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Module 10 · L12 of 20 ~40 min ⚡ +90 XP available

The Mode of a Binomial Distribution

A call centre with a 35% answer rate fields 20 incoming calls in an hour. What number of answered calls is the single most likely outcome? You could plug every value of $k$ from 0 to 20 into the binomial PMF and find the peak, or you could use a one-line formula: $\text{mode} = \lfloor (n+1)p \rfloor$. This lesson explains why that formula works and how to spot the rare case when the distribution has two equally tall peaks.

Today's hook, $X \sim B(20, 0.35)$. Before reading on, guess the single most likely value of $X$, is it 6, 7 or somewhere else? Now consider $X \sim B(11, 0.5)$. Could the mode be two adjacent values at once? Predict, then verify against the formula in card 05.

0/5QUESTS

Recall, your gut answer first

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A biased coin lands heads with probability $0.6$ and is tossed 9 times. Without any formulawhat single number of heads do you expect to be the most likely outcome? Write your reasoning, including what the mean would predict.

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The two moves for finding the mode

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The mode is the value of $k$ that maximises $P(X = k)$. Two habits get you there fast: compute $(n+1)p$ first, then apply the floor and check the integer test. If $(n+1)p$ is itself an integer, the distribution has two adjacent modes otherwise just one.

The mode formula: $\text{mode} = \lfloor (n+1)p \rfloor$. If $(n+1)p$ is an integer, both $(n+1)p$ and $(n+1)p - 1$ are modes (double mode). Otherwise there is a single mode at $\lfloor (n+1)p \rfloor$.

Derivation: $\dfrac{P(X=k)}{P(X=k-1)} = \dfrac{(n-k+1)p}{k(1-p)} \geq 1 \iff k \leq (n+1)p$.

$\text{mode} = \lfloor (n+1)p \rfloor$

Compute $(n+1)p$, not $np$

The mode formula uses $(n+1)p$, that extra $+1$ comes from the derivation. Using $np$ (the mean) instead is the most common error in this lesson.

Floor, not round

$\lfloor x \rfloor$ is the floor function, the largest integer not exceeding $x$. So $\lfloor 3.9 \rfloor = 3$, not 4. Rounding gives wrong answers when $(n+1)p$ is between an integer and the next half.

Integer = two modes

If $(n+1)p$ is exactly an integer $m$, then $P(X = m) = P(X = m - 1)$, two equally likely values. Always check whether $(n+1)p$ comes out as a whole number.

What you'll master

Know

Key facts

Mode of $X \sim B(n, p)$ is $\lfloor (n+1)p \rfloor$
If $(n+1)p$ is an integer, the modes are $(n+1)p$ and $(n+1)p - 1$
The ratio $\dfrac{P(X=k)}{P(X=k-1)} = \dfrac{(n-k+1)p}{k(1-p)}$ rises above 1 iff $k \leq (n+1)p$

Understand

Concepts

Why the mode is close to but not exactly the mean $np$
Why $(n+1)p$ being an integer creates a double mode
How the ratio-of-consecutive-probabilities argument identifies the peak

Can do

Skills

Find the mode of any $B(n, p)$ in seconds using $\lfloor (n+1)p \rfloor$
Identify when a binomial distribution has two modes
Verify a mode by computing $P(X = k)$ and $P(X = k \pm 1)$ to compare

Key terms

ModeThe value of $k$ that maximises $P(X = k)$. For a histogram, the bar with the greatest height. May be a single value or two adjacent values.

Floor function $\lfloor x \rfloor$The largest integer that is less than or equal to $x$. Example: $\lfloor 4.7 \rfloor = 4$, $\lfloor 6 \rfloor = 6$, $\lfloor -2.3 \rfloor = -3$.

Mode formulaFor $X \sim B(n, p)$: mode is $\lfloor (n+1)p \rfloor$. If $(n+1)p$ is an integer, both $(n+1)p$ and $(n+1)p - 1$ are modes.

UnimodalA distribution with exactly one peak (single mode). The typical case for binomial, happens whenever $(n+1)p$ is not an integer.

Bimodal (double mode)In binomial, "double mode" specifically means two adjacent values of $k$ are equally most likely. Occurs precisely when $(n+1)p$ is an integer.

ME12-5NESA outcome: applies appropriate statistical processes to present, analyse and interpret data, including from binomial distributions.

The mode formula and where it comes from

core concept

Why does $\lfloor (n+1)p \rfloor$ give the mode? Form the ratio of consecutive probabilities:

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

The probabilities are rising ($P(X = k) > P(X = k-1)$) iff this ratio exceeds 1:

$$\frac{(n-k+1)p}{k(1-p)} > 1 \iff (n-k+1)p > k(1-p) \iff (n+1)p > k.$$

So $P(X = k)$ increases while $k < (n+1)p$ and decreases once $k > (n+1)p$. The peak therefore sits at $k = \lfloor (n+1)p \rfloor$. If $(n+1)p$ is an integer, equality $k = (n+1)p$ gives ratio $= 1$, meaning $P(X = (n+1)p) = P(X = (n+1)p - 1)$, a double mode.

Worked through the hook:

$X \sim B(20, 0.35)$: $(n+1)p = 21 \times 0.35 = 7.35$. Mode $= \lfloor 7.35 \rfloor = 7$. Single mode at 7.
$X \sim B(11, 0.5)$: $(n+1)p = 12 \times 0.5 = 6$, an integer. Double mode at $k = 5$ and $k = 6$.

Mode versus mean. The mean is $np$ but the mode is $\lfloor (n+1)p \rfloor$. These usually agree to within 1, but they differ when $(n+1)p$ lands on an integer slightly above or below $np$. For $B(20, 0.35)$, mean $= 7$ but mode $= 7$ too. For $B(11, 0.5)$, mean $= 5.5$ but the modes are both 5 and 6.

Why does $\lfloor (n+1)p \rfloor$ give the mode? Form the ratio of consecutive probabilities:

Pause, copy the mode formula $\text{mode}=\lfloor(n+1)p\rfloor$ and the ratio test $P(X=k)>P(X=k-1)\iff k<(n+1)p$ into your book.

Quick check: $X \sim B(15, 0.4)$. Which of the following is the mode?

Spotting a single versus double mode

core concept

We just saw that the mode is $\lfloor(n+1)p\rfloor$, derived from the ratio $P(X=k)/P(X=k-1)=\frac{(n-k+1)p}{k(1-p)}$ which exceeds 1 iff $k<(n+1)p$. That raises a question: what happens when $(n+1)p$ is exactly an integer, does the mode formula give two modes or one? This card answers it → if $(n+1)p$ is an integer, both $(n+1)p$ and $(n+1)p-1$ are modes (equal probability); otherwise the single mode is $\lfloor(n+1)p\rfloor$.

The single-vs-double decision is entirely controlled by whether $(n+1)p$ is an integer.

$(n+1)p$ not an integer. Single mode at $\lfloor (n+1)p \rfloor$.
$(n+1)p$ an integer. Two equally likely adjacent modes: $(n+1)p$ and $(n+1)p - 1$.

Worked examples, single mode:

$B(7, 0.3)$: $(n+1)p = 8 \times 0.3 = 2.4$. Single mode at $\lfloor 2.4 \rfloor = 2$.
$B(50, 0.42)$: $(n+1)p = 51 \times 0.42 = 21.42$. Single mode at $21$.
$B(100, 0.27)$: $(n+1)p = 101 \times 0.27 = 27.27$. Single mode at $27$.

Worked examples, double mode:

$B(9, 0.5)$: $(n+1)p = 10 \times 0.5 = 5$. Two modes at $k = 4$ and $k = 5$.
$B(11, 0.25)$: $(n+1)p = 12 \times 0.25 = 3$. Two modes at $k = 2$ and $k = 3$.
$B(19, 0.1)$: $(n+1)p = 20 \times 0.1 = 2$. Two modes at $k = 1$ and $k = 2$.

$$\text{mode}(B(n,p)) = \begin{cases} \lfloor (n+1)p \rfloor & \text{if } (n+1)p \notin \mathbb{Z}, \\ (n+1)p \text{ and } (n+1)p - 1 & \text{if } (n+1)p \in \mathbb{Z}. \end{cases}$$

Common mistake. Reporting "two modes at $\lfloor (n+1)p \rfloor$ and $\lfloor (n+1)p \rfloor + 1$" whenever $(n+1)p$ is not an integer. The correct rule is the opposite, a non-integer $(n+1)p$ gives exactly one mode. Two modes only occur for the special integer case.

The single-vs-double decision is entirely controlled by whether $(n+1)p$ is an integer.

Pause, copy the single-vs-double mode rule: $(n+1)p$ not an integer gives one mode; $(n+1)p$ an integer gives two adjacent equal-probability modes at $(n+1)p$ and $(n+1)p-1$ into your book.

Did you get this? True or false: $X \sim B(13, 0.5)$ has a single mode because $(n+1)p = 14 \times 0.5 = 7$ is an integer.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · APPLY THE MODE FORMULA

A call centre receives 20 calls in an hour and answers each one with probability $0.35$. Let $X$ count the answered calls. Find the most likely value of $X$.

Identify: $X \sim B(20, 0.35)$. Apply the formula: $(n+1)p = 21 \times 0.35 = 7.35$.

First step is always to compute $(n+1)p$, not $np$.

PROBLEM 2 · IDENTIFY A DOUBLE MODE

Determine whether $X \sim B(11, 0.5)$ has one or two modes and state the values.

$(n+1)p = 12 \times 0.5 = 6$.

Always isolate the value of $(n+1)p$ before deciding the case.

PROBLEM 3 · MODE FROM DATA

A multiple-choice exam has 30 questions, each with a $0.2$ chance of being guessed correctly. Find the most likely number of correct guesses. Then determine which value of $p$ (with $n = 30$) would produce a double mode at exactly $k = 6$.

Part 1: $X \sim B(30, 0.2)$. $(n+1)p = 31 \times 0.2 = 6.2$. Not an integer, so single mode at $\lfloor 6.2 \rfloor = 6$.

Six is the most likely number of correct guesses.

Fill the gap: For $X \sim B(40, 0.3)$, $(n+1)p = 41 \times 0.3 = 12.3$. Since this is not an integer, the distribution has a single mode at $k = $.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using $np$ instead of $(n+1)p$

Students often write "mode = $\lfloor np \rfloor$", which is the mean rounded down. The correct formula is $\lfloor (n+1)p \rfloor$. They agree most of the time but disagree at exactly the cases where the distinction matters, for instance $B(9, 0.5)$ has $np = 4.5$ (floor 4) but $(n+1)p = 5$ (floor 5), giving modes at 4 and 5, not just 4.

Trap 02

Reporting two modes when $(n+1)p$ is not an integer

Some students always report "two modes" because $\lfloor (n+1)p \rfloor$ and $\lfloor (n+1)p \rfloor + 1$ both look like reasonable candidates. The correct rule is the opposite: only when $(n+1)p$ is an integer are there two modes. For $B(20, 0.35)$, $(n+1)p = 7.35$, single mode at 7 only.

Trap 03

Mistaking the mean for the mode

The mean ($np$) and mode ($\lfloor (n+1)p \rfloor$) are typically close but not identical. They differ by up to 1 in either direction. For $B(7, 0.5)$, the mean is $3.5$ (not an integer outcome), while the modes are at $3$ and $4$ (the double-mode case). Always state the mode using the floor formula, not the mean.

Did you get this? True or false: for $X \sim B(7, 0.5)$, the mode is the floor of the mean, namely $\lfloor 3.5 \rfloor = 3$, with no other mode.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the mode of $X \sim B(25, 0.3)$. Show the formula and state whether the mode is single or double.

Determine the mode(s) of $X \sim B(15, 0.4)$.

Show that $X \sim B(19, 0.5)$ has two modes. State the values and verify by computing the ratio $P(X=10)/P(X=9)$.

Find $p$ such that $X \sim B(24, p)$ has a double mode at $k = 8$ and $k = 9$.

A surgeon performs 16 operations with a 75% success rate. Find the most likely number of successes and the probability of that value.

Odd one out: Three of these statements about the mode of a binomial distribution are correct. Which one is NOT?

Revisit your thinking

Earlier you guessed the mode for a biased coin ($p = 0.6$, $n = 9$).

Apply the formula: $(n+1)p = 10 \times 0.6 = 6$. That's an integer, so the distribution has two modes at $k = 5$ and $k = 6$. Your gut answer was probably 5 or 6, and both are correct simultaneously. The mean $np = 5.4$ falls between them, which is why neither is "the" expected value alone.

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Multiple choice

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Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Find the mode of $X \sim B(18, 0.4)$. State whether it is a single or double mode. (2 marks)

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ApplyBand 43 marks

Q2. $X \sim B(15, 0.5)$. Show that the distribution has two modes and identify them. Verify by computing the relevant probabilities. (3 marks)

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AnalyseBand 53 marks

Q3. Starting from the binomial PMF, show that the ratio $\dfrac{P(X = k)}{P(X = k - 1)} = \dfrac{(n - k + 1)p}{k(1 - p)}$. Use this ratio to deduce that the mode of $X \sim B(n, p)$ is $\lfloor (n+1)p \rfloor$ when $(n+1)p$ is not an integer. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $(n+1)p = 26 \times 0.3 = 7.8$. Not integer, single mode at $\lfloor 7.8 \rfloor = 7$.

2. $(n+1)p = 16 \times 0.4 = 6.4$. Not integer, single mode at $\lfloor 6.4 \rfloor = 6$.

3. $(n+1)p = 20 \times 0.5 = 10$ (integer), so two modes at 9 and 10. Ratio $P(X=10)/P(X=9) = \dfrac{(19-10+1)(0.5)}{10(0.5)} = \dfrac{10 \times 0.5}{10 \times 0.5} = 1$, confirming equal probabilities.

4. For a double mode at 8 and 9, set $(n+1)p = 9$: $25p = 9 \Rightarrow p = \dfrac{9}{25} = 0.36$.

5. $X \sim B(16, 0.75)$. $(n+1)p = 17 \times 0.75 = 12.75$, single mode at 12. $P(X = 12) = \binom{16}{12}(0.75)^{12}(0.25)^4 = 1820 \times (0.75)^{12} \times (0.25)^4 \approx 0.2252$.

Q1 (2 marks): $(n+1)p = 19 \times 0.4 = 7.6$ [1]. Not integer, single mode at $\lfloor 7.6 \rfloor = 7$ [1].

Q2 (3 marks): $(n+1)p = 16 \times 0.5 = 8$ (integer) [1]. Two modes at $k = 7$ and $k = 8$ [1]. $P(X = 7) = \binom{15}{7}(0.5)^{15}$ and $P(X = 8) = \binom{15}{8}(0.5)^{15}$; since $\binom{15}{7} = \binom{15}{8} = 6435$, both equal $\dfrac{6435}{32768} \approx 0.1964$ [1].

Q3 (3 marks): $\dfrac{P(X=k)}{P(X=k-1)} = \dfrac{\binom{n}{k}p^k(1-p)^{n-k}}{\binom{n}{k-1}p^{k-1}(1-p)^{n-k+1}}$ [1]. Simplify $\binom{n}{k}/\binom{n}{k-1} = (n-k+1)/k$ and powers give $p/(1-p)$, so ratio $= \dfrac{(n-k+1)p}{k(1-p)}$ [1]. Ratio $> 1 \iff (n-k+1)p > k(1-p) \iff (n+1)p > k$. So probabilities rise while $k < (n+1)p$ and fall once $k > (n+1)p$; the peak is at $k = \lfloor (n+1)p \rfloor$ [1].

Boss battle · The Mode Master

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Five timed questions on locating the mode of a binomial distribution. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering binomial mode questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 · Checkpoint 3