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Question: If \( f\left( x \right)=\dfrac{k}{{{2}^{x}}} \) is a probability distribution of a random variable X...

If f(x)=k2xf\left( x \right)=\dfrac{k}{{{2}^{x}}} is a probability distribution of a random variable X that can take on the values x=0,1,2,3,4x=0,1,2,3,4 . Then kk is equal to
A. 1615\dfrac{16}{15}
B. 1516\dfrac{15}{16}
C. 3116\dfrac{31}{16}
D. None of these

Explanation

Solution

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=0,1,2,3,4x=0,1,2,3,4 to find the G.P. series. We use the sum form of Sn=t11−rn1−r{{S}_{n}}={{t}_{1}}\dfrac{1-{{r}^{n}}}{1-r} to find the value of the variable kk .

Complete step-by-step answer :
f(x)=k2xf\left( x \right)=\dfrac{k}{{{2}^{x}}} is a probability distribution of a random variable X that can take on the values x=0,1,2,3,4x=0,1,2,3,4 .
We know that the sum of the probabilities will be equal to 1.
We put the values x=0,1,2,3,4x=0,1,2,3,4 in f(x)=k2xf\left( x \right)=\dfrac{k}{{{2}^{x}}} .
So, ∑x=04f(x)=∑x=04k2x=1\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 t1=k20=k{{t}_{1}}=\dfrac{k}{{{2}^{0}}}=k and the common ratio be rr where r=12r=\dfrac{1}{2} .
The value of ∣r∣<1\left| r \right|<1 for which the sum of the first n terms of an G.P. is Sn=t11−rn1−r{{S}_{n}}={{t}_{1}}\dfrac{1-{{r}^{n}}}{1-r} .
Therefore, ∑x=04k2x=k1−(12)51−(12)=1\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
k1−(12)51−(12)=1 ⇒k×2[1−125]=1 ⇒31k16=1 ⇒k=1631 \begin{aligned} & 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{aligned}
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, ∑f(x)=1\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.