Question

Suppose $ f(x)=\frac{k}{{{2}^{x}}} $ is a probability distribution of a random variable $X$ that can take on the values $ x=0,1,\text{ }2,\text{ }3,\text{ }4 $ . Then, $k$ is equal to

• A. $ \frac{16}{15} $
• B. $ \frac{15}{16} $
• C. $ \frac{31}{16} $
• D. None of these

Answer 1

Answer: (D)

Given, $ f(x)=\frac{k}{2}, $
Where $ x=0,1,2,3,4, $
We know that, Sum of probability distribution
$ =1 $
$ \frac{k}{{{2}^{0}}}+\frac{k}{{{2}^{1}}}+\frac{k}{{{2}^{2}}}+\frac{k}{{{2}^{3}}}+\frac{k}{{{2}^{4}}}=1 $
$ \Rightarrow $ $ \frac{1}{k}=1+\frac{1}{2}+\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{3}}}+\frac{1}{{{2}^{4}}} $
$ \Rightarrow $ $ \frac{1}{k}=\frac{1.\left( 1-{{\left( \frac{1}{2} \right)}^{5}} \right)}{1-\frac{1}{2}}=\frac{1-\frac{1}{32}}{\frac{1}{2}}=\frac{31}{16} $
$ \Rightarrow $ $ k=\frac{16}{31} $