Question

Suppose $f(x)=\frac{k}{2^x}$ is a probability distribution of a random variable $X$ that can take on the values $x=0,1,2,3,4$ . 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
$=1$
$\frac{k}{2^0}+\frac{k}{2^1}+\frac{k}{2^2}+\frac{k}{2^3}+\frac{k}{2^4}=1$
$\Rightarrow$ $\frac{1}{k}=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}$
$\Rightarrow$ $\frac{1}{k}=\frac{1.\left(1-{\left(\frac{1}{2}\right)}^5\right)}{1-\frac{1}{2}}=\frac{1-\frac{1}{32}}{\frac{1}{2}}=\frac{31}{16}$
$\Rightarrow$ $k=\frac{16}{31}$