Binomial Tail Probabilities

Why Endpoints Matter in Binomial Tails

A binomial point probability describes the chance of one exact count. A tail probability collects the probabilities of several counts, such as no more than two claims or more than four equipment failures. The calculation is a sum of point probabilities, but the event’s wording determines exactly which counts belong in that sum. In particular, “at least” includes its endpoint, while “more than” does not.

This tutorial focuses on writing the count range correctly and summing the relevant binomial masses. The same discipline applies whether the count represents claims, failures, or another set of independent yes-or-no outcomes. The methods and examples are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: A random variable $X$ has a binomial distribution with parameters $n$ and $p$, written $X\sim\operatorname{Bin}(n,p)$, when it counts the successes in $n$ independent trials, each with the same success probability $p$. Its possible counts are $0,1,\ldots,n$, and its probability mass function is $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n. $$ Here $n$ is a nonnegative integer and $0\le p\le 1$. The model does not apply if the trials are not independent, their success probabilities differ, or the number of trials is not fixed.

Translate the Wording Before Calculating

For a binomial count, the endpoint language identifies the first or last count to include. For integer $r$, “at most $r$” means $X\le r$; “less than $r$” means $X\le r-1$. Similarly, “at least $r$” means $X\ge r$, while “more than $r$” means $X\ge r+1$. The support $0,\ldots,n$ still applies: counts below zero or above $n$ cannot occur.

Wording	Event for integer $r$	Counts to sum
At most $r$	$X\le r$	$0$ through $r$
Less than $r$	$X<r$	$0$ through $r-1$
At least $r$	$X\ge r$	$r$ through $n$
More than $r$	$X>r$	$r+1$ through $n$

Therefore, for integer $r$ within the support, the direct tail sums are

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

For strict inequalities, change the endpoint before summing: $P(X<r)$ sums through $r-1$, and $P(X>r)$ starts at $r+1$. If a threshold is not an integer, compare the possible integer counts directly with that threshold. For instance, $X\ge 2.4$ includes counts $3,4,\ldots,n$. This count-range translation is part of the calculation, not a formatting detail.

Why Adding the Masses Gives the Tail

The events $X=0,X=1,\ldots,X=n$ are mutually exclusive: exactly one count occurs. An event such as $X\le r$ is the union of the disjoint events $X=0,\ldots,X=r$. By additivity of probability for disjoint events, its probability is the sum of their point probabilities. Substituting the binomial probability mass function gives the lower-tail formula above. The same argument applies to any selected range of counts, including upper tails.

A useful practical method is to write the event, list its count range, evaluate each included mass, and add. This keeps the endpoint visible throughout the work and makes it easier to detect a missing or extra count.

Identify the model.
Confirm there are $n$ independent trials with a common success probability $p$, so $X\sim\operatorname{Bin}(n,p)$.

Translate the event.
Convert “at most,” “less than,” “at least,” or “more than” into an explicit range of integer counts.

Sum the included masses.
Evaluate $P(X=k)$ for each count in the range and add those values.

Check the range and total.
Verify no count outside $0,\ldots,n$ was included and that the endpoint matches the original wording.

Worked Tail Calculations

Worked Example: No More Than One Claim

A hypothetical portfolio contains five independent short-term policies. For each policy, the probability of at least one claim during the term is $p=0.20$. Let $X$ be the number of policies with a claim. Calculate the probability that no more than one policy has a claim.

Model and method. The number of fixed policies is $n=5$, each policy independently has a claim with probability $p=0.20$, and $X\sim\operatorname{Bin}(5,0.20)$. “No more than one” is inclusive, so the event is $X\le1$, which includes counts $0$ and $1$.

For count zero,

P(X=0)=\binom{5}{0}(0.20)^0(0.80)^5 =1(1)(0.32768)=0.32768.

For count one,

P(X=1)=\binom{5}{1}(0.20)^1(0.80)^4 =5(0.20)(0.4096)=0.40960.

Add the two disjoint cases:

P(X\le1)=0.32768+0.40960=0.73728\approx 0.7373.

Check. The answer is a probability between zero and one, and it includes both zero claims and one claim. The common mistake avoided here is treating “no more than one” as $X<1$, which would omit the one-claim case. The final probability is rounded to four decimal places.

Worked Example: More Than Three Equipment Failures

A hypothetical operator monitors six independent sensors. Each sensor has probability $0.50$ of failing during a specified test. Let $X$ be the number of failed sensors. Find $P(X>3)$, and compare it with the probability of at least three failures.

Model and method. There are $n=6$ fixed, independent trials with common failure probability $p=0.50$, so $X\sim\operatorname{Bin}(6,0.50)$. “More than three” means $X=4,5,6$, not $3,4,5,6$.

The required point probabilities are

\begin{aligned} P(X=4)&=\binom{6}{4}(0.50)^4(0.50)^2=\frac{15}{64}=0.234375,\\ P(X=5)&=\binom{6}{5}(0.50)^5(0.50)^1=\frac{6}{64}=0.093750,\\ P(X=6)&=\binom{6}{6}(0.50)^6(0.50)^0=\frac{1}{64}=0.015625. \end{aligned}

Thus

P(X>3)=\frac{15+6+1}{64}=\frac{22}{64}=0.34375\approx 0.3438.

“At least three” also includes $X=3$. Its additional mass is

P(X=3)=\binom{6}{3}(0.50)^3(0.50)^3=\frac{20}{64}=0.3125, \qquad P(X\ge3)=\frac{20+15+6+1}{64}=\frac{42}{64}=0.65625\approx0.6563.

Check. The difference is exactly $P(X=3)=0.3125$, as expected because that count is included in “at least three” but excluded from “more than three.” The common mistake avoided is using the same starting count for both phrases. Final probabilities are rounded to four decimal places.

Worked Example: At Least Two Claims in Four Exposures

A hypothetical small group has four independently covered exposures. Each exposure has probability $0.25$ of producing a claim during the period. Let $X$ count the exposures with claims. Calculate the probability of at least two claims.

Model and method. The number of trials is $n=4$, the common claim probability is $p=0.25$, and $X\sim\operatorname{Bin}(4,0.25)$. “At least two” includes counts $2,3,4$.

For $X=2$, $3$, and $4$, respectively,

\begin{aligned} P(X=2)&=\binom{4}{2}(0.25)^2(0.75)^2 =6(0.0625)(0.5625)=0.2109375,\\ P(X=3)&=\binom{4}{3}(0.25)^3(0.75) =4(0.015625)(0.75)=0.046875,\\ P(X=4)&=\binom{4}{4}(0.25)^4 =1(0.00390625)=0.00390625. \end{aligned}

Summing the included counts gives

P(X\ge2)=0.2109375+0.046875+0.00390625 =0.26171875\approx0.2617.

Check. The sum includes the largest possible count, four, and excludes counts zero and one. The common mistake avoided is stopping the sum at three and forgetting that “at least two” includes every count above two through the support limit. The final probability is rounded to four decimal places.

Common Endpoint and Support Errors

Replacing a strict inequality with an inclusive one. For integer $r$, $X>r$ starts at $r+1$, whereas $X\ge r$ starts at $r$.
Forgetting the support. A binomial count cannot be negative or exceed $n$. If a requested range extends beyond either boundary, only counts from $0$ through $n$ can contribute.
Using a point probability instead of a tail. $P(X=r)$ includes one count; $P(X\le r)$ or $P(X\ge r)$ includes several.
Changing the model while evaluating the tail. All terms in a binomial tail use the same $n$ and $p$; only the count $k$ changes.
Rounding each mass too early. Keep adequate precision in intermediate calculations and round the final probability to the requested number of decimals.

When a tail seems unexpectedly large or small, first recheck the count range rather than recalculating the binomial formula. A single endpoint error can add or remove an entire point mass. The direct-sum method makes that error visible because each included count is listed explicitly.

Key takeaway: A binomial tail is the sum of the binomial point probabilities over the counts in the event. Translate the wording into an integer range first, preserve inclusive or exclusive endpoints, and restrict the sum to $0,\ldots,n$.

Check Your Understanding

For each problem, identify the binomial count range before evaluating the required probability.

A hypothetical set of four independent devices each has probability $0.40$ of requiring repair. If $X$ is the number requiring repair, calculate $P(X\le1)$. (Answer: $0.4752$)
Five independent policies each have probability $0.50$ of a claim. Find the probability that more than two have claims. (Answer: $0.5000$)
For $X\sim\operatorname{Bin}(6,0.20)$, calculate the probability that the count is at least one and at most three. (Answer: $0.7209$)
Seven independent components each fail with probability $0.30$. Find the probability that at least four fail. (Answer: $0.1260$)
Eight independent contracts each have probability $0.25$ of a claim. Calculate the probability that the number of claims is at least two but less than five. (Answer: $0.6056$)

Wording	Event for integer \(r\)	Counts to sum
At most \(r\)	\(X\le r\)	\(0\) through \(r\)
Less than \(r\)	\(X<r\)	\(0\) through \(r-1\)
At least \(r\)	\(X\ge r\)	\(r\) through \(n\)
More than \(r\)	\(X>r\)	\(r+1\) through \(n\)

What You'll Learn