,11\) each have probability \(1/4\), but each label is represented by two outcomes with unequal individual weights. If \(A,B,C\) are respectively first digit 1, second digit 1, and matching digits, what are the three marginal probabilities and the triple-intersection probability? (Answer: Each marginal is \(1/2\); the triple-intersection probability is the total weight of label \(11\), namely \(1/4\).)
- In the uniform four-outcome model, let \(A=\{1,2\}\), \(B=\{1,3\}\), and \(C=\{1,2\}\). Are the events pairwise independent? (Answer: No; \(P(A\cap C)=1/2\), whereas \(P(A)P(C)=1/4\).)