Adjacent-Probability Ratios in Binomial Calculations

Why Compare Neighboring Counts?

A binomial model can assign probabilities to every count from zero through the number of trials. When the number of trials is large, calculating each probability separately can be slow and error-prone. Often the question is narrower: is a count more likely than the count just below it, and where does the probability reach its maximum? A ratio of consecutive probabilities answers both questions without requiring repeated evaluation of binomial coefficients.

The method applies when the count is binomial: there are a fixed number of independent trials, each trial has the same probability of success, and the random variable counts successes. The examples below use hypothetical situations and state the model parameters explicitly.

Definition: Let \(X\sim\operatorname{Bin}(n,p)\), where \(n\) is a positive integer, \(0<p<1\), and \(q=1-p\). For \(k=0,1,\ldots,n\), write \(P_k=P(X=k)\). The adjacent-probability ratio comparing counts \(k+1\) and \(k\) is \(P_{k+1}/P_k\), defined for \(k=0,1,\ldots,n-1\).

The binomial probability formula gives \(P_k=\binom{n}{k}p^kq^{n-k}\). The ratio is especially useful because the powers of \(p\) and \(q\), as well as most of the binomial coefficients, cancel. This leaves a simple expression whose size relative to 1 tells us which neighboring count is more likely.

Deriving the Adjacent-Probability Ratio

For \(0<p<1\), every binomial probability \(P_k\) is positive, so dividing by it is valid. For any \(k\) from 0 through \(n-1\),

\begin{aligned} \frac{P_{k+1}}{P_k} &=\frac{\binom{n}{k+1}p^{k+1}q^{n-k-1}} {\binom{n}{k}p^kq^{n-k}}\\ &=\frac{\binom{n}{k+1}}{\binom{n}{k}}\frac{p}{q}\\ &=\frac{n-k}{k+1}\frac{p}{q}. \end{aligned}

The last equality follows from the factorial definition of the binomial coefficient: \(\binom{n}{k+1}/\binom{n}{k}=(n-k)/(k+1)\). Therefore the ratio compares the probability at \(k+1\) with the probability at \(k\):

If \(P_{k+1}/P_k>1\), then \(P_{k+1}>P_k\): the probability rises from \(k\) to \(k+1\).
If \(P_{k+1}/P_k<1\), then \(P_{k+1}<P_k\): the probability falls from \(k\) to \(k+1\).
If \(P_{k+1}/P_k=1\), then the two neighboring counts have equal probability.

Theorem: For \(X\sim\operatorname{Bin}(n,p)\) with \(0<p<1\), the adjacent-probability ratio is \(\displaystyle \frac{P(X=k+1)}{P(X=k)}=\frac{n-k}{k+1}\frac{p}{1-p}\), for \(k=0,\ldots,n-1\). Its comparison with 1 determines which of the two neighboring counts is more probable.

Locating the Most Likely Count

To find where the probabilities stop rising and start falling, compare the ratio with 1. Since \(k+1\) and \(q\) are positive, multiplying both sides of the comparison by \((k+1)q\) preserves the inequality:

\frac{n-k}{k+1}\frac{p}{q}\geq 1 \quad\Longleftrightarrow\quad (n-k)p\geq (k+1)q \quad\Longleftrightarrow\quad (n+1)p\geq k+1.

Thus, moving from \(k\) to \(k+1\) does not decrease the probability exactly when \(k+1\leq(n+1)p\). The cutoff depends on \((n+1)p\), not just on the mean \(np\). If \((n+1)p\) is not an integer, the probabilities rise to one count and then fall. If it is an integer, the ratio equals 1 at the step across the two most likely counts.

Theorem: For \(X\sim\operatorname{Bin}(n,p)\), \(0<p<1\), let \(m=(n+1)p\). If \(m\) is not an integer, the unique mode (most likely count) is \(\lfloor m\rfloor\). If \(m\) is an integer, the two modes are \(m-1\) and \(m\), with equal probabilities. Here, \(\lfloor m\rfloor\) is the greatest integer less than or equal to \(m\).

The result follows from the direction of every adjacent comparison, not from assuming that the distribution is symmetric. When \(m\) is not an integer, ratios are greater than 1 up to the step that reaches \(\lfloor m\rfloor\), and are less than 1 after it. When \(m\) is an integer, the ratio across the step from \(m-1\) to \(m\) equals 1; ratios before that step are greater than 1 and ratios after it are less than 1. That proves both the location of the maximum and the tied-mode case.

Worked Examples

Worked Example: Comparing Counts Around the Mode

A hypothetical insurer reviews 12 independent, otherwise comparable applications. Each application has probability \(p=0.25\) of requiring a specified additional review, independently of the others. Let \(X\) be the number requiring review, so \(X\sim\operatorname{Bin}(12,0.25)\), with \(n=12\), \(p=0.25\), and \(q=0.75\). Compare the probabilities of exactly 2, 3, and 4 reviews, and identify the mode.

Method. Use consecutive ratios to move from \(P(X=2)\) to \(P(X=3)\), then to \(P(X=4)\). For the first comparison, \(k=2\):

\frac{P(X=3)}{P(X=2)} =\frac{12-2}{2+1}\frac{0.25}{0.75} =\frac{10}{3}\cdot\frac{1}{3} =\frac{10}{9} \approx 1.1111.

Since \(10/9>1\), exactly 3 reviews is more likely than exactly 2. For the next comparison, \(k=3\):

\frac{P(X=4)}{P(X=3)} =\frac{12-3}{3+1}\frac{0.25}{0.75} =\frac{9}{4}\cdot\frac{1}{3} =\frac{3}{4} =0.75.

Since \(3/4<1\), exactly 4 reviews is less likely than exactly 3. To confirm the mode, calculate \((n+1)p=13(0.25)=3.25\). Its floor is 3, so the unique mode is 3. The neighboring comparisons agree: probabilities rise from 2 to 3 and fall from 3 to 4.

Check: Both ratios compare the probability in the numerator with the probability in the denominator, and their directions agree with a peak at 3. No probability rounding is needed because the requested comparison is determined exactly by the ratios. The common mistake to avoid is reversing the first ratio and then interpreting a value below 1 as evidence that 3 is less likely.

Worked Example: Detecting Two Tied Modes

In a hypothetical quality check, 9 independent components are inspected, each with probability \(p=0.30\) of having a particular feature. Let \(X\sim\operatorname{Bin}(9,0.30)\), with \(n=9\), \(p=0.30\), and \(q=0.70\). Find the most likely count or counts, and verify the tie by calculating their probabilities.

First, \((n+1)p=10(0.30)=3\), an integer. The mode rule predicts two modes, 2 and 3. The ratio across those counts, using \(k=2\), is

\frac{P(X=3)}{P(X=2)} =\frac{9-2}{2+1}\frac{0.30}{0.70} =\frac{7}{3}\cdot\frac{3}{7} =1.

So the probabilities are equal. To check that they are maxima, compare the steps on either side. For \(k=1\), \(\frac{P(X=2)}{P(X=1)}=\frac{8}{2}\frac{3}{7}=\frac{12}{7}>1\). For \(k=3\), \(\frac{P(X=4)}{P(X=3)}=\frac{6}{4}\frac{3}{7}=\frac{9}{14}<1\). Thus probabilities rise up to 2, tie from 2 to 3, and then fall.

The probability values provide a direct numerical check. For 2,

P(X=2)=\binom{9}{2}(0.30)^2(0.70)^7 =36(0.09)(0.0823543) \approx 0.2668.

For 3,

P(X=3)=\binom{9}{3}(0.30)^3(0.70)^6 =84(0.027)(0.117649) \approx 0.2668.

Probabilities are rounded to 4 decimal places; the unrounded values are equal. The common mistake is to report only 3 because \(\lfloor(n+1)p\rfloor=3\). When \((n+1)p\) is exactly an integer, \(m-1\) is also a mode.

Worked Example: Finding a Mode and Comparing Nearby Counts

A hypothetical monitoring team records whether each of 20 independent sensors triggers a specified alert during a test. Each sensor triggers with probability \(p=0.42\). Let \(X\sim\operatorname{Bin}(20,0.42)\), with \(n=20\), \(p=0.42\), and \(q=0.58\). Find the mode and compare the probabilities of 7, 8, 9, and 10 alerts.

Calculate the cutoff: \((n+1)p=21(0.42)=8.82\). Since this is not an integer, the unique mode is \(\lfloor8.82\rfloor=8\). Check the two ratios immediately around that count. For \(k=7\),

\frac{P(X=8)}{P(X=7)} =\frac{20-7}{7+1}\frac{0.42}{0.58} =\frac{13}{8}\frac{21}{29} =\frac{273}{232} \approx 1.1767.

Thus 8 alerts is more likely than 7. For \(k=8\),

\frac{P(X=9)}{P(X=8)} =\frac{20-8}{8+1}\frac{0.42}{0.58} =\frac{12}{9}\frac{21}{29} =\frac{28}{29} \approx 0.9655.

Thus 9 alerts is less likely than 8. To compare 10 with 9, use \(k=9\):

\frac{P(X=10)}{P(X=9)} =\frac{20-9}{9+1}\frac{0.42}{0.58} =\frac{11}{10}\frac{21}{29} =\frac{231}{290} \approx 0.7966.

Since this ratio is also below 1, the probabilities decrease from 8 to 9 to 10. The four counts are therefore ordered \(P(X=8)>P(X=7)\), \(P(X=8)>P(X=9)>P(X=10)\). The common mistake to avoid is rounding \(np=8.4\) to guess the mode; the adjacent-ratio result uses \((n+1)p\), which gives the correct mode here.

Common Mistakes and Why the Method Matters

The ratio method is useful because it focuses on comparisons rather than on the absolute size of the probabilities. A binomial coefficient can be large, and the powers in a binomial probability can be small; evaluating separate probabilities may introduce unnecessary arithmetic and rounding. The ratio cancels most of these terms and directly answers a neighboring-count question.

Reversing the ratio. \(P_{k+1}/P_k>1\) means the upper count is more likely. If you instead calculate \(P_k/P_{k+1}\), the comparison is reversed: a value above 1 then means the lower count is more likely.
Using \(np\) as the exact mode rule. The mean is \(np\), but the mode is located using \((n+1)p\). They may point to different integer counts.
Ignoring an exact integer cutoff. If \((n+1)p\) is an integer, there are two adjacent modes, not one. The probabilities at those counts are tied.
Applying the formula outside its support. The ratio compares \(k\) and \(k+1\) only for \(k=0,\ldots,n-1\). There is no binomial count above \(n\).
Using the result without a binomial model. The derivation assumes a fixed number of independent trials with the same success probability. If those conditions do not hold, the ratio need not have this form.

These comparisons are aligned with the SOA VEE Mathematical Statistics syllabus because they connect the binomial probability formula to shape and the location of the most likely count. The essential habit is to write down which two probabilities the ratio compares before interpreting whether it is above or below 1.

Check Your Understanding

For each problem, state the binomial parameters and use adjacent probabilities or the mode rule as requested.

A hypothetical batch contains 10 independent devices, each with probability \(0.20\) of failing a screening test. Let \(X\) be the number that fail. Find \(P(X=4)/P(X=3)\) and state which count is more likely. (Answer: \(7/16=0.4375\); 3 failures is more likely.)
A hypothetical survey has 15 independent responses, each with probability \(0.40\) of a specified answer. Let \(X\) count those answers. Find the mode and verify the ratios immediately below and above it. (Answer: The unique mode is 6; \(P(X=6)/P(X=5)=10/9\) and \(P(X=7)/P(X=6)=6/7\).)
A hypothetical inspection checks 11 independent items, each with probability \(0.25\) of having a defect. Let \(X\) count defects. Identify all modes and show the ratio between them. (Answer: Modes 2 and 3; \(P(X=3)/P(X=2)=1\).)
A hypothetical system runs 18 independent tests, each with probability \(0.35\) of a positive result. Let \(X\) count positive results. Find the mode and compare the probabilities of 6 and 7 positive results. (Answer: The unique mode is 6; \(P(X=7)/P(X=6)=12/13<1\), so 6 is more likely.)
For \(X\sim\operatorname{Bin}(14,p)\), find the value of \(p\) that makes counts 5 and 6 equally likely. Then determine which of the two is more likely when \(p=0.30\). (Answer: The tie occurs at \(p=0.40\); when \(p=0.30\), \(P(X=6)/P(X=5)=9/14<1\), so 5 is more likely.)

What You'll Learn