Worked Binomial Problems With Exact Counts and Tails

Introduction

Many practical questions ask for the probability of a particular number of occurrences: exactly two faulty items in a batch, no more than two late deliveries, or three or more successes in a fixed set of attempts. These are different count statements, but they can all be handled with the binomial model when the process meets its assumptions.

The key is to define what counts as a success and what the random variable counts before calculating. Then translate the wording carefully: “exactly” names one count, while “at most” and “at least” include a range of counts. A probability for a range is found by adding the probabilities of its possible exact counts, or sometimes by using the complementary event.

Definition: A count $X$ has a binomial distribution with parameters $n$ and $p$, written $X\sim\operatorname{Bin}(n,p)$, when it counts successes in $n$ trials and: (1) $n$ is fixed; (2) each trial has two possible outcomes, success or failure; (3) the trials are independent; and (4) every trial has the same probability $p$ of success. Here $0\leq p\leq1$, and $X$ can take integer values from 0 to $n$.

The Exact-Count Formula

For a binomial variable, the probability of exactly $k$ successes is

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

The factor $p^k(1-p)^{n-k}$ is the probability of one particular sequence with $k$ successes and $n-k$ failures. The binomial coefficient $\binom{n}{k}$ counts how many different sequences have exactly $k$ successes. Multiplying gives the probability of all such sequences.

This formula gives an exact-count probability, not a cumulative probability. To find a range of counts, add the relevant exact-count probabilities. For example, “at most $r$” includes all counts from zero through $r$, while “at least $r$” includes $r$ itself and all larger possible counts.

Key result: For $X\sim\operatorname{Bin}(n,p)$, $$ P(X\leq r)=\sum_{k=0}^{r}\binom{n}{k}p^k(1-p)^{n-k} $$ and $$ P(X\geq r)=\sum_{k=r}^{n}\binom{n}{k}p^k(1-p)^{n-k}. $$ The inequalities are inclusive: $X\leq r$ includes $r$, and $X\geq r$ includes $r$.

Why the Formula and Tail Sums Work

For any fixed arrangement of $k$ successes among $n$ trials, independence means the probability of that arrangement is the product of the trial probabilities. Each success contributes a factor $p$, and each failure contributes a factor $1-p$. Therefore, that arrangement has probability $p^k(1-p)^{n-k}$.

There are $\binom{n}{k}$ ways to choose the $k$ trial positions that will be successes. These arrangements cannot happen simultaneously: on any one run, the sequence has only one arrangement. Their probabilities therefore add. This gives the exact-count formula. Similarly, the events $X=0,X=1,\ldots,X=r$ are mutually exclusive and together describe $X\leq r$, so their probabilities add. The same reasoning applies to $X\geq r$.

A complement can make a tail shorter to calculate. Since $X\leq r-1$ and $X\geq r$ are disjoint and cover all possible values of $X$,

P(X\geq r)=1-P(X\leq r-1).

The cutoff changes from $r$ to $r-1$ because the event being removed is $X\geq r$, including $r$. Likewise, $P(X\leq r)=1-P(X\geq r+1)$. These relationships are useful checks against off-by-one errors.

Worked Binomial Problems

Worked Example: Exactly Two Faulty Sensors

Suppose, hypothetically, that a production line’s sensors are assessed one at a time. Each sensor has probability $0.08$ of being faulty, and the outcomes for the 12 sensors in a batch are independent. What is the probability that exactly two sensors in the batch are faulty?

Check the model. The batch contains a fixed $n=12$ sensors. Each sensor is classified as faulty or not faulty, the probability of a faulty sensor is the same $p=0.08$ for each, and the problem states that outcomes are independent. Thus the faulty-sensor count $X$ follows $\operatorname{Bin}(12,0.08)$.

“Exactly two” means $X=2$. Using the binomial formula,

\begin{aligned} P(X=2) &=\binom{12}{2}(0.08)^2(1-0.08)^{12-2}\\ &=66(0.0064)(0.92)^{10}\\ &\approx 66(0.0064)(0.4344)\\ &\approx 0.1835. \end{aligned}

The probability is about $0.1835$, or $18.35\%$. In this hypothetical process, about 18.35% of batches of 12 would be expected to contain exactly two faulty sensors over many repetitions. This is a probability for a batch, not a prediction that any particular batch must have two faulty sensors.

Worked Example: At Most Two Late Packages

Suppose, hypothetically, that each of eight packages sent under the same conditions has probability $0.30$ of arriving late, independently of the others. What is the probability that at most two packages are late?

Check the model. There are a fixed eight packages. Each outcome is late or not late, every package has the same late probability $p=0.30$, and the outcomes are assumed independent. Therefore, the number $X$ of late packages follows $\operatorname{Bin}(8,0.30)$.

“At most two” means $X\leq2$, so the possible counts are 0, 1, and 2. Add their exact-count probabilities:

\begin{aligned} P(X\leq2) &=P(X=0)+P(X=1)+P(X=2)\\ &=\binom{8}{0}(0.30)^0(0.70)^8 +\binom{8}{1}(0.30)^1(0.70)^7\\ &\qquad+\binom{8}{2}(0.30)^2(0.70)^6\\ &=0.057648+0.197650+0.296475\\ &\approx0.5518. \end{aligned}

Thus the probability that no more than two of the eight packages are late is about $0.5518$, or $55.18\%$. The count of two is included; “at most two” does not mean fewer than two.

Worked Example: At Least Three Successful Attempts

Suppose, hypothetically, that a technician completes ten independent diagnostic attempts, each with probability $0.20$ of finding the issue being sought. What is the probability of finding it on at least three attempts?

Check the model. The number of attempts is fixed at ten, each attempt has two relevant outcomes (find or do not find the issue), the success probability is the same $p=0.20$ each time, and the attempts are assumed independent. The number of successful attempts $X$ therefore follows $\operatorname{Bin}(10,0.20)$.

“At least three” means $X\geq3$. Directly adding $P(X=3)$ through $P(X=10)$ is possible, but it is shorter to subtract the complementary event $X\leq2$ from 1:

\begin{aligned} P(X\geq3) &=1-P(X\leq2)\\ &=1-\bigl[P(X=0)+P(X=1)+P(X=2)\bigr]\\ &=1-\left[ \binom{10}{0}(0.20)^0(0.80)^{10} +\binom{10}{1}(0.20)^1(0.80)^9 +\binom{10}{2}(0.20)^2(0.80)^8 \right]\\ &=1-(0.107374+0.268435+0.301990)\\ &\approx0.3222. \end{aligned}

The probability of success on at least three attempts is about $0.3222$, or $32.22\%$. The complementary calculation includes zero, one, and two successes, leaving every possible count from three through ten in the requested event.

Translating the Wording Into Counts

Before substituting into a formula, rewrite the requested event using $X$. The following translations help keep the endpoints clear.

Wording	Event	Counts included
Exactly $r$	$X=r$	Only $r$
At most $r$	$X\leq r$	0 through $r$
At least $r$	$X\geq r$	$r$ through $n$
Fewer than $r$	$X<r$, or $X\leq r-1$	0 through $r-1$
More than $r$	$X>r$, or $X\geq r+1$	$r+1$ through $n$

The probability of any event must be between 0 and 1. If a calculation gives a value outside this range, recheck the terms and arithmetic. It is also useful to check that the event and its complement partition all possible counts. For instance, $X\leq2$ and $X\geq3$ cover all integer counts without overlap.

Common Mistakes and Why Careful Setup Matters

Omitting the binomial coefficient. The expression $p^k(1-p)^{n-k}$ gives the probability of one particular arrangement, not all arrangements with $k$ successes. Multiply by $\binom{n}{k}$ to count those arrangements.
Leaving out the endpoint. “At most two” includes two, and “at least two” includes two. Translate the phrase into an inequality before calculating.
Using the wrong complement. $P(X\geq r)=1-P(X\leq r-1)$, not $1-P(X\leq r)$. The latter incorrectly removes the count $r$ from the desired event.
Using a binomial formula without checking the process. A fixed number of trials alone is not enough. If success probabilities vary or outcomes are dependent, the binomial assumptions may fail, and the calculation may not describe the process.
Calling the probability an expected count. A result such as $P(X\geq3)=0.3222$ is a probability of an event, not a claim that 0.3222 successes will occur. The count itself is an integer; a probability summarizes uncertainty about that count.

These calculations matter because they convert a model of repeated outcomes into a probability statement about a count that can be understood in context. Their usefulness depends on both parts being sound: the binomial assumptions must reasonably describe the process, and the event must be translated accurately. A precise decimal cannot repair a mistaken model or an incorrectly chosen tail.

Key takeaway: Define the success, identify $n$ and $p$, and check the binomial conditions first. Then express the question as an exact count or an inclusive range, calculate using the binomial probability mass function, and state the result in context.

Check Your Understanding

For each question, identify the event carefully before deciding which binomial probabilities to calculate.

A hypothetical process has $n=9$ independent trials, each with success probability $p=0.15$. Write the expression for the probability of exactly four successes.
In a binomial setting with $n=7$, which exact counts are included in the event “at most three successes”?
For a binomial variable $X$ with $n=12$, express $P(X\geq5)$ using a complementary event. Which count is the upper endpoint of the complementary range?
Explain why $\binom{n}{k}$ appears in the exact-count formula rather than using only $p^k(1-p)^{n-k}$.
A process has a fixed number of trials, but the probability of success differs from trial to trial. Which binomial condition is not met, and why should the binomial formula not be used without further justification?

Wording	Event	Counts included
Exactly \(r\)	\(X=r\)	Only \(r\)
At most \(r\)	\(X\leq r\)	0 through \(r\)
At least \(r\)	\(X\geq r\)	\(r\) through \(n\)
Fewer than \(r\)	\(X<r\), or \(X\leq r-1\)	0 through \(r-1\)
More than \(r\)	\(X>r\), or \(X\geq r+1\)	\(r+1\) through \(n\)