Generating Count Probabilities from \(p_0\)
When a count distribution is specified by its parameters, its probabilities can be calculated directly from a probability mass function. For Poisson and binomial counts, there is another efficient method: start with \(p_0=P(N=0)\), then generate each later probability from the one immediately before it. This is useful when a question asks for several probabilities or when a computed sequence needs to be checked for indexing errors.
The method depends on both the distribution and its support. A Poisson count can take any nonnegative integer value. A binomial count with \(m\) trials cannot exceed \(m\); its probability sequence therefore ends at \(p_m\). In both cases, the initial probability is essential: the recurrence gives ratios between probabilities, not their absolute scale.
Why the Recurrences Work
The recurrence follows by taking the ratio of adjacent probabilities. This also shows why the starting probability and the correct index matter: each new term is obtained from a specific ratio, not from an independent approximation.
For each distribution, express \(p_n/p_{n-1}\) from its probability mass function.
Multiply the ratio by \(p_{n-1}\) to obtain \(p_n\), then use the specified \(p_0\) as the starting value.
For a binomial count, calculate through \(n=m\); terms beyond \(m\) are zero. For a Poisson count, continue for every nonnegative integer as needed.
For a Poisson count with parameter \(\lambda>0\), the probability mass function is \(p_n=e^{-\lambda}\lambda^n/n!\). Dividing by \(p_{n-1}\) gives
For a binomial count, \(p_n=\binom{m}{n}q^n(1-q)^{m-n}\) for \(0\leq n\leq m\). For \(1\leq n\leq m\), the adjacent-probability ratio is
Thus \(p_n=((m-n+1)/n)(q/(1-q))p_{n-1}\). The factor \(m-n+1\) decreases with \(n\); it is what makes the sequence terminate. At \(n=m+1\), the corresponding factor is zero, so the recurrence gives \(p_{m+1}=0\). There are no binomial outcomes above \(m\).
Poisson Probabilities: Repeated Multiplication
For a Poisson count, each step uses \(p_n=(\lambda/n)p_{n-1}\). This avoids calculating a new factorial for every probability. It also provides a straightforward check: when only a portion of the sequence is calculated, its sum is less than one because the remaining tail has not yet been included.
Worked Example: Generating a Poisson Sequence and Checking Its Total
A hypothetical repair desk records the number \(N\) of urgent equipment requests received in a fixed shift. Assume \(N\sim\operatorname{Poisson}(2)\), so the mean count per shift is \(\lambda=2\). Generate \(p_0\) through \(p_{10}\), then check the total probability represented by these terms.
Method. Start with \(p_0=e^{-2}=0.1353352832\) and apply \(p_n=(2/n)p_{n-1}\). The intermediate recurrence factors and probabilities, rounded to six decimals for display, are:
| \(n\) | Factor \(2/n\) | Calculation from previous term | \(p_n\) |
|---|---|---|---|
| 0 | — | \(e^{-2}\) | 0.135335 |
| 1 | 2 | \(2(0.1353352832)\) | 0.270671 |
| 2 | 1 | \(1(0.2706705665)\) | 0.270671 |
| 3 | \(2/3\) | \((2/3)(0.2706705665)\) | 0.180447 |
| 4 | \(1/2\) | \((1/2)(0.1804470443)\) | 0.090224 |
| 5 | \(2/5\) | \((2/5)(0.0902235221)\) | 0.036089 |
| 6 | \(1/3\) | \((1/3)(0.0360894089)\) | 0.012030 |
| 7 | \(2/7\) | \((2/7)(0.0120298030)\) | 0.003437 |
| 8 | \(1/4\) | \((1/4)(0.0034370866)\) | 0.000859 |
| 9 | \(2/9\) | \((2/9)(0.0008592716)\) | 0.000191 |
| 10 | \(1/5\) | \((1/5)(0.0001909492)\) | 0.000038 |
Adding the unrounded probabilities from \(p_0\) through \(p_{10}\) gives \(0.999991692\). The remaining probability is \(P(N>10)=0.000008308\), so the included terms plus the omitted tail total \(1.000000000\), apart from any displayed rounding. The full sequence sums to one because its terms are \(e^{-2}2^n/n!\), and the exponential series gives \(e^{-2}\sum_{n=0}^{\infty}2^n/n!=1\). This checks the recurrence-generated sequence without treating the partial sum as a complete probability distribution.
Common mistake avoided: The probabilities through \(p_{10}\) do not sum to exactly one by themselves. A partial Poisson sequence omits the positive tail \(P(N>10)\); that is not evidence that the recurrence is incorrect.
Binomial Probabilities: A Finite Sequence
For a binomial count, the initial probability is \(p_0=(1-q)^m\), and each later term uses the number of trials still represented in the ratio. Unlike the Poisson case, a complete binomial sequence can be generated in finitely many steps and checked by adding \(p_0,\ldots,p_m\).
Worked Example: A Complete Binomial Sequence
A hypothetical quality check samples \(m=5\) independent components. Each component independently has probability \(q=0.20\) of requiring rework. Let \(N\) be the number requiring rework, so \(N\sim\operatorname{Binomial}(5,0.20)\). Generate every probability and find \(P(N\geq4)\).
Method. Here \(p_0=(1-0.20)^5=0.8^5=0.32768\), and \(q/(1-q)=0.20/0.80=0.25\). For \(n=1,\ldots,5\), use \(p_n=((6-n)/n)(0.25)p_{n-1}\).
For \(n=1\), the factor is \((5/1)(0.25)=1.25\), so \(p_1=1.25(0.32768)=0.40960\). For \(n=2\), the factor is \((4/2)(0.25)=0.50\), so \(p_2=0.50(0.40960)=0.20480\). For \(n=3\), the factor is \((3/3)(0.25)=0.25\), so \(p_3=0.25(0.20480)=0.05120\). For \(n=4\), the factor is \((2/4)(0.25)=0.125\), so \(p_4=0.125(0.05120)=0.00640\). For \(n=5\), the factor is \((1/5)(0.25)=0.05\), so \(p_5=0.05(0.00640)=0.00032\).
The check is \(0.32768+0.40960+0.20480+0.05120+0.00640+0.00032=1.00000\). The requested probability is \(P(N\geq4)=p_4+p_5=0.00640+0.00032=0.00672\). Thus the probability is \(0.0067\) to four decimal places.
Common mistake avoided: The recurrence multiplier is not \(m/n\) times \(q/(1-q)\). It is \((m-n+1)/n\) times \(q/(1-q)\), with the trial count adjusted at each step.
Worked Example: Using the Binomial Endpoint and Checking Exactly
A hypothetical inspection examines \(m=6\) sealed packages, each independently having probability \(q=0.25\) of a labeling defect. Let \(N\sim\operatorname{Binomial}(6,0.25)\). Generate the full sequence and find the probability of between two and four defects, inclusive.
Method. The initial probability is \(p_0=0.75^6=729/4096\). The odds factor is \(q/(1-q)=1/3\), so \(p_n=((7-n)/n)(1/3)p_{n-1}\). Keeping exact numerators over the common denominator \(4096\) makes the total check transparent.
For \(n=1\), the factor is \(6/3=2\), giving \(p_1=1458/4096\). For \(n=2\), the factor is \((5/2)/3=5/6\), giving \(p_2=1215/4096\). For \(n=3\), the factor is \((4/3)/3=4/9\), giving \(p_3=540/4096\). For \(n=4\), the factor is \((3/4)/3=1/4\), giving \(p_4=135/4096\). For \(n=5\), the factor is \((2/5)/3=2/15\), giving \(p_5=18/4096\). For \(n=6\), the factor is \((1/6)/3=1/18\), giving \(p_6=1/4096\).
The numerator check is \(729+1458+1215+540+135+18+1=4096\), so the probabilities sum to one. The requested probability is \(P(2\leq N\leq4)=(1215+540+135)/4096=1890/4096=0.46142578125\), or \(0.4614\) to four decimal places.
Common mistake avoided: Stopping after \(p_5\) would omit the possible outcome of six defects. At the endpoint \(n=m=6\), the recurrence still produces \(p_6\); only outcomes above six have probability zero.
Reading the Total as a Diagnostic
A probability-total check is especially useful for catching arithmetic and indexing mistakes. For a binomial distribution, calculate all terms from \(p_0\) to \(p_m\) and add them; the sum must be one. A total different from one can indicate an incorrect initial probability, a misplaced factor, or an omitted endpoint. When rounding is used, a small discrepancy may be due to rounded terms, so perform the check with unrounded values where possible.
For a Poisson distribution, any finite list \(p_0,\ldots,p_k\) has sum \(P(N\leq k)\), not one. The difference \(1-\sum_{n=0}^{k}p_n\) is the probability of values greater than \(k\). The sum approaches one as \(k\) increases. Do not add an artificial endpoint or renormalize a truncated list unless the problem explicitly defines a different model.
The recurrences also have parameter restrictions. The Poisson recurrence above assumes \(\lambda>0\); for \(\lambda=0\), the count is zero with probability one. The binomial ratio form assumes \(0<q<1\); if \(q=0\), all probability is at zero, and if \(q=1\), all probability is at \(m\). These degenerate cases are easiest to handle directly rather than by using a ratio with a zero denominator.
Check Your Understanding
For each count model, state the recurrence you use, begin from \(p_0\), and keep enough precision to check your final result.
- A hypothetical dispatch center has \(N\sim\operatorname{Poisson}(3)\) calls in a defined interval. Use the recurrence to obtain \(p_0,p_1,p_2\) and calculate \(P(N\leq2)\). (Answer: \(0.4232\))
- A hypothetical test has \(N\sim\operatorname{Binomial}(4,0.25)\) defective units. Generate \(p_0,p_1,p_2\) and calculate \(P(N\leq2)\). (Answer: \(0.9492\))
- For a hypothetical count \(N\sim\operatorname{Poisson}(1.5)\), generate \(p_0\) through \(p_3\) and calculate \(P(1\leq N\leq3)\). (Answer: \(0.7112\))
- A hypothetical audit has \(N\sim\operatorname{Binomial}(7,0.20)\). Generate the probabilities from \(p_0\) through \(p_3\) and calculate \(P(N\leq3)\). (Answer: \(0.9667\))
- A hypothetical shipment has \(N\sim\operatorname{Binomial}(5,0.30)\) mislabeled packages. Generate the sequence through \(p_5\), check its total, and find \(P(N\geq4)\). (Answer: \(0.0308\))