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Exam FAM · Frequency Probabilities and Approximations · Tutorial 533 of 988

Frequency Probabilities Over Integer Intervals

Calculate the probability of a claim count falling between inclusive integer bounds under binomial, Poisson, and negative binomial models.

Intermediate 10 min read

What You'll Learn

  • Translate a stated count interval into inclusive integer endpoints.
  • Identify the support and parameters needed for each frequency model.
  • Calculate interval probabilities by summing the relevant probability masses.
  • Apply the negative binomial parameterization for failures before a fixed number of successes.
  • Check interval calculations for omitted endpoints and inconsistent model assumptions.

Why Integer-Interval Probabilities Matter

Many frequency questions ask whether a claim count falls within a range rather than taking one exact value. A hypothetical insurer might need the probability that a policy group generates between two and five claims during a year, including both endpoints. The calculation is direct once the model, its parameters, and the meaning of the interval are clear: identify every integer in the range, calculate its probability under the model, and add those probabilities.

The main risks are not complicated algebra. They are misreading an inclusive endpoint, using a count outside the model’s support, or carrying over a parameterization from another model. The methods below apply to binomial, Poisson, and negative binomial counts, with assumptions stated before each calculation.

Definition: Let \(N\) be a claim count taking integer values. For integer bounds \(L\) and \(U\), with \(L \le U\), the inclusive interval event is \(L \le N \le U\). Its probability is the sum of the probabilities for all integer values from \(L\) through \(U\), subject to the support of \(N\).

The General Calculation

A probability mass function gives \(P(N=k)\), the probability of exactly \(k\) claims. The interval probability follows by adding the mass at each admissible count. The addition is valid because the events \(N=L\), \(N=L+1\), and so on through \(N=U\) are mutually exclusive and together make up the interval event.

$$ P(L \le N \le U)=\sum_{k=L}^{U}P(N=k). $$

For a count with support starting at zero, a negative lower bound does not create negative claim counts: begin the sum at zero instead. For a binomial count with maximum possible value \(n\), an upper bound greater than \(n\) contributes no additional terms beyond \(n\). If the stated range contains no possible values, its probability is zero.

Use the model’s probability mass function for each term. For a binomial model, \(N\sim\operatorname{Bin}(n,p)\), where \(n\) is the fixed number of independent trials and \(p\) is the probability of a claim on each trial:

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

For a Poisson model, \(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda\) is the expected count over the specified period or exposure:

$$ P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!}, \qquad k=0,1,2,\ldots. $$

For the negative binomial calculations here, \(N\) counts failures before the \(r\)th success, with success probability \(p\) on each independent trial. Under this convention, \(r\) is a positive integer, \(0<p\le 1\), and:

$$ P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k, \qquad k=0,1,2,\ldots. $$

A negative binomial model can describe overdispersed claim counts, but the parameterization must be specified: other conventions use different meanings for the count and parameters. The formula above is for failures before \(r\) successes; do not apply it to a differently defined variable without translating the parameters.

A Reliable Procedure

A short sequence of checks prevents most interval-probability errors. In particular, write the event with inequality signs before substituting any numbers. This makes it harder to omit an endpoint or confuse an interval such as “at least two and fewer than five” with “from two through five.”

1
Translate the wording.
Write the event as an inequality, and determine whether each endpoint is included. Convert “fewer than” or “more than” to the corresponding integer boundary.
2
Check the model and support.
State the distribution and parameters. Determine which values of the count are possible under that distribution.
3
Calculate each required mass.
Substitute each admissible integer into the appropriate probability mass function, keeping enough precision during the calculation.
4
Add and check.
Sum every required term exactly once. Confirm that the result is between zero and one and that the included values match the stated event.

Worked Interval Calculations

Worked Example: A Binomial Count Between Two Bounds

A hypothetical service program monitors eight independent devices for one year. Each device has probability \(p=0.25\) of generating a claim during that year, independently of the other devices. Let \(N\) be the number of devices generating a claim. Find the probability that the program receives between two and four claims, inclusive.

Method and assumptions. There are a fixed \(n=8\) independent trials, each with the same claim probability \(p=0.25\), so \(N\sim\operatorname{Bin}(8,0.25)\). The event is \(2\le N\le4\); both endpoints are included. The binomial support is \(0,1,\ldots,8\), so all three counts in the interval are possible.

Calculate each of the three masses:

$$ \begin{aligned} P(N=2)&=\binom{8}{2}(0.25)^2(0.75)^6 =28(0.0625)(0.177978515625) =0.3114624023,\\ P(N=3)&=\binom{8}{3}(0.25)^3(0.75)^5 =56(0.015625)(0.2373046875) =0.2076416016,\\ P(N=4)&=\binom{8}{4}(0.25)^4(0.75)^4 =70(0.00390625)(0.31640625) =0.0865173340. \end{aligned} $$

Add the masses:

$$ P(2\le N\le4)=0.3114624023+0.2076416016+0.0865173340 =0.6056213379. $$

Rounded to four decimal places, the probability is 0.6056. The method is to sum the binomial masses for the integer values in the inclusive range. A common mistake this avoids is stopping at three and leaving out the probability of four claims. As a check, all three terms are nonnegative, and their sum is a valid probability less than one.

Worked Example: A Poisson Count Over a Range

A hypothetical block of equipment is monitored for six machine-years. Assume claims follow a Poisson model with rate \(1/3\) claim per machine-year, so the count \(N\) across the block is Poisson with parameter \(\lambda=(1/3)(6)=2\). Find the probability of one, two, or three claims.

Method and assumptions. The stated Poisson model assumes the expected count scales with exposure, giving \(\lambda=2\) for six machine-years. The event is \(1\le N\le3\), and all three values are in the Poisson support of nonnegative integers.

Use the Poisson probability mass function for each count:

$$ \begin{aligned} P(N=1)&=e^{-2}\frac{2^1}{1!}=2e^{-2}=0.2706705665,\\ P(N=2)&=e^{-2}\frac{2^2}{2!}=2e^{-2}=0.2706705665,\\ P(N=3)&=e^{-2}\frac{2^3}{3!}=\frac{4}{3}e^{-2}=0.1804470443. \end{aligned} $$

Therefore,

$$ P(1\le N\le3)=0.2706705665+0.2706705665+0.1804470443 =0.7217881773. $$

Rounded to four decimal places, the answer is 0.7218. This calculation uses the exposure-adjusted parameter \(\lambda=2\), not the per-machine-year rate \(1/3\) as though it were the parameter for the whole block. The three displayed masses sum to the requested event’s probability; none of the count values from one through three is omitted.

Worked Example: A Negative Binomial Count Between Two Bounds

For a hypothetical portfolio, suppose the annual claim count \(N\) follows a negative binomial model with parameters \(r=3\) and \(p=0.5\), using the convention that \(N\) counts failures before the third success. Find the probability that the count is between two and four claims, inclusive.

Method and assumptions. The model is the specified failures-before-\(r\)-successes negative binomial distribution. Thus its support is \(0,1,2,\ldots\), and the event is \(2\le N\le4\). For each count \(k\), use \(\binom{k+r-1}{k}p^r(1-p)^k\).

$$ \begin{aligned} P(N=2)&=\binom{2+3-1}{2}(0.5)^3(0.5)^2 =\binom{4}{2}(0.125)(0.25) =6(0.03125)=0.1875,\\ P(N=3)&=\binom{3+3-1}{3}(0.5)^3(0.5)^3 =\binom{5}{3}(0.125)(0.125) =10(0.015625)=0.15625,\\ P(N=4)&=\binom{4+3-1}{4}(0.5)^3(0.5)^4 =\binom{6}{4}(0.125)(0.0625) =15(0.0078125)=0.1171875. \end{aligned} $$

Add the required terms:

$$ P(2\le N\le4)=0.1875+0.15625+0.1171875=0.4609375. $$

Rounded to four decimal places, the probability is 0.4609. The method is to apply the specified negative binomial mass function at each integer in the interval. The parameterization matters: using a formula in which the random variable counts successes, or in which \(r\) has a different role, would not calculate this event’s probability. The final probability is nonnegative and less than one, as required.

Why the Sum Works—and Where Errors Enter

The interval formula is an application of addition for mutually exclusive events. For example, when \(L\le U\), the event \(L\le N\le U\) can be written as the union of the exact-count events \(N=L\), \(N=L+1\), through \(N=U\). A count cannot equal two different integers at once, so these events do not overlap. Their probabilities therefore add to give the probability of their union.

This reasoning does not depend on whether the count is binomial, Poisson, or negative binomial. The model determines each point probability and the possible values; the interval rule itself is the same. If a stated bound falls outside the support, first remove impossible values and then sum the remaining masses. Do not extend a binomial sum above \(n\), or include a negative count for a model whose support begins at zero.

A useful exam check is to list the included integers before calculating. For “at least two and at most four,” the list is \(2,3,4\). For “more than two but less than five,” it is also \(3,4\). Similar wording can produce different endpoints, even though the numerical bounds sound close.

Common Mistake: Dropping an Inclusive Endpoint

Return to the negative binomial example. The requested event was \(2\le N\le4\). A mistaken calculation might include only \(N=2\) and \(N=3\):

$$ P(N=2)+P(N=3)=0.1875+0.15625=0.34375. $$

That is the probability of \(2\le N\le3\), not \(2\le N\le4\). The correct calculation also includes \(P(N=4)=0.1171875\), giving:

$$ 0.34375+0.1171875=0.4609375. $$

The correction is not a change to the negative binomial formula; it is a correction to the event being summed. Writing the inequality and listing its integer values before substituting numbers prevents this endpoint error.

What to Check Before Reporting an Answer

An interval probability is not usually a new kind of frequency model calculation. It is a careful assembly of point probabilities. Before reporting the result, confirm that the model’s assumptions fit the setup: fixed independent trials with a common probability for the binomial model; a Poisson count with the stated expected count for the Poisson model; or the correctly parameterized negative binomial model. If those assumptions do not describe the situation, the arithmetic may be correct while the model-based answer is not appropriate.

  • Translate the wording into an inequality before doing arithmetic.
  • Include each integer endpoint when the wording says “from,” “through,” “at least,” or “at most” as appropriate.
  • Use only values in the model’s support.
  • Keep precision through the sum and round the final probability to four decimal places when reporting probabilities.

Interval calculations are useful whenever a decision depends on a band of possible claim counts—for example, a hypothetical staffing plan that is adequate only when the count remains in a specified range. Correctly identifying the event is just as important as selecting the right distribution.

Check Your Understanding

For each problem, identify the interval values and sum the appropriate point probabilities. Probabilities in the answers are rounded to four decimal places.

  1. A hypothetical group has five independent policies, each with probability \(0.3\) of producing a claim during the year. If \(N\sim\operatorname{Bin}(5,0.3)\), find \(P(1\le N\le2)\). (Answer: 0.6689)
  2. A hypothetical count follows a Poisson model with \(\lambda=1.5\). Find the probability of zero, one, or two claims. (Answer: 0.8088)
  3. Let \(N\) follow the negative binomial convention of failures before \(r\) successes, with \(r=2\) and \(p=0.4\). Find the probability of one through three failures, inclusive. (Answer: 0.5030)
  4. A hypothetical count follows \(\operatorname{Bin}(10,0.1)\). Find the probability of at most one claim, making clear which integers are included. (Answer: 0.7361)
  5. A hypothetical portfolio count is Poisson with \(\lambda=4\). Find the probability of between three and six claims, inclusive. (Answer: 0.6512)
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