Finding the Mode of a Binomial Distribution

Why Adjacent Probabilities Locate the Mode

For a discrete distribution, the mode is the value where probability mass is largest. A binomial probability mass function can involve large binomial coefficients, so comparing every support value directly is often inconvenient. Comparing each probability with its neighbor gives a simpler route: the sign of the change tells us whether the probability sequence is rising or falling.

For \(X\sim\operatorname{Bin}(n,p)\), the support is the finite set \(0,1,\ldots,n\). We will compare the mass at \(k+1\) with the mass at \(k\). The resulting ratio also reveals exactly when two adjacent values tie for the maximum.

Definition: A mode of a discrete random variable \(X\) is any support value \(k\) at which \(P(X=k)\) attains its maximum over the support. A distribution may have more than one mode.

The Adjacent-Probability Ratio

For \(n\geq 1\) and \(0<p<1\), the binomial probability mass function is

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

All these probabilities are positive, so ratios between adjacent probabilities are defined. For \(k=0,\ldots,n-1\), use the binomial coefficient identity \(\binom{n}{k+1}/\binom{n}{k}=(n-k)/(k+1)\):

\frac{P(X=k+1)}{P(X=k)} = \frac{n-k}{k+1}\frac{p}{1-p}.

This ratio is greater than 1 when the mass increases from \(k\) to \(k+1\), equal to 1 when the two masses tie, and less than 1 when the mass decreases. Since \(k+1\) and \(1-p\) are positive, comparison with 1 is equivalent to

\begin{aligned} \frac{n-k}{k+1}\frac{p}{1-p} &\geq 1\\ (n-k)p &\geq (k+1)(1-p)\\ (n+1)p &\geq k+1. \end{aligned}

Thus the direction of each adjacent comparison is controlled by the single number \(a=(n+1)p\). The transition from \(k\) to \(k+1\) is increasing when \(k+1<a\), is a tie when \(k+1=a\), and is decreasing when \(k+1>a\).

Theorem: Let \(X\sim\operatorname{Bin}(n,p)\), where \(n\geq1\). If \(0<p<1\) and \((n+1)p\) is not an integer, the unique mode is \(\lfloor(n+1)p\rfloor\). If \(0<p<1\) and \((n+1)p=r\) is an integer, the two modes are \(r-1\) and \(r\). At \(p=0\), the unique mode is 0; at \(p=1\), the unique mode is \(n\).

To prove the interior cases, write \(q_k=P(X=k)\) and \(a=(n+1)p\). Because \(0<p<1\), we have \(0<a<n+1\). If \(a\) is not an integer, set \(m=\lfloor a\rfloor\). For each transition from \(k\) to \(k+1\) with \(k+1\leq m\), we have \(k+1<a\), so \(q_{k+1}>q_k\). For each transition with \(k+1\geq m+1\), we have \(k+1>a\), so \(q_{k+1}<q_k\). Consequently the sequence rises through \(q_m\) and then falls, making \(m\) the unique mode. This includes \(m=0\), when every transition decreases.

If \(a=r\) is an integer, then \(1\leq r\leq n\). Transitions with \(k+1<r\) increase the mass, the transition with \(k+1=r\) ties, and transitions with \(k+1>r\) decrease it. Hence \(q_{r-1}=q_r\), and both are larger than all other masses. At the parameter boundaries, the binomial distribution is degenerate: \(P(X=0)=1\) when \(p=0\), and \(P(X=n)=1\) when \(p=1\). These give the stated modes directly. If \(n=0\), the support consists only of 0, which is its mode for every \(p\). \(\square\)

Worked Examples

Worked Example: A Unique Interior Mode

Let \(X\sim\operatorname{Bin}(10,0.3)\). Find the mode and check the probabilities on either side of it.

First calculate the adjusted parameter:

$$ (n+1)p=11(0.3)=3.3. $$

This is not an integer, so the theorem gives the unique mode \(k=\lfloor3.3\rfloor=3\). The adjacent ratios around 3 confirm the change in direction:

\frac{P(X=3)}{P(X=2)} =\frac{10-2}{3}\frac{0.3}{0.7} =\frac{8}{3}\frac{3}{7} =\frac{8}{7}>1, \qquad \frac{P(X=4)}{P(X=3)} =\frac{10-3}{4}\frac{0.3}{0.7} =\frac{7}{4}\frac{3}{7} =\frac{3}{4}<1.

For a direct numerical check, \(P(X=3)=\binom{10}{3}(0.3)^3(0.7)^7\approx0.26683\), while \(P(X=2)\approx0.23347\) and \(P(X=4)\approx0.20012\). The ratio argument establishes that probabilities increase up to 3 and decrease after 3; the numerical values agree with it.

Worked Example: Two Adjacent Modes

Let \(X\sim\operatorname{Bin}(9,0.4)\). Determine whether there is one mode or a tie.

Here

$$ (n+1)p=10(0.4)=4. $$

Because this value is an integer, the modes are \(4-1=3\) and \(4\). Check the tie and the changes on either side:

\frac{P(X=3)}{P(X=2)} =\frac{7}{3}\frac{0.4}{0.6} =\frac{7}{3}\frac{2}{3} =\frac{14}{9}>1, \qquad \frac{P(X=4)}{P(X=3)} =\frac{6}{4}\frac{2}{3}=1, \qquad \frac{P(X=5)}{P(X=4)} =\frac{5}{5}\frac{2}{3} =\frac{2}{3}<1.

The equal probabilities can also be checked directly: \(P(X=3)=\binom{9}{3}(0.4)^3(0.6)^6=84(0.064)(0.046656)=0.250822656\), and \(P(X=4)=\binom{9}{4}(0.4)^4(0.6)^5=126(0.0256)(0.07776)=0.250822656\). The sequence increases to this tied pair and decreases after it, so neither value is merely a local tie: both are global modes.

Worked Example: A Mode at the Lower End of the Support

Let \(X\sim\operatorname{Bin}(4,0.1)\). The adjusted parameter is

$$ (n+1)p=5(0.1)=0.5, $$

so the mode is \(\lfloor0.5\rfloor=0\). In fact, the first adjacent ratio is

\frac{P(X=1)}{P(X=0)} =\frac{4}{1}\frac{0.1}{0.9} =\frac{4}{9}<1.

Every later transition also decreases, because each integer \(k+1\) is greater than \(0.5\). As a numerical check, \(P(X=0)=(0.9)^4=0.6561\) and \(P(X=1)=4(0.1)(0.9)^3=0.2916\). Thus the probability is already falling at the first step, in agreement with the mode at 0.

Worked Example: Boundary Parameter Values

Let \(X\sim\operatorname{Bin}(6,p)\) and consider \(p=0\) and \(p=1\). At \(p=0\), all six trials are failures with probability 1, so \(P(X=0)=1\) and \(P(X=k)=0\) for \(k=1,\ldots,6\). The unique mode is 0. At \(p=1\), all six trials are successes with probability 1, so \(P(X=6)=1\) and \(P(X=k)=0\) for \(k=0,\ldots,5\). The unique mode is 6.

The interior adjacent-ratio formula is not used for these cases. It assumes \(0<p<1\), which makes every support probability positive and validates the cancellations in deriving the ratio. At the endpoints some probabilities in a ratio's denominator are zero, so those ratios may be undefined. Directly identifying the degenerate distribution avoids that problem.

Common Mistakes and Why the Mode Matters

A frequent error is to round \(np\) to the nearest integer and call the result the mode. The relevant quantity is \((n+1)p\), not \(np\). For example, when \(n=9\) and \(p=0.4\), \(np=3.6\), but the distribution has two modes, 3 and 4, because \((n+1)p=4\) is an integer. The adjacent-ratio argument gives both the correct location and the tie condition.

Do not assume there is always only one mode. When \((n+1)p\) is an integer strictly between 0 and \(n+1\), the probabilities at the two adjacent integers are equal maxima.
Do not apply the ratio formula at \(p=0\) or \(p=1\). At those endpoints, handle the distribution directly; zero probabilities can make adjacent ratios undefined.
Do not confuse a mode with a mean. The mode identifies a most probable support value. It need not equal the mean \(np\), and it does not summarize the entire probability distribution.

Finding the mode is useful when a question asks for the most probable count of successes or when checking the shape of a binomial probability sequence. The essential calculation is short: find \((n+1)p\), then decide whether it lies between integers or exactly on one. The separate boundary check ensures that the argument is being used only where its probability ratios are valid.

Key takeaway: For \(0<p<1\), adjacent binomial masses rise while the next support value is below \((n+1)p\), tie when it equals \((n+1)p\), and fall when it is above. This gives one mode at \(\lfloor(n+1)p\rfloor\), except at an integer value, where there are two adjacent modes.

Check Your Understanding

Use the adjacent-ratio criterion where appropriate, and explain how you decide whether a tie occurs.

For \(X\sim\operatorname{Bin}(12,0.2)\), find the mode or modes.
For \(X\sim\operatorname{Bin}(7,0.5)\), determine whether the mode is unique. State the value of \((n+1)p\).
Why is the probability ratio \(P(X=k+1)/P(X=k)\) guaranteed to be defined for every adjacent pair when \(0<p<1\), but not necessarily at the boundary parameter values?
For \(X\sim\operatorname{Bin}(5,0)\), identify the mode and explain why the interior ratio formula is unnecessary.
State the distinction between the mode and the mean of a binomial distribution.