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X is a random variable that follows a binomial distribution, that is\(X \sim Bin\left( {n,p} \right)\). Therefore, the pdf of X is:

\(f\left( x \right) = {}^n{C_r}{p^r}{\left( {1 - p} \right)^{n - r}}\;for\;x = 0,1, \cdots n\;and\;0 \le p \le 1\)

Let us denote

\(\begin{align}{a_k} &= P\left( {X = k} \right)\\ &= {}^n{C_k}{p^k}{\left( {1 - p} \right)^{n - k}}\\{a_{k + 1}} &= P\left( {X = k + 1} \right)\\ &= {}^n{C_{k + 1}}{p^{k + 1}}{\left( {1 - p} \right)^{n - k - 1}}\end{align}\)

Let us calculate the ratio and reduce the form to a simpler form.

\(\begin{align}\frac{{{a_{k + 1}}}}{{{a_k}}} &= \frac{{{}^n{C_{k + 1}}}}{{{}^n{C_k}}}\frac{{{p^{k + 1}}{{\left( {1 - p} \right)}^{n - k - 1}}}}{{{p^k}{{\left( {1 - p} \right)}^{n - k}}}}\\ &= \frac{{n - k}}{{k + 1}} \cdot \frac{p}{{1 - p}}\end{align}\)

\(\begin{align}When\;\frac{{{a_{k + 1}}}}{{{a_k}}} < 1\\ \Rightarrow {a_{k + 1}} < {a_k}\\ \Rightarrow k > \left( {n + 1} \right)p - 1\end{align}\)

\(\begin{align}When\;\frac{{{a_{k + 1}}}}{{{a_k}}} = 1\\ \Rightarrow {a_{k + 1}} = {a_k}\\ \Rightarrow k = \left( {n + 1} \right)p - 1\end{align}\)

\(\begin{align}When\;\frac{{{a_{k + 1}}}}{{{a_k}}} < 1\\ \Rightarrow {a_{k + 1}} > {a_k}\\ \Rightarrow k < \left( {n + 1} \right)p - 1\end{align}\)

With the following equations, we can infer the following:

The equality at

\({a_{k + 1}} = {a_k}\)

Shows that

\(k = \left( {n + 1} \right)p - 1\).

Two cases arise here, k could be a non-integer number or an integer.

Case 1: If\({\bf{k = }}\left( {{\bf{n + 1}}} \right){\bf{p - 1}}\)is not an integer:

This is the case of a unimodal binomial distribution.

The mode, in this case, is, \(k = \left( {n + 1} \right)p - 1\)

Case 2: If\({\bf{k = }}\left( {{\bf{n + 1}}} \right){\bf{p - 1}}\)it is an integer:

This is the case of a bimodal binomial distribution.

The mode, in this case, is, \(k = \left( {n + 1} \right)p - 1\) and \(k = np + p\).

Let us plug in certain values to check the areas where equality doesn鈥檛 hold.

At \(p = 0,k = \left( {n + 1} \right)p - 1\) becomes -1. This is an integer value.

This is negative; in this case, the mode is single-mode, k=0. Otherwise, the rule of k being non鈥搃nteger is given the value is non-negative holds good and yields single mode.

At \(p = 1,k = \left( {n + 1} \right)p - 1\) becomes an integer value; hence it is not a bimodal case. Here, the mode occurs \(k = n\) after solving the equation.

Therefore, when

1) If k is a non-integer value.

\(k = \left( {n + 1} \right)p - 1\) and it is unimodal.

2) If k is an integer value

\(k = \left( {n + 1} \right)p - 1\)and \(k = np + p\) are the two modes, it is bimodal.

3) At p=0, mode is 0

4) At p=1, the mode is n.