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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