Why the Observation Rule Determines the Model
A count of claims can have different probability models depending on how the count is collected. If an insurer observes a fixed number of policies, the number of claims is bounded by that number of policies. If observation continues until a specified number of successes has occurred, the number of failures before stopping has no fixed upper bound. The probabilities of individual outcomes matter, but the rule defining when observation ends is the key distinction.
In this tutorial, a trial has two possible outcomes: a success, with probability \(p\), or a failure, with probability \(1-p\). Trials are assumed independent and to have the same success probability. Which outcome is called a success depends on the problem. For example, it might mean “a claim occurred” in a fixed group of policies, or “a period was claim-free” in a process that ends after a target number of claim-free periods.
The parameters have different roles in the two models. In the binomial model, \(n\) is the fixed number of trials and \(p\) is the probability of the outcome being counted as a success. In the negative binomial model, \(r\) is the fixed target number of successes, while \(K\) counts failures accumulated before reaching that target. Do not assume that a negative binomial count is automatically the number of insurance claims: first establish which trial outcome is a success and which is a failure.
A Quick Structural Test
Before calculating a probability, describe the observation process in one sentence. “We inspect all 12 items” indicates a fixed number of trials. “We continue until the fourth acceptable item is found” indicates a stopping count. Then check that the model’s assumptions match the process: the trials must be independent, and the success probability must remain constant.
| Feature | Binomial count | Negative binomial failure count |
|---|---|---|
| What is fixed? | Number of trials \(n\) | Number of successes \(r\) |
| What is counted? | Successes in the \(n\) trials | Failures before the \(r\)th success |
| Support | \(0,\ldots,n\) | \(0,1,2,\ldots\) |
| When does observation end? | After the specified \(n\) trials | When the \(r\)th success occurs |
A count with a finite maximum is not enough, by itself, to establish a binomial model. The model also requires a fixed number of independent Bernoulli trials with a common success probability. Likewise, an unbounded count is not automatically negative binomial: the process must have the stated stopping rule and trial assumptions. A changing success probability, dependent outcomes, or trials that can produce multiple claims do not fit these basic models without further modelling.
Why the Probability Formulas Differ
For \(N\sim\operatorname{Bin}(n,p)\), exactly \(j\) of the \(n\) trials must be successes. There are \(\binom{n}{j}\) ways to choose their positions. Each such sequence has probability \(p^j(1-p)^{n-j}\), giving the binomial formula.
For \(K\sim\operatorname{NB}(r,p)\), a sequence with \(k\) failures before the \(r\)th success must end in a success: that final trial is the one that reaches the target. Among the preceding \(k+r-1\) trials, exactly \(k\) are failures and \(r-1\) are successes. There are \(\binom{k+r-1}{k}\) ways to place those failures. Every resulting sequence, including its final success, has probability \((1-p)^k p^r\). Therefore,
This derivation also explains why the negative binomial count has no fixed upper bound: any number \(k\) of failures could occur before the target success, provided the trials continue. The formula does not apply if the process is stopped after a fixed number of trials regardless of whether the target has been reached.
Worked Examples
Worked Example: Claims in a Fixed Group of Policies
A hypothetical portfolio contains 8 policies observed over one year. Each policy independently has probability \(0.10\) of producing a claim during the year. Let \(N\) be the number of policies that produce a claim. Find the probability of exactly 2 claims, and state the support of \(N\).
Method and model. The number of policies is fixed at \(n=8\), each policy contributes one claim indicator, and the indicators are independent with the same claim probability \(p=0.10\). Thus \(N\sim\operatorname{Bin}(8,0.10)\). The support is \(0,1,\ldots,8\).
For exactly 2 claims, \(j=2\). The binomial coefficient is \(\binom{8}{2}=8\cdot7/(2\cdot1)=28\). The probability is
Rounded to four decimal places, \(P(N=2)=0.1488\). The result is between 0 and 1, and the event is possible because 2 lies within the support \(0,\ldots,8\).
Common mistake avoided: Do not use a negative binomial formula just because the number of claims could have been larger in a different portfolio. Here there are exactly 8 fixed claim opportunities, so the count cannot exceed 8.
Worked Example: Claims Before a Target Number of Claim-Free Periods
A hypothetical monitoring program observes successive months for one insured device. In each month, independently, the month is claim-free with probability \(0.80\); otherwise, it has a claim, with probability \(0.20\). Observation ends at the third claim-free month. Let \(K\) be the number of months with claims before the program ends. Find \(P(K=2)\) and \(P(K\leq1)\).
Method and model. Here the success is a claim-free month, so \(p=0.80\), and the target is \(r=3\) successes. A claim month is a failure. Since the process ends at the third claim-free month, \(K\sim\operatorname{NB}(3,0.80)\), with support \(0,1,2,\ldots\).
For exactly 2 claim months before the third claim-free month, the last month must be claim-free. The two failures and the first two successes occupy the preceding four months in \(\binom{4}{2}=6\) possible orders. Thus
For \(K\leq1\), add the probabilities for zero and one claim month. With zero failures, the first three months are all claim-free:
With one failure, the final month is claim-free and the failure can occupy any of the first three positions:
Therefore \(P(K\leq1)=0.512+0.3072=0.8192\). Rounded to four decimal places, the answers are \(P(K=2)=0.1229\) and \(P(K\leq1)=0.8192\).
Common mistake avoided: The target is three claim-free months, not three total months. The number of observed months is random, so treating it as a fixed number of trials would use the wrong model.
Worked Example: Same Trial Probabilities, Different Observation Rules
A hypothetical review process has independent trials, each successful with probability \(0.70\) and failed with probability \(0.30\). Compare two protocols: Protocol A conducts exactly 6 trials and counts failures; Protocol B continues until 4 successes and counts failures before the fourth success. Find the probability of exactly 2 failures under each protocol.
Method and model. In Protocol A, the fixed number of trials is 6. Counting failures is equivalent to counting successes with probability \(0.30\), so the failure count has a binomial distribution with \(n=6\) and failure probability \(0.30\). In Protocol B, the target is \(r=4\) successes, each with probability \(0.70\), and the count is failures before that target; it has a negative binomial distribution.
For Protocol A, exactly 2 failures means 4 successes and 2 failures among the 6 trials:
For Protocol B, the sequence must end in the fourth success. The two failures are placed among the first \(2+4-1=5\) trials, giving
Rounded to four decimal places, the probabilities are 0.3241 for Protocol A and 0.2161 for Protocol B. Both protocols use the same trial probabilities, but the possible sequences differ: Protocol A always has 6 trials, while Protocol B stops as soon as its fourth success occurs.
Common mistake avoided: Do not substitute the same number of trials into both formulas. The stopping rule, not just the success probability, determines the count distribution.
Common Exam Trap: Applying the Stopping Rule That Is Not There
Suppose the 8 fixed policies in the first example are mistakenly treated as a negative binomial process with \(r=2\), \(p=0.10\), and \(k=2\). That calculation gives
This is not the probability of exactly 2 claims among 8 policies. It describes a different event: exactly 2 failures before the second success, with the process ending at that second success. In the fixed group, all 8 policies are observed whether or not two claims have already occurred. The correct calculation is the binomial probability \(0.1488\) found above.
A reliable check is to write down the stopping rule before selecting a formula. If the problem says to observe all members of a specified group, use a fixed-trial model when its Bernoulli assumptions hold. If it says to continue until the \(r\)th specified success, consider the negative binomial failure count. If neither description fits, neither model should be selected without further justification.
Check Your Understanding
For each problem, identify the model from the observation rule before computing or interpreting the count.
- A hypothetical block of 6 policies has independent annual claim probability \(0.15\) per policy. Find the probability of exactly one claim. (Answer: Binomial; \(0.3993\), to four decimal places.)
- Independent trials have success probability \(0.60\). Trials continue until the second success. Find the probability of exactly 3 failures before stopping. (Answer: Negative binomial; \(0.0922\), to four decimal places.)
- A hypothetical group of 9 devices has independent failure probability \(0.20\) per device. Find the probability of at most one failure. (Answer: Binomial; \(0.4362\), to four decimal places.)
- Independent trials have success probability \(0.75\), and observation ends at the third success. Find the probability that the process ends within 5 trials. (Answer: Negative binomial failure count with \(K\leq2\); \(0.8965\), to four decimal places.)
- For independent trials with success probability \(0.70\), compare the probability of exactly 2 failures in exactly 6 trials with the probability of exactly 2 failures before the fourth success. (Answer: Binomial probability \(0.3241\); negative binomial probability \(0.2161\), to four decimal places.)