Why Small Setup Errors Change the Answer
Many incorrect Bernoulli and binomial answers are not arithmetic errors. They begin with a model that does not match the process, a count event translated incorrectly, or a probability for one particular sequence mistaken for a probability for a total number of successes. These errors are avoidable if the model and event are stated before calculating.
The checks in this tutorial apply to a fixed number of trials. A binomial model requires both independence and a constant success probability; having one condition does not make up for the other. Once the model is appropriate, careful wording and counting determine the correct probability.
For \(X\sim\operatorname{Bin}(n,p)\), the probability of exactly \(k\) successes is
The formula is not a general formula for any count of events. It applies only when the fixed-trial, independence, and constant-\(p\) conditions hold.
A Quick Model-and-Event Check
State what counts as one trial, what outcome is called success, and how many trials are performed.
Ask whether the outcome of one trial changes the probability distribution of another. Sampling without replacement from a finite group is a common reason trials are dependent.
Each trial must have the same success probability \(p\). Independent trials with different success probabilities do not have a binomial count.
Write the possible integer values of \(X\) before evaluating a point probability or adding probabilities.
For exactly \(k\) successes, include all distinct ways to place those successes among the \(n\) trials.
Why the Binomial Coefficient Is Needed
The binomial coefficient is a count of arrangements, not an adjustment for the size of the probability. To see why it appears, consider a particular ordered sequence of \(n\) outcomes with exactly \(k\) successes. Independence and the common success probability imply that the sequence has \(k\) success factors of \(p\) and \(n-k\) failure factors of \(1-p\). Its probability is therefore \(p^k(1-p)^{n-k}\).
There are \(\binom{n}{k}\) different ways to choose which \(k\) of the \(n\) trial positions are successes. These sequences are mutually exclusive and each has the same probability just found. Adding their probabilities gives the binomial formula. This reasoning also pinpoints two common errors: omitting the coefficient counts only one sequence, while using the formula when trial probabilities differ incorrectly assumes all sequences with \(k\) successes have the same probability.
Worked Example: Sampling Without Replacement
A sealed tray contains 6 sensor units, of which 2 fail a calibration check and 4 pass. A technician selects 3 units at random without replacement. Find the probability that exactly 1 selected unit fails. Determine whether a binomial model with \(n=3\) and \(p=2/6\) is appropriate.
Method and model check. Define success as selecting a unit that fails. The first selection has failure probability \(2/6=1/3\). But if the first unit fails, only 1 failing unit remains among 5, so the second-selection failure probability is \(1/5\), not \(1/3\). The selections are dependent, and the binomial model does not apply.
Calculate the probability using the tray composition. There are \(\binom{6}{3}=20\) equally likely sets of 3 units. To have exactly 1 failure, choose 1 of the 2 failing units and 2 of the 4 passing units:
Common mistake avoided. Treating the selections as independent would give \(\binom{3}{1}(1/3)(2/3)^2=4/9\), about 0.4444. That answer uses a binomial model whose independence condition is false. The correct probability is 0.6000, as verified by counting all 20 equally likely sets and finding 12 favorable sets.
Worked Example: Independent Trials With Different Probabilities
A hypothetical monitoring system makes three independent checks. The probability that check 1 flags an issue is 0.20, while checks 2 and 3 each flag an issue with probability 0.50. Find the probability that exactly one check flags an issue. Is the number of flags binomial?
Method and model check. The checks are independent, but their success probabilities are not all equal: they are 0.20, 0.50, and 0.50. Thus, the count is not binomial. Calculate the probability by adding the three mutually exclusive ways exactly one check can flag an issue.
If only check 1 flags, the probability is \(0.20(0.50)(0.50)=0.050\). If only check 2 flags, it is \(0.80(0.50)(0.50)=0.200\). If only check 3 flags, it is \(0.80(0.50)(0.50)=0.200\). Therefore,
Common mistake avoided. Averaging the probabilities gives \((0.20+0.50+0.50)/3=0.40\), but that average does not make the trials identically distributed. A binomial calculation with \(n=3\) and \(p=0.40\) would give \(3(0.40)(0.60)^2=0.432\), not the correct 0.450. The direct enumeration checks the result using the actual probabilities for each check.
Translate Bounds Before Calculating
The count \(X\) takes integer values. Consequently, small wording differences change which outcomes belong to an event. “At least 2” means \(X\geq 2\), “more than 2” means \(X\geq 3\), “at most 2” means \(X\leq 2\), and “fewer than 4” means \(X\leq 3\). When two conditions are joined, both must hold.
Worked Example: Reading Inclusive and Strict Bounds
A hypothetical quality screen checks 5 items independently, each with probability \(p=0.20\) of failing. Let \(X\) be the number of failures. Find the probability of at least 2 but fewer than 4 failures.
Method and model. The number of trials is fixed at \(n=5\), each trial has the same failure probability \(p=0.20\), and the checks are independent. Therefore \(X\sim\operatorname{Bin}(5,0.20)\). The event “at least 2 but fewer than 4” is \(2\leq X<4\), or \(X=2\) or \(X=3\).
For exactly 2 failures,
For exactly 3 failures,
The outcomes are disjoint, so
Common mistake avoided. Replacing “at least 2” with “more than 2” and “fewer than 4” with “at most 4” would give \(P(X=3)+P(X=4)\). Since \(P(X=4)=\binom{5}{4}(0.20)^4(0.80)=0.0064\), that incorrect event has probability \(0.0512+0.0064=0.0576\). The stated event includes 2 and excludes 4, so 0.2560 is the correct answer.
Single-Trial Probability Versus Count Probability
In a Bernoulli trial, \(p\) is the probability of success on that trial. In a binomial model, \(P(X=k)\) is the probability of a total count across \(n\) trials. These are different quantities. For example, \(P(X=1)\) is generally not \(p\): it includes every arrangement with exactly one success.
Worked Example: Count All Arrangements of Two Successes
A hypothetical device completes 4 independent operations, each with success probability \(p=0.30\). Let \(X\) be the number of successful operations. Find \(P(X=2)\), and compare it with the probability of one specified arrangement in which the first two operations succeed and the last two fail.
Method and model. The number of trials is fixed at 4, the success probability is constant at 0.30, and the operations are independent. Thus \(X\sim\operatorname{Bin}(4,0.30)\). For exactly 2 successes, there are \(\binom{4}{2}=6\) arrangements.
The probability of the specified arrangement is
All 6 arrangements have this probability, so
As a further check on the distinction, the probability that exactly one operation succeeds is \(P(X=1)=\binom{4}{1}(0.30)(0.70)^3=4(0.30)(0.343)=0.4116\), not 0.30.
Common mistake avoided. Omitting the factor \(\binom{4}{2}=6\) gives only the probability of one specified arrangement, 0.0441, rather than the probability of any arrangement with 2 successes, 0.2646. The six arrangements are disjoint, so multiplying by 6 is justified.
Common Error Checklist
- Assuming independence from equal probabilities. Equal marginal success probabilities do not establish independence. Check whether knowing one outcome changes the chance of another.
- Assuming a binomial model because trials are independent. Independent trials with different success probabilities do not satisfy the common-\(p\) condition.
- Ignoring whether endpoints are inclusive. Convert the wording to inequalities and then list the integer values of \(X\) before calculating.
- Leaving out the binomial coefficient. The product \(p^k(1-p)^{n-k}\) is the probability of one specified arrangement. Multiply by \(\binom{n}{k}\) to include every arrangement.
- Using \(p\) as the probability of a count event. \(p\) describes one trial; \(P(X=k)\) describes the total across all trials.
Check Your Understanding
For each problem, identify the model conditions before calculating. Probabilities in the answers are rounded to four decimal places where needed.
- A reliability team performs 6 independent checks, each with probability 0.25 of detecting a fault. Find the probability that none detects a fault. (Answer: 0.1780)
- Five independent applications are reviewed, each with probability 0.40 of requiring additional documentation. Find the probability that at most 2 require it. (Answer: 0.6826)
- A box contains 8 components, 3 of which are scratched. Two are selected without replacement. Find the probability exactly one selected component is scratched, and say whether the count is binomial. (Answer: \(15/28\approx0.5357\); not binomial because the selections are dependent.)
- Three independent alerts have probabilities 0.10, 0.40, and 0.70 of activating, respectively. Find the probability exactly one activates. Is the count binomial? (Answer: 0.5040; not binomial because the success probabilities differ.)
- Five independent trials each have success probability 0.20. Find the probability of exactly 3 successes and the probability of the specified sequence success, success, success, failure, failure. (Answer: \(P(X=3)=0.0512\); specified sequence probability \(=0.00512\).)