Common Errors in Events and Probability Assignments

Why Probability Errors Often Begin Before the Calculation

Many incorrect probability calculations use arithmetic that is perfectly correct on the wrong model. The difficulty is often earlier: the outcomes have not been specified, two events overlap, or an assumption of equal likelihood has been added without justification. A calculation can be checked only after the experiment and its probability assignments are clear.

This tutorial develops a practical audit for those errors. The main questions are: What counts as one outcome? Do the listed outcomes cover every possibility? What mechanism assigns probabilities to them? And, when several events are combined, have any outcomes been counted more than once?

Outcomes, Events, and Probability Assignments

Definition: A sample space \(\Omega\) is the set of elementary outcomes for a specified experiment. An event \(A\) is a subset of \(\Omega\). A probability assignment \(P\) gives each event a number between 0 and 1, with \(P(\Omega)=1\) and countable additivity for disjoint events. On a finite or countable sample space, this can be specified by outcome probabilities \(p(\omega)=P(\{\omega\})\), which must be nonnegative and sum to 1; an event’s probability is the sum of the probabilities of its outcomes.

These definitions make the phrase “the probability of an outcome” depend on the model: the experiment must say what is recorded as an elementary result, and the assignment must describe how those results occur. A sample space can be refined or summarized, but its outcomes cannot be assigned probabilities arbitrarily. In particular, writing down a set of possible labels does not make those labels equally likely.

For a finite sample space, equal likelihood means \(P(\{\omega\})=1/|\Omega|\) for every \(\omega\in\Omega\). This conclusion is justified only when the experimental mechanism supports it. The uniform probability model and symmetry can provide such a justification; merely having a short list of outcomes cannot.

Worked Example: Two Dice and Two Different Sample Spaces

Roll two fair six-sided dice and ask for the probability that their sum is 7. One description of the result is the ordered pair \((i,j)\), where \(i\) is the first die’s face and \(j\) is the second die’s face. There are \(6\cdot 6=36\) such pairs, and fairness and independence make each pair have probability \(1/36\). Exactly six pairs, \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\), sum to 7. Therefore,

P(\text{sum is }7)=\frac{6}{36}=\frac{1}{6}.

A different description records only the sum, so the possible recorded values are \(2,3,\ldots,12\). There are 11 possible sums, but they are not equally likely. For example, sum 2 comes only from \((1,1)\), while sum 7 comes from six pairs. Thus assigning probability \(1/11\) to each sum would describe a different experiment, not the sum of two fair dice.

The two descriptions are compatible when the probability of a sum is calculated by combining the probabilities of all pairs that produce it. The mistake is not using a summarized sample space; it is treating its distinct values as equally likely without checking the underlying mechanism.

Overlapping Events: Count the Intersection Only Once

If events overlap, adding their probabilities counts outcomes in their intersection twice. The correct relationship follows by splitting their union into disjoint pieces.

Theorem: For any two events \(A\) and \(B\),

P(A\cup B)=P(A)+P(B)-P(A\cap B).

Proof. The events \(A\setminus B\), \(A\cap B\), and \(B\setminus A\) are pairwise disjoint, and their union is \(A\cup B\). By additivity, \(P(A\cup B)=P(A\setminus B)+P(A\cap B)+P(B\setminus A)\). Also, \(A\) is the disjoint union of \(A\setminus B\) and \(A\cap B\), so \(P(A)=P(A\setminus B)+P(A\cap B)\); likewise \(P(B)=P(B\setminus A)+P(A\cap B)\). Adding these last two equalities counts \(P(A\cap B)\) twice. Subtracting it once gives the stated formula. \(\square\)

Worked Example: The Overlap When Rolling Two Dice

Continue with two fair dice and the 36 equally likely ordered pairs. Let \(A\) be the event that the sum is at least 8, and let \(B\) be the event that the sum is even. The sums 8 through 12 have respectively 5, 4, 3, 2, and 1 ordered pairs, so \(P(A)=15/36\). There are 18 pairs with an even sum, so \(P(B)=18/36\).

The intersection consists of the even sums that are at least 8: sums 8, 10, and 12. They have 5, 3, and 1 ordered pairs, respectively. Thus \(P(A\cap B)=9/36\). Applying the union formula,

P(A\cup B)=\frac{15}{36}+\frac{18}{36}-\frac{9}{36} =\frac{24}{36}=\frac{2}{3}.

The direct count checks the result: the event “sum at least 8 or sum even” contains 24 ordered pairs. Simply adding \(15/36\) and \(18/36\) would give \(33/36\), because the nine pairs in the intersection would be counted twice.

Equal Likelihood Must Come From the Mechanism

A frequent error is to divide by the number of named outcomes even though the experiment does not select those outcomes with equal probabilities. “There are three possible labels” does not imply that each has probability \(1/3\). To justify equal likelihood, specify what is selected and why the selection mechanism treats the elementary outcomes symmetrically.

Worked Example: An Unequal Spinner

A spinner has three sectors labelled A, B, and C. Their proportions of the circle are \(1/2\), \(1/3\), and \(1/6\), respectively. Assume the spinner is fair in the sense that the landing position is uniform over the circle. What is the probability of landing on A or C?

The sectors are not equal in size, so the three labels are not equally likely. The probability of landing in a sector is its fraction of the circle. The sectors A and C are disjoint, hence

P(\text{A or C})=P(\text{A})+P(\text{C}) =\frac{1}{2}+\frac{1}{6}=\frac{2}{3}.

Assigning probability \(1/3\) to each label would give \(2/3\) for A or C as well, but only by coincidence: the correct calculation gives \(1/2+1/6=2/3\), whereas the claimed model would assign \(2/3\) to A alone instead of \(1/2\). A result for one event that happens to match does not validate an incorrect probability assignment.

Omitted Outcomes and Probability Totals

A proposed list of outcomes must describe every result the experiment can produce. If a possible outcome is missing, then the listed probabilities may fail to sum to 1, or the model may silently describe a different experiment. A basic audit is to check both coverage and total mass:

Model check: For a finite or countable list \(\Omega\) of elementary outcomes, require \(p(\omega)\geq 0\) for every \(\omega\in\Omega\) and \(\sum_{\omega\in\Omega}p(\omega)=1\). These numerical conditions do not replace checking that \(\Omega\) actually contains all possible outcomes of the stated experiment.

Worked Example: A Missing Face on a Ten-Sided Die

A fair ten-sided die has faces labelled 1 through 10. Its outcome space is \(\{1,2,\ldots,10\}\), with probability \(1/10\) for each face. Suppose a calculation omits face 10 and then assigns probability \(1/9\) to each of 1 through 9.

The listed values sum to 1, but they do not model the stated die: face 10 can occur and has been left out. For example, the true probability of a result at most 3 is \(3/10\), while the incomplete model gives \(3/9=1/3\). The false model has effectively changed the experiment to a die conditioned not to show 10.

A sum of 1 is necessary but not sufficient: the list may be normalized and still omit possible outcomes by assigning the remaining probability incorrectly. First specify all outcomes; then assign and check their probabilities.

Sometimes numbers are supplied as weights rather than probabilities. Nonnegative weights \(w(\omega)\) with finite positive total can be normalized by dividing each by their sum. But if numbers are already claimed to be probabilities and their sum is not 1, they do not form a probability assignment as given.

Worked Example: Repairing Weights, Not Probabilities

Suppose three outcomes have stated weights 2, 3, and 6. Their total weight is \(2+3+6=11\), so they can be converted into probabilities by dividing by 11:

\left(\frac{2}{11},\frac{3}{11},\frac{6}{11}\right), \qquad \frac{2}{11}+\frac{3}{11}+\frac{6}{11}=1.

The third outcome then has probability \(6/11\). This repair is justified because the original numbers were weights, meaning relative amounts that were not required to sum to 1. If instead the proposal claimed that the three probabilities were \(0.2,0.3,0.6\), their sum would be \(1.1\), which is impossible for a probability distribution. One must correct the model using information about the experiment; simply calling invalid probabilities “weights” after the fact does not justify a repair.

A Practical Audit and Common Mistakes

For a proposed model, use this sequence before calculating event probabilities:

Specify the experiment and recorded outcome.
Say what is observed on one trial. Distinguish elementary results from summaries such as a sum, category, or count.

Check the sample space.
Confirm that the outcomes are distinct and that every possible result is included.

Justify the probabilities.
Use the mechanism or a stated symmetry argument; do not infer equal likelihood from the number of labels.

Check event relationships and totals.
Identify intersections before adding event probabilities, and verify nonnegativity and a total of 1.

A common wrong argument is: “There are 11 possible sums, so each sum has probability \(1/11\).” The correction is that the number of possible values does not establish their probabilities. For two fair dice, the sums arise from different numbers of equally likely ordered pairs, so the sum outcomes have unequal probabilities.

Another wrong argument is: “Add the probabilities of the two events; each probability is correct on its own.” The correction is to check whether the events are disjoint. If not, subtract the intersection once, as in the union formula. Conversely, when a list of outcomes gives a total below or above 1, do not proceed as if the numbers were probabilities: investigate missing outcomes, duplicate outcomes, or incorrect assignments first.

Key takeaway: A sound calculation needs a complete sample space, a probability assignment justified by the experiment, and event arithmetic that respects overlap. A numerical total of 1 is an essential check, but it cannot by itself establish that the model matches the experiment.

Check Your Understanding

For each question, identify the modeling issue before attempting a calculation.

Two fair dice are rolled. Why are the 11 possible sums not equally likely? Give the number of ordered pairs producing sums 2 and 7.
If \(P(A)=0.4\), \(P(B)=0.5\), and \(P(A\cap B)=0.2\), what is \(P(A\cup B)\)? What incorrect result comes from adding \(P(A)\) and \(P(B)\) directly?
A proposed model assigns probabilities \(0.3,0.3,0.3\) to three outcomes. What must be checked before using it, and what basic numerical condition does it fail?
A fair ten-sided die is described using only outcomes 1 through 9, each with probability \(1/9\). Why is this not a model of the stated die?
Three nonnegative weights are 1, 4, and 5. Convert them to probabilities and verify the total.