
If $ f\left( x \right)=\dfrac{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. $ \dfrac{16}{15} $
B. $ \dfrac{15}{16} $
C. $ \dfrac{31}{16} $
D. None of these
Answer
507.3k+ views
Hint: We first try to use the condition for probability distribution that the sum of the values is 1. We use the values of $ x=0,1,2,3,4 $ to find the G.P. series. We use the sum form of $ {{S}_{n}}={{t}_{1}}\dfrac{1-{{r}^{n}}}{1-r} $ to find the value of the variable $ k $ .
Complete step-by-step answer:
$ f\left( x \right)=\dfrac{k}{{{2}^{x}}} $ is a probability distribution of a random variable X that can take on the values $ x=0,1,2,3,4 $ .
We know that the sum of the probabilities will be equal to 1.
We put the values $ x=0,1,2,3,4 $ in $ f\left( x \right)=\dfrac{k}{{{2}^{x}}} $ .
So, $ \sum\limits_{x=0}^{4}{f\left( x \right)}=\sum\limits_{x=0}^{4}{\dfrac{k}{{{2}^{x}}}}=1 $ . We have a G.P. series.
The first term be $ {{t}_{1}}=\dfrac{k}{{{2}^{0}}}=k $ and the common ratio be $ r $ where $ r=\dfrac{1}{2} $ .
The value of $ \left| r \right|<1 $ for which the sum of the first n terms of an G.P. is $ {{S}_{n}}={{t}_{1}}\dfrac{1-{{r}^{n}}}{1-r} $ .
Therefore, $ \sum\limits_{x=0}^{4}{\dfrac{k}{{{2}^{x}}}}=k\dfrac{1-{{\left( \dfrac{1}{2} \right)}^{5}}}{1-\left( \dfrac{1}{2} \right)}=1 $ . Simplifying we get
$ \begin{align}
& k\dfrac{1-{{\left( \dfrac{1}{2} \right)}^{5}}}{1-\left( \dfrac{1}{2} \right)}=1 \\
& \Rightarrow k\times 2\left[ 1-\dfrac{1}{{{2}^{5}}} \right]=1 \\
& \Rightarrow \dfrac{31k}{16}=1 \\
& \Rightarrow k=\dfrac{16}{31} \\
\end{align} $
The correct option is D.
So, the correct answer is “Option D”.
Note: The probability function can be of two types where we have probability mass function and probability density function. In both cases we get the sum as 1. So, $ \sum{f\left( x \right)}=1 $ . A probability distribution can be described in various forms, such as by a probability mass function or a cumulative distribution function. One of the most general descriptions, which applies for continuous and discrete variables.
Complete step-by-step answer:
$ f\left( x \right)=\dfrac{k}{{{2}^{x}}} $ is a probability distribution of a random variable X that can take on the values $ x=0,1,2,3,4 $ .
We know that the sum of the probabilities will be equal to 1.
We put the values $ x=0,1,2,3,4 $ in $ f\left( x \right)=\dfrac{k}{{{2}^{x}}} $ .
So, $ \sum\limits_{x=0}^{4}{f\left( x \right)}=\sum\limits_{x=0}^{4}{\dfrac{k}{{{2}^{x}}}}=1 $ . We have a G.P. series.
The first term be $ {{t}_{1}}=\dfrac{k}{{{2}^{0}}}=k $ and the common ratio be $ r $ where $ r=\dfrac{1}{2} $ .
The value of $ \left| r \right|<1 $ for which the sum of the first n terms of an G.P. is $ {{S}_{n}}={{t}_{1}}\dfrac{1-{{r}^{n}}}{1-r} $ .
Therefore, $ \sum\limits_{x=0}^{4}{\dfrac{k}{{{2}^{x}}}}=k\dfrac{1-{{\left( \dfrac{1}{2} \right)}^{5}}}{1-\left( \dfrac{1}{2} \right)}=1 $ . Simplifying we get
$ \begin{align}
& k\dfrac{1-{{\left( \dfrac{1}{2} \right)}^{5}}}{1-\left( \dfrac{1}{2} \right)}=1 \\
& \Rightarrow k\times 2\left[ 1-\dfrac{1}{{{2}^{5}}} \right]=1 \\
& \Rightarrow \dfrac{31k}{16}=1 \\
& \Rightarrow k=\dfrac{16}{31} \\
\end{align} $
The correct option is D.
So, the correct answer is “Option D”.
Note: The probability function can be of two types where we have probability mass function and probability density function. In both cases we get the sum as 1. So, $ \sum{f\left( x \right)}=1 $ . A probability distribution can be described in various forms, such as by a probability mass function or a cumulative distribution function. One of the most general descriptions, which applies for continuous and discrete variables.
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