Calculating an Exact Number of Successes
A binomial probability answers a specific question: if a fixed number of independent trials have the same probability of success, what is the probability of observing exactly a stated number of successes? The calculation combines the probability of one particular success-and-failure pattern with the number of patterns that have the required number of successes.
The earlier material on recognizing when a binomial model applies establishes the conditions for using this model. Before calculating, verify that the number of trials is fixed, each trial has two outcomes for the purpose of the question, the trials are independent, and the probability of success is the same on every trial. A binomial formula is not appropriate if these assumptions do not hold.
What Each Part of the Formula Represents
The formula has three components, each with a distinct role. The factor \(p^k\) is the probability contribution from the \(k\) successes, and \((1-p)^{n-k}\) is the contribution from the remaining \(n-k\) failures. Their product is the probability of one particular arrangement of those outcomes. The coefficient \(\binom{n}{k}\) counts how many different arrangements contain exactly \(k\) successes.
| Symbol or factor | Meaning | Question to ask |
|---|---|---|
| \(n\) | Number of trials | How many trials are performed? |
| \(p\) | Probability of success on each trial | Which outcome is defined as success? |
| \(k\) | Required number of successes | How many successes must occur exactly? |
| \(\binom{n}{k}\) | Number of arrangements with \(k\) successes | How many positions can hold the successes? |
| \(p^k(1-p)^{n-k}\) | Probability of one such arrangement | Are the success and failure factors in the right powers? |
The word exactly is important. This formula gives the probability of one specified count \(k\), not a probability for several possible counts. First translate the event into the required count; then substitute that count into the formula.
Why the Binomial Probability Formula Holds
To justify the formula, first fix one arrangement of \(k\) successes and \(n-k\) failures. Independence means the probability of that particular arrangement is the product of the trial probabilities. Since every success has probability \(p\) and every failure has probability \(1-p\), that product is \(p^k(1-p)^{n-k}\). For example, the arrangement success, failure, success has probability \(p(1-p)p=p^2(1-p)\).
The number of arrangements is the number of ways to select which \(k\) of the \(n\) trial positions are successes. That number is \(\binom{n}{k}\). The arrangements are mutually exclusive: a sequence of trial outcomes cannot be two different arrangements at once. They all have the same probability, so adding their probabilities gives
This argument relies on independent trials with a common success probability. If outcomes are dependent or the success probability changes from trial to trial, the arrangements need not have the same probability, and this binomial formula does not apply as written.
A Reliable Calculation Process
State which of the two trial outcomes counts as a success. This fixes the meaning of \(p\).
Write \(n\) for the number of trials, \(p\) for the success probability, and \(k\) for the exact count requested.
Use \(\binom{n}{k}p^k(1-p)^{n-k}\). Check that the failure count is \(n-k\) and the failure probability is \(1-p\).
Evaluate the coefficient and powers, multiply, and state the probability in context. If rounding is required, round only the final result.
Worked Example: Faults Found in Four Independent Checks
In a hypothetical equipment-monitoring exercise, each of four independent checks has probability \(0.10\) of finding a fault. Let \(X\) be the number of checks that find a fault. Find the probability that exactly two checks find a fault. Give the probability to four decimal places.
Method and parameters. A success is a check finding a fault. The trials are independent, the success probability is constant, and the number of trials is fixed, so \(X\sim\operatorname{Bin}(4,0.10)\). Thus \(n=4\), \(p=0.10\), and \(k=2\).
Substitute. There are \(4-2=2\) checks that do not find a fault, each with probability \(1-0.10=0.90\).
Answer and check. The probability is \(0.0486\), or \(4.86\%\). One particular arrangement, such as fault, fault, no fault, no fault, has probability \(0.01(0.81)=0.0081\). There are six ways to choose the two checks that find faults, so \(6(0.0081)=0.0486\), agreeing with the formula. The method avoids counting only one arrangement.
Worked Example: Exactly Three Positive Results in a Hypothetical Screen
A hypothetical screening exercise records results from eight independent tests. For each test, the probability of a positive result is \(0.25\). Let \(X\) count the positive results. Calculate the probability of exactly three positives, to four decimal places.
Method and parameters. A positive result is defined as a success, so \(n=8\), \(p=0.25\), \(k=3\), and \(X\sim\operatorname{Bin}(8,0.25)\). The number of negative results in the event is \(8-3=5\), with probability \(1-0.25=0.75\) per test.
Substitute and evaluate. The number of ways to select the three positive-result positions is \(\binom{8}{3}=\frac{8\cdot7\cdot6}{3\cdot2\cdot1}=56\). Also, \((0.25)^3=0.015625\) and \((0.75)^5=0.2373046875\).
Answer and check. The probability is \(0.2076\) to four decimal places. The powers represent three positive results and five negative results, and the coefficient counts the possible positions of the three positives. The method avoids using \(0.25\) for the negative results or omitting the five failures.
Worked Example: Exactly Four Claims in Six Hypothetical Policies
Suppose a hypothetical portfolio review considers six independent policies for a specified period. Each policy has probability \(0.40\) of producing a claim during that period. Let \(X\) be the number of policies that produce a claim. Find the probability of exactly four claims, to four decimal places.
Method and parameters. A claim is the success outcome, so \(n=6\), \(p=0.40\), \(k=4\), and \(X\sim\operatorname{Bin}(6,0.40)\). The event also requires two policies without claims; their per-policy probability is \(0.60\).
Count arrangements and calculate. There are \(\binom{6}{4}=\binom{6}{2}=15\) ways to select the four policies that produce claims. The probability of one such arrangement is \((0.40)^4(0.60)^2\).
Answer and check. The probability is \(0.1382\) to four decimal places. The four-claim factor is \(0.40^4\), not \(0.40^2\); the remaining two policies contribute \(0.60^2\). The method avoids confusing the number of claims with the number of non-claims.
Common Mistakes and Checks
An exact-count calculation is short, but several errors can alter it substantially. Check the event, the parameter assignments, and the arithmetic separately rather than treating the formula as a single substitution.
- Leaving out the binomial coefficient. The product \(p^k(1-p)^{n-k}\) is for one specific arrangement, not all arrangements with \(k\) successes. Multiply by \(\binom{n}{k}\).
- Using \(p\) for both outcomes. The probability of a failure is \(1-p\), not \(p\), unless \(p=0.5\). The failure factor must be \((1-p)^{n-k}\).
- Reversing the powers. The exponent on \(p\) is the number of successes \(k\); the exponent on \(1-p\) is the number of failures \(n-k\).
- Changing the definition of success mid-calculation. If the question asks for exactly two checks that do not find a fault, then success for the count is “does not find a fault,” and the success probability for that count is \(0.90\), not \(0.10\).
- Rounding intermediate values. Keep sufficient precision through the multiplication, then round the final probability as requested. Also check that the result is between zero and one.
For example, in the four-check fault scenario, calculating only \((0.10)^2(0.90)^2=0.0081\) is a wrong answer for exactly two faults: it accounts for only one arrangement. The correction is to count all \(\binom{4}{2}=6\) arrangements, giving \(6(0.0081)=0.0486\).
Check Your Understanding
For each problem, assume the stated trials are independent and have a common success probability. Calculate the probability of the exact count requested and round to four decimal places.
- A hypothetical batch has four independent units, each with probability \(0.30\) of passing a stress check. What is the probability that exactly one unit passes? (Answer: 0.4116)
- Five independent service calls each have probability \(0.20\) of requiring a specialist. What is the probability that exactly zero calls require a specialist? (Answer: 0.3277)
- In six independent trials, success has probability \(0.50\). What is the probability of exactly four successes? (Answer: 0.2344)
- Each of seven hypothetical devices independently signals an alert with probability \(0.10\). What is the probability that exactly two devices signal an alert? (Answer: 0.1240)
- Nine independent applications each have probability \(0.20\) of meeting a stated criterion. What is the probability that exactly three applications meet it? (Answer: 0.1762)