Why Tail Cutoffs Need Care
A hypergeometric tail probability adds the probabilities of all possible counts on one side of a cutoff. The arithmetic is usually straightforward; the main risk is selecting the wrong endpoint. In particular, “at least,” “more than,” and “no more than” do not describe the same event for an integer-valued count.
Complements can reduce the number of terms to calculate. To use them correctly, first identify exactly which support values belong to the event, then identify which values belong to its complement. This tutorial focuses on those tail boundaries and on checking that the resulting probabilities cover the intended outcomes.
The denominator counts all samples of size \(n\). For a sample to contain exactly \(x\) successes, choose \(x\) from the \(K\) successes and \(n-x\) from the \(N-K\) non-successes. The bounds \(L\) and \(U\) matter: a count cannot be negative, exceed the sample size, exceed the number of successes in the population, or require more non-successes than exist.
Translate the Words Into Integer Endpoints
The following translations apply because \(X\) takes integer values. They determine the first or last support value to include before any probability is calculated.
| Event | Equivalent integer event | Support values included |
|---|---|---|
| At least \(r\), \(X\ge r\) | \(X\ge r\), when \(r\) is an integer | \(\max(L,r),\ldots,U\) |
| More than \(r\), \(X>r\) | \(X\ge r+1\), when \(r\) is an integer | \(\max(L,r+1),\ldots,U\) |
| At most \(r\), \(X\le r\) | \(X\le r\), when \(r\) is an integer | \(L,\ldots,\min(U,r)\) |
| Fewer than \(r\), \(X<r\) | \(X\le r-1\), when \(r\) is an integer | \(L,\ldots,\min(U,r-1)\) |
For an integer cutoff \(r\), complements give
The integer condition is important. If \(r\) is not an integer, first translate the event to an integer cutoff. For any real \(r\),
Here \(\lceil r\rceil\) is the smallest integer at least \(r\), and \(\lfloor r\rfloor\) is the largest integer at most \(r\). These formulas reflect the fact that an integer-valued count cannot take a value strictly between consecutive integers. For example, \(X\ge 2.4\) means \(X\ge3\), while \(X\le2.4\) means \(X\le2\).
Why the Complement Formula Works
For any event \(A\), the event \(A^c\) contains exactly the outcomes that are not in \(A\). The two events are disjoint and together contain every possible outcome, so their probabilities sum to one. For a hypergeometric count and integer \(r\), the event \(X\ge r\) and its complement \(X\le r-1\) partition every integer value in the support.
This is useful when the complement requires fewer probability terms. It does not change the distribution or the sample space; it changes only which side of the cutoff is summed. If the cutoff lies outside the support, the same logic applies, but the event may be certain or impossible. For instance, if \(r>U\), then \(P(X\ge r)=0\), and if \(r\le L\), then \(P(X\ge r)=1\).
Worked Examples
Worked Example: An Upper Tail by Complement
A hypothetical inspection selects 4 components without replacement from a lot of 12. Exactly 5 components in the lot have a specified feature. Find the probability that the sample contains at least 3 featured components. The model is \(X\sim\operatorname{Hypergeometric}(N=12,K=5,n=4)\); the interest rate and mortality assumption are not applicable to this sampling problem.
Method and support. The support is \(L=\max(0,4-7)=0\) through \(U=\min(4,5)=4\). “At least 3” means \(X\ge3\), so the direct upper-tail values are 3 and 4. The complement is \(X\le2\), which has three values; here direct summation of the upper tail is shorter.
Calculate the denominator. \(\binom{12}{4}=495\), the number of possible samples.
Calculate the included terms. For \(x=3\), the numerator is \(\binom{5}{3}\binom{7}{1}=10(7)=70\). For \(x=4\), it is \(\binom{5}{4}\binom{7}{0}=5(1)=5\). Thus \(P(X\ge3)=(70+5)/495=75/495=5/33\).
Check by complement. The numerator for \(X\le2\) is \(\binom{5}{0}\binom{7}{4}+\binom{5}{1}\binom{7}{3}+\binom{5}{2}\binom{7}{2} =35+175+210=420\). Therefore \(1-420/495=75/495\), agreeing with the direct calculation. To four decimal places, the answer is \(0.1515\). The method avoids treating “at least 3” as “more than 3,” which would omit \(X=3\).
Worked Example: A Lower Tail and Its Upper Complement
A hypothetical archive has 15 files, of which 4 require a special review. A sample of 5 files is chosen without replacement. Find the probability that at most 1 selected file requires review. The model is \(X\sim\operatorname{Hypergeometric}(N=15,K=4,n=5)\); no interest rate or mortality assumption applies.
Method and support. The support runs from \(L=\max(0,5-11)=0\) to \(U=\min(5,4)=4\). “At most 1” means \(X\le1\), so the event includes \(X=0\) and \(X=1\). Its complement is \(X\ge2\).
Calculate the lower tail. The total number of samples is \(\binom{15}{5}=3003\). For \(x=0\), the numerator is \(\binom{4}{0}\binom{11}{5}=1(462)=462\). For \(x=1\), it is \(\binom{4}{1}\binom{11}{4}=4(330)=1320\). The combined numerator is \(462+1320=1782\), so \(P(X\le1)=1782/3003=0.5934\) to four decimal places.
Check by complement. The terms for \(X\ge2\) have numerators \(\binom{4}{2}\binom{11}{3}=990\), \(\binom{4}{3}\binom{11}{2}=220\), and \(\binom{4}{4}\binom{11}{1}=11\). Their sum is \(990+220+11=1221\), and \(1-1221/3003=1782/3003\). The check confirms both the endpoint and the arithmetic. The common mistake avoided is replacing “at most 1” with “less than 1,” which would leave out \(X=1\).
Worked Example: A Strict Cutoff With a Short Complement
A hypothetical batch contains 10 sensors, exactly 3 of which have a particular calibration flag. Four sensors are selected without replacement. Find the probability that more than 1 selected sensor has the flag. The model is \(X\sim\operatorname{Hypergeometric}(N=10,K=3,n=4)\); interest and mortality assumptions are not applicable.
Method and support. Here \(L=\max(0,4-7)=0\) and \(U=\min(4,3)=3\). Since \(X\) is integer-valued, “more than 1” means \(X\ge2\), not \(X\ge1\). The event includes \(X=2,3\); its complement \(X\le1\) has the two values 0 and 1.
Calculate the upper tail. The denominator is \(\binom{10}{4}=210\). For \(x=2\), the numerator is \(\binom{3}{2}\binom{7}{2}=3(21)=63\). For \(x=3\), it is \(\binom{3}{3}\binom{7}{1}=1(7)=7\). Thus \(P(X>1)=P(X\ge2)=(63+7)/210=70/210=1/3\).
Check by complement. The numerator for \(X=0\) is \(\binom{3}{0}\binom{7}{4}=35\), and for \(X=1\) it is \(\binom{3}{1}\binom{7}{3}=3(35)=105\). Therefore \(1-(35+105)/210=1-140/210=70/210=1/3\), or \(0.3333\) to four decimal places. The direct and complementary calculations agree.
Common Off-by-One Mistakes
A common incorrect calculation is to write \(P(X>2)=1-P(X\le2)\). That subtracts the event \(X\le2\), leaving \(X\ge3\), which is correct for \(X>2\). But the error arises when this rule is confused with \(P(X\ge2)\): writing \(1-P(X\le2)\) for that event leaves out \(X=2\). The correct identity is
Always translate the original words before using a complement. In particular, “at least \(r\)” includes \(r\); “more than \(r\)” starts at \(r+1\) for integer \(r\); “at most \(r\)” includes \(r\); and “fewer than \(r\)” ends at \(r-1\). Then intersect the listed values with the actual support from \(L\) to \(U\). Values outside the support contribute no probability.
A Reliable Tail-Probability Check
Decide whether the condition is inclusive or strict, such as \(X\ge r\) versus \(X>r\).
Use the integer-valued nature of the count to identify the first or last included count.
Retain only values from \(L\) through \(U\); if none remain, the probability is zero.
Use the method with fewer terms, while ensuring that the complement ends immediately before or starts immediately after the desired cutoff.
Confirm the event and its complement cover the entire support without overlap, and that their probabilities sum to one.
This approach is useful whenever a problem asks for a chance of “at least,” “no more than,” “exceeding,” or “fewer than” a specified count. It is aligned with the SOA VEE Mathematical Statistics syllabus: the essential skills are recognizing the discrete support, translating event language into correct bounds, and using complements without losing or adding a boundary value.
Check Your Understanding
For each hypothetical sampling problem, identify the support and translate the event into integer values before calculating. Probabilities in the answers are rounded to four decimal places.
- A sample of 3 items is drawn without replacement from 9, of which 2 have a warning label. Find the probability that at least 1 selected item has the label, using a complement. (Answer: \(7/12=0.5833\))
- From 11 records, 4 are flagged; 3 records are selected without replacement. Find the probability that at most 1 is flagged, and state the upper-tail event used as its complement. (Answer: \(119/165=0.7212\); complement \(X\ge2\))
- A sample of 4 is drawn from 12 units, 5 of which meet a criterion. Find the probability that more than 2 selected units meet it. (Answer: \(5/33=0.1515\))
- Six of 16 packages have a seal defect. Five packages are selected without replacement. Find the probability that fewer than 2 selected packages have a defect, using the correct strict-inequality boundary. (Answer: \(9/26=0.3462\))
- Five of 13 archived records are marked for review, and 6 are selected without replacement. Find the probability that between 2 and 4 selected records, inclusive, are marked for review. (Answer: \(350/429=0.8159\))