Question

If $f\left(x\right)=\frac{k}{2^x}$ 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

The correct option is D

None of these

Explanation for the correct option.

Find the value of $k$.

As the probability distribution $f\left(x\right)=\frac{k}{2^x}$ takes the values $x=0,1,2,3,4$. So the sum of probability distribution for each value must be $1$ that is

$f\left(0\right)+f\left(1\right)+f\left(2\right)+f\left(3\right)+f\left(4\right)=1$.

Now, using the function $f\left(x\right)=\frac{k}{2^x}$, the equation can be formed and solved for $k$ as:

$$\begin{gathered} \frac{k}{2^0}+\frac{k}{2^1}+\frac{k}{2^2}+\frac{k}{2^3}+\frac{k}{2^4}=1 \\ \Rightarrow \frac{k}{1}+\frac{k}{2}+\frac{k}{4}+\frac{k}{8}+\frac{k}{16}=1 \\ \Rightarrow \frac{16k+8k+4k+2k+k}{16}=1 \\ \Rightarrow \frac{31k}{16}=1 \\ \Rightarrow k=\frac{16}{31} \end{gathered}$$

Hence, the correct option is D.