Counting Successes Through Complementation

Why Count Failures?

A binomial model counts successes in a fixed number of independent trials, each with the same success probability. But an event described in terms of successes can sometimes be simpler to calculate by counting failures instead. If success is very likely, for instance, “at least nine successes out of ten” is equivalent to “at most one failure.” The second description may involve fewer probability terms.

The key is to transform both the count and the event carefully. The failure count has a complementary probability parameter, and inequalities reverse their direction when a count is replaced by the number of trials minus that count.

Definition: Let \(X\sim\operatorname{Bin}(n,p)\) count successes in \(n\) independent Bernoulli trials, each with success probability \(p\). Define the number of failures by \(Y=n-X\). Then \(Y\) counts failures, and its probability parameter is \(q=1-p\).

Here \(n\) is a nonnegative integer and \(0\le p\le 1\). The complementary parameter \(q\) is the probability of failure on one trial. It is not a new assumption about the experiment: each trial has either a success, with probability \(p\), or a failure, with probability \(q\).

The Complementation Result

The relationship between the two counts is exact, not an approximation. It follows directly by recoding each trial as a failure indicator.

Theorem: If \(X\sim\operatorname{Bin}(n,p)\) and \(Y=n-X\), then \(Y\sim\operatorname{Bin}(n,1-p)\). For every integer \(r\), \[ P(X\ge r)=P(Y\le n-r), \qquad P(X\le r)=P(Y\ge n-r). \] More generally, \(X=r\) if and only if \(Y=n-r\).

Proof. Write \(B_i=1\) if trial \(i\) is a success and \(B_i=0\) otherwise, for \(i=1,\ldots,n\). Then \(X=\sum_{i=1}^{n}B_i\). Define \(C_i=1-B_i\), so \(C_i\) indicates failure on trial \(i\). Each \(C_i\) is Bernoulli with probability \(P(C_i=1)=1-p\), and the \(C_i\) remain independent because each is a function of a separate independent trial. Therefore \(Y=\sum_{i=1}^{n}C_i=n-X\) is binomial with parameters \(n\) and \(1-p\).

For the event transformations, \(X\ge r\) is equivalent to \(n-Y\ge r\), which is equivalent to \(Y\le n-r\). Likewise, \(X\le r\) is equivalent to \(n-Y\le r\), or \(Y\ge n-r\). Finally, \(X=r\) is equivalent to \(Y=n-r\). These equivalences hold because the counts are integers, including when \(p=0\) or \(p=1\). \(\square\)

For an integer threshold, strict inequalities require extra care. The complete set of translations is:

\begin{aligned} X\ge r &\iff Y\le n-r, & X>r &\iff Y\le n-r-1,\\ X\le r &\iff Y\ge n-r, & X<r &\iff Y\ge n-r+1,\\ X=r &\iff Y=n-r. \end{aligned}

For example, “more than \(r\) successes” means at least \(r+1\) successes. It therefore leaves at most \(n-r-1\) failures, not \(n-r\). A one-unit error here changes the event.

Worked Examples

Worked Example: At Least One Rare Success

Each of 20 independent trials succeeds with probability \(0.03\). Find the probability of at least one success.

Let \(X\) be the number of successes. The event is \(X\ge1\), which becomes \(Y\le19\) failures. It is simpler to use the complementary event: \(X\ge1\) fails exactly when there are no successes, or \(Y=20\).

The probability of 20 failures is \(0.97^{20}\), since the failure probability on each trial is \(1-0.03=0.97\). Thus

P(X\ge1)=1-P(Y=20) =1-0.97^{20} \approx 1-0.543794343 =0.456205657.

So the probability of at least one success is approximately \(0.4562\). The complement is useful because it replaces a sum over many possible positive success counts with one probability for zero successes.

Worked Example: At Least Nine Successes in Ten Trials

A trial succeeds with probability \(0.8\), independently of the other trials. Find the probability of at least nine successes in ten trials.

Let \(X\sim\operatorname{Bin}(10,0.8)\) count successes and let \(Y=10-X\) count failures. The failure probability is \(0.2\), so \(Y\sim\operatorname{Bin}(10,0.2)\). The event translation is

X\ge9 \iff Y\le10-9=1.

Thus we add the probabilities of zero and one failure. Using the binomial probability formula,

\begin{aligned} P(X\ge9) &=P(Y\le1)\\ &=P(Y=0)+P(Y=1)\\ &=\binom{10}{0}(0.2)^0(0.8)^{10} +\binom{10}{1}(0.2)^1(0.8)^9\\ &=0.8^{10}+10(0.2)(0.8)^9\\ &=0.1073741824+0.268435456\\ &=0.3758096384. \end{aligned}

The success-count version would add the probabilities of nine and ten successes. Complementation gives the same two terms, expressed as zero or one failure.

Worked Example: A High-Probability Success Rate

Suppose 12 independent trials each succeed with probability \(0.98\). Find the probability of at least nine successes.

Counting failures is particularly convenient: the failure probability is \(0.02\), and \(X\ge9\) is equivalent to \(Y\le3\), where \(Y=12-X\). Therefore the required probability is the sum of the probabilities of zero, one, two, and three failures:

\begin{aligned} P(X\ge9) &=P(Y\le3)\\ &=\sum_{j=0}^{3}\binom{12}{j}(0.02)^j(0.98)^{12-j}\\ &=(0.98)^{12} +12(0.02)(0.98)^{11}\\ &\quad+66(0.02)^2(0.98)^{10} +220(0.02)^3(0.98)^9\\ &\approx 0.784716724 +0.192175524 +0.021570722 +0.001467396\\ &\approx 0.999930366. \end{aligned}

For the last term, the coefficient is \(\binom{12}{3}=220\), so its value is \(220(0.02)^3(0.98)^9\approx0.001467396\). The four displayed rounded terms sum to \(0.999930366\). The probability is very high because the expected number of failures is small; the complementary representation makes clear that the excluded cases require at least four failures.

A Reliable Translation Procedure

When an event is stated in terms of successes, write down the relationship \(Y=n-X\) before changing the probability calculation. This separates two tasks: translating the event correctly and deciding whether the translated event is convenient to evaluate.

Define the success and failure counts.
Set \(X\) to be the number of successes and \(Y=n-X\) to be the number of failures.

Change the probability parameter.
Use \(p\) for success and \(q=1-p\) for failure, so \(Y\sim\operatorname{Bin}(n,q)\).

Translate the event algebraically.
Replace \(X\) by \(n-Y\), then solve the resulting inequality for \(Y\). Keep the integer threshold in the calculation.

Choose the useful representation.
Compare the success-count and failure-count versions. Use the one with fewer terms or more convenient probabilities; they describe the same event probability.

For integer \(r\) between \(0\) and \(n\), the translations can be written as sums when needed:

\begin{aligned} P(X\ge r) &=P(Y\le n-r) =\sum_{j=0}^{n-r}\binom{n}{j}(1-p)^j p^{n-j},\\ P(X\le r) &=P(Y\ge n-r) =\sum_{j=n-r}^{n}\binom{n}{j}(1-p)^j p^{n-j}. \end{aligned}

These formulas use the binomial probability mass function for the failure count. They also show why complementation is a choice of representation rather than a new probability rule: every term for a failure count \(j\) corresponds to the success count \(n-j\).

Common Mistakes and When Complementation Helps

The most common error is to complement the probability parameter but not the event threshold. If \(X\ge r\), the correct failure event is \(Y\le n-r\), not \(Y\ge r\). The count of failures is constrained by how many trial positions remain after accounting for the required successes.

A second error is mishandling strict inequalities. For example, \(X>r\) means \(X\ge r+1\), so \(Y\le n-r-1\). Replacing this by \(Y\le n-r\) incorrectly includes the case \(X=r\).

Finally, complementing is not guaranteed to make a calculation shorter. If an event is \(X\le r\) and \(p\) is small, the success-count representation may already involve only a few terms. Its failure version, \(Y\ge n-r\), might instead involve many terms. The appropriate representation depends on the event and on \(p\), not simply on whether the count is called successes or failures.

Key takeaway: For a binomial count, \(Y=n-X\) is the corresponding failure count and has probability parameter \(1-p\). Translate events using this identity first; then use whichever count gives the clearest or shortest calculation.

Check Your Understanding

For each question, define the failure count as \(Y=n-X\) and translate the event before evaluating or comparing probabilities.

If \(X\) counts successes in 15 trials, express the event \(X\ge12\) as an event about \(Y\).
In 8 trials, what failure-count event is equivalent to \(X>5\)?
If \(X\sim\operatorname{Bin}(n,p)\), state the distribution of \(n-X\), including its probability parameter.
For 20 trials with success probability \(0.04\), write a complement expression for the probability of at least one success.
Explain why \(X\le r\) is equivalent to \(Y\ge n-r\), rather than \(Y\le n-r\).