Explanation
{\textbf{Step - 1: Finding n, p and q}}
{\text{We know that, mean = np and variance = np(1 - p)}}
\therefore {\text{ np = 2 and np(1 - p) = 1}}
{\text{Dividing, }}\dfrac{{{\text{np(1 - p)}}}}{{{\text{np}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}
\Rightarrow {\text{ 1 - p = }}\dfrac{{\text{1}}}{{\text{2}}}
\Rightarrow {\text{ p = }}\dfrac{{\text{1}}}{{\text{2}}}
\because {\text{ q = 1 - p}}
\Rightarrow {\text{ q = 1 - }}\dfrac{{\text{1}}}{{\text{2}}}
\Rightarrow {\text{ q = }}\dfrac{{\text{1}}}{{\text{2}}}
{\text{Substituting value of p in np = 2}}
\Rightarrow {\text{ n }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = 2}}
\Rightarrow {\text{ n = 4}}
{\textbf{Step - 2: Calculating probability}}
{\text{Probability of value greater than 1, P(X > 1) = 1 - P(X = 0) - P(X = 1)}}
\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{1}}}{{{{\text{2}}^{\text{4}}}}}{\text{ - }}{{\text{ }}^{\text{4}}}{{\text{C}}_{\text{1}}}\dfrac{{\text{1}}}{{{{\text{2}}^{\text{4}}}}}
\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{1}}}{{{\text{16}}}}{\text{ - }}\dfrac{{\text{4}}}{{{\text{16}}}}
\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{5}}}{{{\text{16}}}}
\Rightarrow {\text{ P(X > 1) = }}\dfrac{{{\text{11}}}}{{{\text{16}}}}{\text{ }}
\textbf{Hence option A is correct}
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