Counting Subsets for Discrete Uniform Probabilities

Why Counting Gives the Probability

When a discrete random variable can take only finitely many values, an event can be described by listing the support values that satisfy its condition. If every support value is equally likely, the probability is the fraction of those values that satisfy the condition. The key is to count distinct support points, not the number of ways you happened to describe them.

This method applies to any finite support, not only consecutive integers. It also applies when the outcomes are subsets, provided the random experiment gives every subset in the stated collection the same probability. Before counting, identify exactly what counts as one outcome and verify that the outcomes are equally likely.

Definition: A random variable $X$ is discrete uniform on a finite support $S$ if $S$ is a finite set of possible values and $P(X=x)=1/|S|$ for every $x\in S$. For any event $A$ specified by a condition on $X$, $$ P(X\in A)=\frac{|A\cap S|}{|S|}. $$ Here $|A\cap S|$ is the number of distinct support values satisfying the condition, and $|S|$ is the total number of support values.

The numerator is not the number of descriptions of the event, nor the number of calculations used to find it. Each support value is counted once, even if it satisfies more than one part of the event. If the event is a union of conditions, the overlap must therefore be handled correctly.

A Short Justification

Suppose the support has $N$ points. Since the distribution is discrete uniform, each point has probability $1/N$. An event containing $r$ support points is the union of $r$ mutually exclusive outcomes, so additivity of probability gives its probability as $r(1/N)$. This yields the counting rule:

P(X\in A)=\sum_{x\in A\cap S}P(X=x) =\sum_{x\in A\cap S}\frac{1}{|S|} =\frac{|A\cap S|}{|S|}.

For a subset-valued outcome, the support consists of the allowed subsets. If one subset is selected uniformly from all $k$-element subsets of $n$ distinct objects, there are $\binom{n}{k}$ support points. A favorable count can often be found by choosing objects from separate categories and multiplying the counts of independent choices. This counts subsets, not ordered lists.

Counting rule for subsets: If each $k$-element subset of $n$ distinct objects is equally likely, then an event has probability $$ \frac{\text{number of favorable }k\text{-element subsets}}{\binom{n}{k}}. $$ This rule requires uniform selection among subsets. If the experiment instead assigns different probabilities to different subsets, the favorable subsets cannot simply be divided by their total number.

Worked Examples

Worked Example: A Union of Conditions on a Finite Support

A hypothetical calibration device selects one value uniformly from $S=\{-8,-3,1,4,7,10,13,16,19,22,25,31\}$. Find the probability that the selected value is negative or satisfies $(X+3)(X-4)=0$.

Method. Count the support points satisfying each condition, then subtract any overlap. The support has 12 distinct values, each with probability $1/12$.

Let $A$ be the event $X<0$. The support values satisfying it are $-8$ and $-3$, so $|A|=2$.

Let $B$ be the event $(X+3)(X-4)=0$. The equation has roots $-3$ and $4$, both in the support, so $|B|=2$. The value $-3$ satisfies both conditions, so $|A\cap B|=1$.

By inclusion-exclusion, the number satisfying at least one condition is $2+2-1=3$. Therefore $$ P(A\cup B)=\frac{3}{12}=0.2500. $$ The result is between 0 and 1, and exactly three listed support points, $-8,-3,4$, satisfy the event. This check confirms the count. Common mistake avoided: adding the two counts without subtracting the shared value would incorrectly give $4/12$.

Worked Example: A Committee with a Category Requirement

A hypothetical committee is selected uniformly from all 4-person committees formed from 3 qualified analysts, $Q_1,Q_2,Q_3$, and 4 other specialists, $N_1,N_2,N_3,N_4$. Among the other specialists, $N_1$ and $N_2$ are designated technicians. Find the probability that the committee has exactly 2 qualified analysts and at least 1 designated technician.

Method. Count all possible committees, then count favorable committees by first choosing the qualified analysts and then the other specialists. Since each committee is an unordered subset, order within a committee does not create a new outcome.

The total number of committees is $\binom{7}{4}=35$.

To have exactly 2 qualified analysts, choose 2 of the 3 qualified analysts: $\binom{3}{2}=3$ choices. The remaining 2 committee members must come from the 4 other specialists. There are $\binom{4}{2}=6$ pairs of other specialists, but one pair, $\{N_3,N_4\}$, has no designated technician. Thus $6-1=5$ pairs satisfy the requirement.

The favorable count is $3\times5=15$. Hence $$ P(\text{exactly 2 qualified and at least 1 designated}) =\frac{15}{35} =\frac{3}{7} =0.4286. $$ The count is no greater than the 35 possible committees, as it must be. Common mistake avoided: choosing the two remaining specialists from all six pairs would include the one pair that fails the designated-technician condition.

Worked Example: Counting by Disjoint Cases

A hypothetical inspection team selects 4 components uniformly from 9 distinct components: 3 priority components, 2 reserve components, and 4 ordinary components. Find the probability that the selection contains at least 2 priority components and at least 1 reserve component.

Method. Count the total 4-component subsets, then split the favorable selections into disjoint cases according to the numbers of priority and reserve components selected. This prevents any subset from being counted twice.

There are $\binom{9}{4}=126$ possible selections. Write $k$ for the number of priority components and $j$ for the number of reserve components. The requirements are $k\geq2$, $j\geq1$, and the total selected is 4.

If $k=2$ and $j=1$, one ordinary component is also needed. The number of selections is $\binom{3}{2}\binom{2}{1}\binom{4}{1}=3\times2\times4=24$.

If $k=2$ and $j=2$, no ordinary component is needed. The count is $\binom{3}{2}\binom{2}{2}=3\times1=3$.

If $k=3$, exactly one more component must be selected. It must be a reserve component to meet the reserve requirement, giving $\binom{3}{3}\binom{2}{1}=1\times2=2$ selections. These cases exhaust the possibilities: selecting 3 priority components leaves room for only 1 other component.

The favorable total is $24+3+2=29$, so $$ P(\text{at least 2 priority and at least 1 reserve}) =\frac{29}{126} =0.2302. $$ The disjoint cases account for 29 of the 126 possible selections, which is consistent with the probability being less than $1/2$. Common mistake avoided: counting only selections with exactly 2 priority components would omit the valid case with 3.

Checks Before You Divide

A correct fraction depends on using the right support and counting its points once. These checks are especially useful when the wording describes an experiment first and a random variable second.

State one outcome clearly. For a committee, one outcome is a particular subset of people, not an ordered sequence of names, unless order is part of the experiment.
Verify equal likelihood. Counting outcomes works directly only when each support point has the same probability.
Count only outcomes in the support. A solution to an equation is favorable only if it is one of the possible values.
Check overlaps and cases. For a union, subtract overlap when adding counts. For a case split, make the cases disjoint and exhaustive.
Keep a derived variable separate from its original outcomes. If equally likely outcomes are mapped to values of $X$, different values of $X$ may have different numbers of preimages. In that situation $X$ need not be uniform, so counting its distinct values is not enough.

For example, suppose a fair six-sided die is rolled twice and $X$ is the sum. The 36 ordered roll pairs are equally likely, but the possible sums are not: a sum of 7 has six pairs, whereas a sum of 2 has one. The support values therefore do not have equal probabilities. The counting rule can still be used on the 36 equally likely roll pairs, but not by treating the 11 distinct sums as equally likely.

Key takeaway: For a discrete uniform variable on a finite support, count the distinct support points satisfying the event and divide by the total number of support points. For subset outcomes, count the subsets themselves, and check that the experiment makes those subsets equally likely.

Check Your Understanding

For each problem, identify the equally likely outcomes before counting favorable ones.

A variable is uniform on $S=\{-9,-4,0,3,6,9,12,15\}$. Find the probability that $X<0$ or $X^2=81$. (Answer: $3/8=0.3750$)
Two objects are selected uniformly from 6 distinct objects, including designated objects $A$ and $B$. Find the probability that exactly one of $A$ and $B$ is selected. (Answer: $8/15=0.5333$)
A 3-person group is selected uniformly from 8 people, of whom 3 are flagged for a hypothetical review. Find the probability that exactly 2 selected people are flagged. (Answer: $15/56=0.2679$)
A 4-component selection is made uniformly from 8 components: 3 priority, 2 reserve, and 3 ordinary. Find the probability of selecting at least 2 priority components and at least 1 reserve component. (Answer: $23/70=0.3286$)
A 3-object subset is selected uniformly from 7 objects, 2 of which are marked. Find the probability that exactly one marked object is selected. (Answer: $20/35=4/7=0.5714$)

What You'll Learn