Complementary Events for Binomial Tails

Why Complementary Events Help

A binomial probability can be calculated by adding the probabilities of the individual success counts in the event. For an upper tail such as “at least one success,” however, it can be shorter to calculate the probability of the opposite event and subtract it from one. The key is to translate the wording into the exact event first: a one-unit error in the cutoff changes the probability.

Suppose \(X\) counts the successes in \(n\) independent trials, each with success probability \(p\). Each trial has the same two outcomes, and \(X\sim\operatorname{Bin}(n,p)\). Write \(q=1-p\). The binomial probability mass function is

P(X=k)=\binom{n}{k}p^kq^{n-k},\qquad k=0,1,\ldots,n.

The support consists only of integers from 0 through \(n\). The complement method does not change the binomial model or its assumptions; it changes which probabilities are added. It is especially useful when the desired event contains most possible counts but its complement contains only a few.

Definition: For any event \(A\), its complement \(A^c\) is the event that \(A\) does not occur. Since \(A\) and \(A^c\) are disjoint and together cover all outcomes, \(P(A)=1-P(A^c)\). For a binomial count, translate \(A^c\) using the integer support before calculating.

Translate the Threshold Before Calculating

For an integer threshold \(r\), “at least \(r\)” means \(X\ge r\), while “more than \(r\)” means \(X>r\), or equivalently \(X\ge r+1\). Their complements are not the same: the first includes counts up to \(r-1\), and the second includes counts up to \(r\).

\begin{aligned} r\in\mathbb Z:\quad (X\ge r)^c&=\{X\le r-1\},\\ (X>r)^c&=\{X\le r\}. \end{aligned}

If a threshold is not an integer, do not apply those integer formulas to it as though it were one. For real \(r\), the complements are \(X<r\) for \(X\ge r\), and \(X\le r\) for \(X>r\). Because \(X\) is integer-valued, these can be written \(X\le\lceil r\rceil-1\) and \(X\le\lfloor r\rfloor\), respectively. For example, \(X>2.4\) means \(X\ge3\), and its complement is \(X\le2\).

Calculation rule: For integer \(r\), calculate \(P(X\ge r)\) as \(1-P(X\le r-1)\), and calculate \(P(X>r)\) as \(1-P(X\le r)\). Expand the lower-tail probability using the binomial probability mass function, including every integer in the stated range.

A Short Justification of the Complement Formula

The complement rule follows from a partition of the sample space. Every outcome either belongs to \(A\) or to \(A^c\), and no outcome belongs to both. Therefore their probabilities add to one. For the binomial event \(A=\{X\ge r\}\), when \(r\) is an integer, the only possible counts outside \(A\) are \(0,1,\ldots,r-1\). Thus

P(X\ge r) =1-P(X\le r-1) =1-\sum_{k=0}^{r-1}\binom{n}{k}p^k(1-p)^{n-k}.

This expression is useful when the lower sum has fewer terms than the direct sum from \(r\) through \(n\). The formula remains valid when the sum is empty: if \(r\le0\), then \(X\ge r\) is certain. At the other extreme, if \(r>n\), the event is impossible. In routine calculations, first check whether the threshold lies within the support.

State the model and the event.
Confirm that the count is binomial with fixed \(n\), common success probability \(p\), and independent trials. Rewrite the verbal condition as an inequality for \(X\).

Find the complement on integer support.
Identify the largest count included in the complement. Check whether the original wording says “at least” or “more than.”

Calculate and subtract.
Evaluate the needed binomial probabilities, add them, and subtract that sum from one. Keep sufficient precision until the final rounding.

Worked Examples

Worked Example: At Least One Successful Inspection

A hypothetical quality-control team independently inspects 8 components. For each component, the probability of finding a specified defect is 0.10, the same for every inspection. Let \(X\) be the number of components with the defect. Find the probability that at least one defect is found.

Method. This is a binomial count with \(n=8\), \(p=0.10\), and \(q=0.90\). The event is \(X\ge1\). Its complement is \(X=0\), so one binomial probability is needed rather than the probabilities for \(X=1,\ldots,8\).

\begin{aligned} P(X\ge1) &=1-P(X=0)\\ &=1-\binom{8}{0}(0.10)^0(0.90)^8\\ &=1-0.43046721\\ &=0.56953279. \end{aligned}

Thus the probability is \(0.5695\) to four decimal places. As a check, the zero-defect probability \(0.43046721\) and the at-least-one probability \(0.56953279\) add to 1. The common mistake avoided here is adding the probabilities of all positive counts when the single zero-count probability gives the answer by complementation.

Worked Example: More Than Two Claims

In a hypothetical group of 10 independently issued short-term policies, each policy has probability \(0.20\) of generating a claim during the observation period. Let \(X\) be the number of claims. Find the probability of more than 2 claims.

Method. Here \(n=10\), \(p=0.20\), and \(q=0.80\). “More than 2” means \(X>2\), or \(X\ge3\). Its complement is \(X\le2\), which includes exactly 0, 1, and 2 claims.

\begin{aligned} P(X>2) &=1-\{P(X=0)+P(X=1)+P(X=2)\}\\ &=1-\{(0.80)^{10} +10(0.20)(0.80)^9 +45(0.20)^2(0.80)^8\}\\ &=1-\{0.1073741824+0.2684354560+0.3019898880\}\\ &=1-0.6777995264\\ &=0.3222004736. \end{aligned}

The probability is \(0.3222\) to four decimal places. The calculation checks because the two disjoint events \(X\le2\) and \(X>2\) have probabilities summing to \(0.6777995264+0.3222004736=1\). A common mistake is to treat “more than 2” as “at least 2,” which would include \(X=2\) in the target event. The complement boundary \(X\le2\) confirms that \(X=2\) is not in the requested event.

Worked Example: A Noninteger Cutoff and an Upper Tail

A hypothetical monitoring system records whether each of 8 independent sensor readings exceeds a fixed alarm level. Each reading exceeds the level with probability \(0.25\). Let \(X\) count exceedances. Find the probability that the count is at least \(2.5\), and express the result to four decimal places.

Method. Since \(X\) is an integer, \(X\ge2.5\) is equivalent to \(X\ge3\). Its complement is \(X\le2\), not \(X\le1.5\) interpreted as a new integer cutoff. Here \(n=8\), \(p=0.25\), and \(q=0.75\).

\begin{aligned} P(X\ge2.5) &=P(X\ge3)\\ &=1-\{P(X=0)+P(X=1)+P(X=2)\}\\ &=1-\{(0.75)^8 +8(0.25)(0.75)^7 +28(0.25)^2(0.75)^6\}\\ &=1-\{0.1001129150390625 +0.2669677734375 +0.31146240234375\}\\ &=1-0.6785430908203125\\ &=0.3214569091796875. \end{aligned}

Therefore, the probability is \(0.3215\) to four decimal places. As an arithmetic check, the three lower probabilities sum to \(0.6785430908203125\), and adding the upper-tail result gives 1. The common mistake avoided is applying an integer-threshold rule directly to a noninteger cutoff. Equivalently, the general rule gives the complement \(X\le\lfloor2.5\rfloor=2\).

Choosing and Checking the Complement

Complementation is an efficiency choice, not a requirement. For \(X\ge r\), compare the number of terms in the direct sum, \(r\) through \(n\), with the number in the complement, 0 through \(r-1\). Use the shorter sum when that makes the arithmetic simpler. For “at least one,” the complement always has one term, \(X=0\). For a threshold near the lower end of the support, the direct sum may be as short as or shorter than the complement.

A useful check is to confirm that the boundary count appears on the correct side. For integer \(r\), the event \(X\ge r\) includes \(r\), whereas \(X>r\) does not. Accordingly, the complement of the first stops at \(r-1\), and the complement of the second stops at \(r\). For real thresholds, use the floor or ceiling conversion before summing. Finally, a probability and its complement must both lie between 0 and 1 and sum to 1, apart from any rounding.

“At least \(r\)”: include \(r\) in the event; for integer \(r\), subtract \(P(X\le r-1)\) from 1.
“More than \(r\)”: exclude \(r\); for integer \(r\), subtract \(P(X\le r)\) from 1.
Noninteger cutoff: convert using integer support; \(X>r\) has complement \(X\le\lfloor r\rfloor\).
At least one: subtract the probability of zero successes, \(q^n\), from 1.

Key takeaway: Translate the event precisely, determine its complement on the integer support, and then calculate \(1-P(A^c)\). The method is most effective when the complement requires fewer binomial terms than the original tail.

Check Your Understanding

For each problem, identify the complementary event before calculating. Give probabilities to four decimal places.

Six independent applications each have probability \(0.10\) of being approved. What is the probability that at least one is approved? (Answer: \(0.4686\))
Seven independent devices each fail during a specified test with probability \(0.20\). What is the probability that more than one fails? (Answer: \(0.4233\))
Nine independent seedlings each survive a trial with probability \(0.30\). What is the probability that at least three survive? (Answer: \(0.5372\))
For five independent readings, each has probability \(0.40\) of exceeding a reference level. If \(X\) is the number that exceed it, what is \(P(X>1.5)\)? (Answer: \(0.6630\))
Eight independent shipments each have probability \(0.40\) of arriving late. What is the probability that more than four shipments arrive late? (Answer: \(0.1737\))

What You'll Learn