Binomial Point Probabilities

When a Binomial Point Probability Applies

An insurance frequency question may ask for the probability that exactly a specified number of exposures experience a claim during a defined period. The binomial model is appropriate when the number of exposure trials is fixed, each trial has the same probability of success, and the trials are independent. The model then gives the probability of exactly $k$ successes among $n$ trials.

A success must be defined in the context of the problem. For example, a trial could be one hypothetical policy during one year, with “success” meaning that the policy has at least one covered claim in that year. The resulting binomial count is the number of claim-bearing policies, not necessarily the total number of claims: a policy that can produce multiple claims does not automatically correspond to a single-event binomial count.

Definition: A random variable $N$ has a binomial distribution with parameters $n$ and $p$, written $N\sim\operatorname{Bin}(n,p)$, if it counts successes in $n$ independent trials, each of which has the same success probability $p$. Here $n$ is a fixed nonnegative integer and $0\le p\le 1$. The count can take the values $0,1,\ldots,n$. Its point probability is $$ P(N=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n, $$ where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.

The word point means the probability at one particular count, such as $P(N=2)$. The formula requires the probability of exactly $k$, not at least $k$, at most $k$, or a range of counts. Those different events require adding or otherwise combining appropriate point probabilities.

Why the Formula Has a Combination Factor

First consider one particular arrangement of the $n$ trial outcomes in which exactly $k$ trials succeed and the remaining $n-k$ fail. Independence and the common success probability give this particular arrangement probability $p^k(1-p)^{n-k}$. But the event “exactly $k$ successes” includes every arrangement with $k$ successes, not just one chosen arrangement.

There are $\binom{n}{k}$ ways to choose which $k$ of the $n$ trials succeed. These arrangements are mutually exclusive: a trial outcome cannot be two different arrangements at once. Adding their equal probabilities therefore gives

P(N=k) =\underbrace{\binom{n}{k}}_{\text{ways to choose the successful trials}} \underbrace{p^k(1-p)^{n-k}}_{\text{probability of each arrangement}}.

This also explains a frequent error: $p^k(1-p)^{n-k}$ alone is the probability of one specified arrangement, such as the first $k$ trials succeeding and all remaining trials failing. It is not generally the probability of exactly $k$ successes.

A Reliable Calculation Process

Before substituting into the formula, name the trial and success in context. Then confirm that the model assumptions fit, identify $n$, $p$, and $k$, and check that $k$ is in the support $0,\ldots,n$. A target count outside this range has probability zero. The following process keeps the count and the probability of failure aligned.

Define success and the trial.
Specify what one exposure represents and what event makes it a success.

Check the binomial assumptions.
There must be a fixed number of trials, independent outcomes, and a common success probability for every trial.

Identify the point event.
Translate “exactly $k$” into $P(N=k)$, and verify that $0\le k\le n$.

Substitute and evaluate.
Calculate the combination count and the success and failure factors, then multiply. Round the probability only at the final step.

Worked Example: Exactly Two Claim-Bearing Policies

A hypothetical insurer considers 9 independent one-year policies. For each policy, the probability of at least one covered claim during the year is 0.20. What is the probability that exactly 2 policies have at least one covered claim?

Method and model. Let one trial be one policy-year and define success as at least one covered claim during that policy-year. The number of trials is fixed at $n=9$; the trials are assumed independent, and each has the same success probability $p=0.20$. Thus $N\sim\operatorname{Bin}(9,0.20)$, and the requested event is $N=2$. The failure probability for each trial is $1-p=0.80$.

Substitute. There are $\binom{9}{2}=9!/(2!7!)=36$ ways for 2 of the 9 policies to be claim-bearing. The probability of any one such arrangement is $(0.20)^2(0.80)^7$.

\begin{aligned} P(N=2) &=\binom{9}{2}(0.20)^2(0.80)^7\\ &=36(0.04)(0.2097152)\\ &=1.44(0.2097152)\\ &=0.301989888\\ &\approx 0.3020. \end{aligned}

The probability is 0.3020, or about 30.20%. As a check, the combination factor counts 36 possible pairs of claim-bearing policies; using only $0.04(0.2097152)$ would count just one specified pair. A common mistake is to omit the factor 36 and thereby understate the probability.

Worked Example: Exactly One Claim in a Small Policy Group

For a hypothetical group of 12 independent one-year policies, suppose each policy has probability 0.05 of at least one covered claim in the year. Find the probability that exactly one policy is claim-bearing.

Method and model. Each policy-year is one trial, success means at least one claim, and the assumptions specify a fixed $n=12$, independent trials, and common $p=0.05$. Therefore $N\sim\operatorname{Bin}(12,0.05)$. Here $k=1$, and the failure probability is $1-0.05=0.95$.

Substitute and compute. The one successful policy can be selected in $\binom{12}{1}=12$ ways. For any one selection, that policy succeeds and the other 11 policies fail.

\begin{aligned} P(N=1) &=\binom{12}{1}(0.05)(0.95)^{11}\\ &=12(0.05)(0.5688000923)\\ &=0.6(0.5688000923)\\ &=0.3412800554\\ &\approx 0.3413. \end{aligned}

The probability is 0.3413, or about 34.13%. The intermediate value $0.5688000923$ is the probability that 11 specified policies all have no claim; the factor $12(0.05)$ accounts for choosing which policy is the sole claim-bearing one. A common error is to use $0.05(0.95)^{11}$ as the answer, which includes only one specified choice of that policy.

Worked Example: Exactly Three Claim-Bearing Exposures

A hypothetical portfolio has 10 independent exposures for a defined period. Each exposure has probability 0.20 of the event being counted. Find the probability of exactly 3 successes, and explain what the value counts.

Method and model. Treat each exposure as one trial and the stated event as success. The problem supplies a fixed $n=10$, independence, and common $p=0.20$, so $N\sim\operatorname{Bin}(10,0.20)$. The target $k=3$ is within the possible count range, and the failure probability is 0.80.

Calculate the point probability. The number of ways to select the 3 successful exposures is $\binom{10}{3}=10!/(3!7!)=120$. For each selection, the three chosen exposures succeed and the other seven fail.

\begin{aligned} P(N=3) &=\binom{10}{3}(0.20)^3(0.80)^7\\ &=120(0.008)(0.2097152)\\ &=0.96(0.2097152)\\ &=0.201326592\\ &\approx 0.2013. \end{aligned}

The probability of exactly 3 successes is 0.2013, or about 20.13%. It includes all 120 ways to select the three successful exposures. The calculation does not mean there are exactly three individual claims if a single exposure can generate multiple claims; the count here is of exposures meeting the defined success condition. A common mistake is to interpret the binomial count as a total-claim count without confirming what one trial and one success represent.

Assumption Checks and Exam Traps

The arithmetic is useful only when the model describes the scenario. In an insurance setting, check the following before treating a count as binomial:

Fixed number of trials: $n$ is set before the outcomes are observed. The model counts successes among these trials.
One consistent success definition: The event counted as success must have the same meaning on each trial. If success means “at least one claim,” the count is claim-bearing exposures, not the total number of individual claims.
Common probability: Each trial has the same probability $p$. If different exposures have different probabilities, the binomial point formula does not apply as written.
Independence: The outcome on one trial does not change the probability distribution of another trial. Dependence between exposures conflicts with this assumption.
Valid count: $k$ must be an integer from 0 through $n$. If $k<0$, $k>n$, or $k$ is not an integer, the event is impossible under this model.

Another trap is confusing “exactly $k$” with a more inclusive event. The expression $\binom{n}{k}p^k(1-p)^{n-k}$ gives only the probability at $k$. For example, it does not by itself give the probability of at least $k$ successes. Also check exponents: exactly $k$ successes means $k$ factors of $p$ and $n-k$ factors of $1-p$, because all remaining trials must fail.

A point probability is never a probability for one particular trial unless the problem itself has $n=1$. For $n>1$, the combination factor is usually essential because the successful trials can occur in many positions. When $k=0$ or $k=n$, there is just one possible arrangement, since $\binom{n}{0}=\binom{n}{n}=1$.

Key takeaway: For a fixed number $n$ of independent, equally likely trials, the probability of exactly $k$ successes is $\binom{n}{k}p^k(1-p)^{n-k}$. Define the trial and success carefully, include the number of possible arrangements, and use the formula only when its assumptions hold. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each question, identify the binomial parameters and calculate the requested point probability. Round probabilities to four decimal places unless an exact value is requested.

A hypothetical group has 7 independent policy-years, each with probability 0.10 of at least one claim. What is the probability that none has a claim? (Answer: $0.9^7=0.4783$)
Eight independent exposures each have probability 0.25 of success. Find the probability of exactly 2 successes. (Answer: $\binom{8}{2}(0.25)^2(0.75)^6=0.3115$)
Ten independent policies each have probability 0.08 of at least one claim in the period. Find the probability that exactly one policy has a claim. (Answer: $\binom{10}{1}(0.08)(0.92)^9=0.3777$)
Six independent trials each have success probability 0.30. Find the probability of exactly 4 successes. (Answer: $\binom{6}{4}(0.30)^4(0.70)^2=0.0595$)
Five independent exposures each have probability 0.20 of success. Find the probability that all five succeed, and state why there is no combination factor greater than 1. (Answer: $\binom{5}{5}(0.20)^5=0.00032$, which rounds to $0.0003$; there is only one way for all five to succeed.)

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