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Exam ASTAM · Claim Frequency Models · Tutorial 104 of 979

Deriving The Binomial Probability Function

Learn why the binomial formula counts claim patterns, how to calculate exact claim-count probabilities, and how to check that the model fits the problem.

Beginner 10 min read

What You'll Learn

  • State the assumptions required for a binomial claim count.
  • Derive the probability of exactly k claims by counting claim patterns.
  • Use combinations to calculate probabilities for small portfolios.
  • Distinguish one specified claim pattern from any pattern with the same number of claims.
  • Check endpoint cases and identify common errors in binomial calculations.

Why the Binomial Formula Counts Claim Patterns

When a portfolio contains a fixed number of risks, each of which either has a claim or does not during a specified period, the total number of risks with claims is a count. To find the probability of exactly \(k\) claims, two ingredients are needed: the probability of one particular pattern of \(k\) claims, and the number of distinct patterns that contain \(k\) claims. The binomial probability function combines these ingredients.

The distinction between a particular pattern and any pattern is central. For example, “risks A and C claim, while all others do not” describes one particular pattern. “Exactly two risks claim” includes every pattern in which any two of the risks claim.

Definition: Let \(N\) be the number of risks with a claim during a specified period. The binomial model applies when there are a fixed number \(n\) of risks, each risk has the same probability \(p\) of a claim, and claim occurrences are independent from risk to risk. Each risk contributes either 0 or 1 to \(N\). Under these assumptions, \(N\) has a binomial distribution with parameters \(n\) and \(p\), written \(N\sim\operatorname{Bin}(n,p)\).

Write \(q=1-p\) for the probability that a single risk has no claim. The possible values of \(N\) are the integers \(0,1,\ldots,n\). In particular, \(N\) cannot be negative or exceed the number of risks. The model concerns claim occurrence for each risk in the period; it does not describe claim severity, and it is not the right model for a count in which one risk can contribute multiple claims.

The assumptions are substantive, not just convenient parts of a formula. A fixed \(n\) rules out a random number of exposed risks. Equal \(p\) requires the same claim probability for each risk in the modeled group. Independence means knowing whether one risk claims does not change the probability that another does. If these conditions do not fit the situation, the binomial probability function may not apply.

Deriving the Probability of Exactly \(k\) Claims

First consider one specified arrangement of outcomes among the \(n\) risks. If that arrangement has claims on exactly \(k\) risks and no claims on the other \(n-k\), independence gives its probability as \(p^kq^{n-k}\). This value is the same for every arrangement containing exactly \(k\) claims because all risks share the same \(p\).

It remains to count how many such arrangements there are. Choosing the \(k\) risks that claim from the \(n\) risks gives \(\binom{n}{k}\) choices. Each choice identifies one distinct arrangement: the chosen risks claim, and the others do not. Those arrangements are mutually exclusive, so their probabilities add.

$$ P(N=k)=\binom{n}{k}p^kq^{n-k} =\binom{n}{k}p^k(1-p)^{n-k}, \qquad k=0,1,\ldots,n. $$
Derivation: There are \(\binom{n}{k}\) mutually exclusive patterns with exactly \(k\) claims. Each has probability \(p^k(1-p)^{n-k}\), by independence and the common claim probability. Adding the equal probabilities of all these patterns gives \(\binom{n}{k}p^k(1-p)^{n-k}\). This is why the combination count multiplies the probability of one pattern.

The combination \(\binom{n}{k}\) counts selections without regard to order. A useful expression is

$$ \binom{n}{k}=\frac{n!}{k!(n-k)!}. $$

The division by \(k!\) and \((n-k)!\) removes the orderings within the group of claim risks and within the group of nonclaim risks. For small portfolios, it is often simplest to list the choices or evaluate the combination directly.

The endpoint cases provide quick checks. Exactly zero claims has probability \(q^n\), since all \(n\) risks must have no claim. Exactly \(n\) claims has probability \(p^n\), since all risks must claim. When \(p=0\), only zero claims is possible; when \(p=1\), only \(n\) claims is possible. These checks also help reveal an incorrectly placed exponent.

Worked Applications

Worked Example: Exactly Two Claims Among Five Risks

A hypothetical portfolio has five independent risks. Each has probability \(0.20\) of at least one claim during the year. Find the probability that exactly two risks have a claim.

Method and model check. The number of risks is fixed at \(n=5\); each contributes either a claim or no claim; and the problem specifies independence and a common probability \(p=0.20\). Thus the binomial probability function applies. The no-claim probability for one risk is \(q=1-0.20=0.80\).

Count the patterns. The two risks with claims can be chosen in \(\binom{5}{2}=\frac{5!}{2!3!}=\frac{120}{12}=10\) ways. One particular pattern has probability \(0.20^2(0.80)^3\).

Calculate. Since \(0.20^2=0.04\) and \(0.80^3=0.512\), one pattern has probability \(0.04(0.512)=0.02048\). Therefore,

$$ P(N=2)=10(0.02048)=0.2048. $$

The probability that exactly two risks have claims is \(0.2048\), or 20.48%. As a check on the interpretation, the probability \(0.02048\) is for one specified pair; the answer for any pair is ten times as large. Common mistake avoided: using the probability of one particular pair as though it covered every possible pair.

Worked Example: Comparing the Two Endpoint Counts

A hypothetical group contains four independent equipment risks, each with probability \(0.30\) of a claim during a quarter. Find the probabilities of no claims and of claims on all four risks.

Model values. Here \(n=4\), \(p=0.30\), and \(q=1-0.30=0.70\). The stated independence and common probability support a binomial model.

No claims. There is only one arrangement with no claims: all four risks have no claim. Equivalently, \(\binom{4}{0}=1\). Thus

$$ P(N=0)=\binom{4}{0}(0.30)^0(0.70)^4 =1(1)(0.2401)=0.2401. $$

Claims on all four. There is also only one arrangement with claims on all four risks, so \(\binom{4}{4}=1\). Hence

$$ P(N=4)=\binom{4}{4}(0.30)^4(0.70)^0 =1(0.0081)(1)=0.0081. $$

The results are 0.2401 for no claims and 0.0081 for claims on all four. The no-claim probability is larger because \(0.70>0.30\), and the respective calculations use \(q^4\) and \(p^4\). Common mistake avoided: applying the claim probability to the no-claim outcome, or retaining a factor with exponent zero as if it changed the value.

Worked Example: Any Two Claims Versus a Named Pair

A hypothetical six-risk portfolio has independent claim occurrences with common probability \(0.15\) per risk for the period. Find (a) the probability of exactly two claims on any of the six risks and (b) the probability that two named risks claim and the other four do not.

Model values. The fixed portfolio size is \(n=6\), the claim probability is \(p=0.15\), and the no-claim probability is \(q=0.85\). The independence assumption means the probability of a specified pattern is the product of the six individual outcome probabilities.

(b) One named pair first. The probability that the two named risks claim and the other four do not is \(p^2q^4=(0.15)^2(0.85)^4\). Calculate \((0.15)^2=0.0225\) and \((0.85)^4=0.52200625\), so

$$ (0.0225)(0.52200625)=0.011745140625. $$

(a) Any pair. There are \(\binom{6}{2}=\frac{6!}{2!4!}=15\) possible pairs of claiming risks. The patterns are disjoint, and each has the same probability just calculated. Therefore,

$$ P(N=2)=15(0.011745140625)=0.176177109375. $$

To four decimal places, the probability of exactly two claims is \(0.1762\), and the probability for the particular named pair is \(0.0117\). The first is fifteen times the second before rounding because it includes all fifteen possible pairs. Common mistake avoided: confusing the probability of a specified arrangement with the probability of the whole count event.

Checks, Boundaries, and Exam Traps

A careful calculation begins by translating the event into a count. “Exactly \(k\) risks claim” calls for the count \(N=k\), not for a single arrangement unless the problem names which risks claim. After identifying \(n\), \(p\), and \(q\), check that \(p+q=1\) and that \(0\leq k\leq n\). Then count patterns and multiply by the probability of one pattern.

The binomial theorem gives a useful normalization check: adding the probabilities for all possible counts must give 1.

$$ \sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k} =\bigl(p+(1-p)\bigr)^n=1. $$

This identity checks the whole probability function, but it does not by itself prove that a model is appropriate. A portfolio with different claim probabilities, dependent outcomes, or risks that can each generate multiple claims does not meet the stated binomial assumptions merely because a calculation produces a number between 0 and 1.

  • Omitting the combination count: \(p^kq^{n-k}\) is the probability of one specified pattern, not generally the probability of exactly \(k\) claims.
  • Counting ordered sequences: the risks are distinct, but their outcomes are not being arranged in time order. Use \(\binom{n}{k}\), not an extra factor for permutations.
  • Using the wrong complement: the probability of no claim is \(q=1-p\). Exactly \(k\) claims includes \(k\) factors of \(p\) and \(n-k\) factors of \(q\).
  • Using the formula outside its assumptions: check fixed portfolio size, independence, common claim probability, and at most one claim indicator per risk in the period.

These checks matter because the combination coefficient can be large even when a particular claim pattern is unlikely. The binomial calculation accounts for that multiplicity exactly; it does not assume that all possible counts are equally likely.

Key takeaway: For a fixed set of \(n\) independent risks with common claim probability \(p\), the probability of exactly \(k\) claims is the number of possible \(k\)-risk claim patterns multiplied by the probability of one pattern: \(\binom{n}{k}p^k(1-p)^{n-k}\). Use this only when the model assumptions fit.

Check Your Understanding

For each problem, identify the count event, then distinguish any set of claiming risks from a named set when needed. Probabilities in the answers are rounded to four decimal places.

  1. A hypothetical portfolio has three independent risks, each with claim probability \(0.40\). What is the probability of exactly one claim? (Answer: 0.4320)
  2. Five independent risks each have claim probability \(0.10\). Find the probability of no claims during the period. (Answer: 0.5905)
  3. Five independent risks each have claim probability \(0.20\). Find the probability that exactly three risks have claims, allowing any three risks to be the claimants. (Answer: 0.0512)
  4. Seven independent risks each have claim probability \(0.05\). Find the probability of exactly one claim. (Answer: 0.2573)
  5. Eight independent risks each have claim probability \(0.25\). Find the probability that exactly two have claims, allowing any two risks to be the claimants. (Answer: 0.3115)

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