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A discrete random variable is one that can only take on a finite number of different values.

\(\begin{array}{c}{\rm{E(X) = np;}}\\{\rm{V(X) = np(1 - p)}}\\{\rm{ = npq}}\\^{\rm{\sigma }}{\rm{x = }}\sqrt {{\rm{npq}}} \end{array}\)

As a result, from

\({\rm{V(X) = np(1 - p)}}\)

If p\({\rm{ = 0}}\)or \({\rm{1}}\), we can see that the value of V(X)will also be zero.

When p\({\rm{ = 0}}\), every trial fails, and the variability in X is \({\rm{0}}\) in the first scenario.

When p \({\rm{ = 1}}\), the converse occurs: every trial succeeds, implying that X is not variable.

Therefore, the value is: \(\begin{array}{c}{\rm{p = 0}}\\{\rm{p = 1}}\end{array}\).

As, the value of V(X) is a function of p, the following statement is true.

\(\begin{array}{c}\frac{{\rm{d}}}{{{\rm{dp}}}}{\rm{V(X) = }}\frac{{\rm{d}}}{{{\rm{dp}}}}{\rm{(np(1 - p))}}\\{\rm{ = }}\frac{{\rm{d}}}{{{\rm{dp}}}}\left( {{\rm{np - n}}{{\rm{p}}^{\rm{2}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{n}}\frac{{\rm{d}}}{{{\rm{dp}}}}{\rm{p - n}}\frac{{\rm{d}}}{{{\rm{dp}}}}{{\rm{p}}^{\rm{2}}}\\{\rm{ = n \times 1 - n \times 2 \times p}}\\{\rm{ = n(1 - 2p); }}\end{array}\)

(\({\rm{1}}\)): derivative qualities.

Since,

\(\frac{{\rm{d}}}{{{\rm{dp}}}}{\rm{n(1 - 2p) = - 2n < 0}}\)

The value p is:

\({\rm{n(1 - 2p) = 0}}\)

The highest possible value is:

\({\rm{n(1 - 2p) = 0}}\)

It is for:

\({\rm{1 - 2p = 0}}\)

Or is equally:

\({\rm{p = }}\frac{{\rm{1}}}{{\rm{2}}}\)

Therefore, the maximum value is: \({\rm{p = }}\frac{{\rm{1}}}{{\rm{2}}}\).