Question

If $P\left( X=x \right)=C{{\left( \dfrac{2}{3} \right)}^{x}}$; $x=1,2,3,4,......$ is a probability mass function, the value of C is
(A) $\dfrac{1}{4}$
(B) $\dfrac{1}{3}$
(C) $\dfrac{1}{2}$
(D) $\dfrac{1}{6}$

Answer 1

Hint: We start solving this problem by first considering a property that the sum of probabilities of all possible values of X is always equal to 1, which is \[\sum\limits_{x\in X}{P\left( X=x \right)}=1\]. Then we use this property and substitute the given value of probability mass function and then equate it to 1. Then we use the formula for the sum of infinite GP, $a+ar+a{{r}^{2}}+...........$ is equal to $\dfrac{a}{1-r}$ and find the value of the sum and thereby we find the value of C.

Complete step by step answer:
We are given that X has the probability mass function, $P\left( X=x \right)=C{{\left( \dfrac{2}{3} \right)}^{x}}$.
Now let us consider a property of Probability mass function.
For a random variable X following a probability mass function $P\left( X=x \right)$, the sum of probabilities of all possible values of X is always equal to 1, that is
\[\sum\limits_{x\in X}{P\left( X=x \right)}=1\]
Here we are given that x takes the values $x=1,2,3,4,......$.
So, from the above-discussed property of probability mass function, we can say that when their probabilities are added the value must be equal to 1.
So, we get
$\Rightarrow \sum\limits_{i=1}^{\infty }{P\left( X=x \right)}=1$
Now, substituting the value of $P\left( X=x \right)=C{{\left( \dfrac{2}{3} \right)}^{x}}$ in the above equation we get,
$\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{\infty }{C{{\left( \dfrac{2}{3} \right)}^{x}}}=1 \\
 & \Rightarrow C\sum\limits_{i=1}^{\infty }{{{\left( \dfrac{2}{3} \right)}^{x}}}=1 \\
 & \Rightarrow C\left( {{\left( \dfrac{2}{3} \right)}^{1}}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+............ \right)=1 \\
 & \Rightarrow C\left( \dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+............ \right)=1............\left( 1 \right) \\
\end{align}$
Now let us consider the sum $\dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+...........$.
As we see $\dfrac{2}{3},{{\left( \dfrac{2}{3} \right)}^{2}},{{\left( \dfrac{2}{3} \right)}^{3}},............$ is an infinite G.P with common ratio and first term both equal to $\dfrac{2}{3}$. So, we need to find the value of sum of infinite GP.
Now let us consider the formula, sum of infinite G.P $a+ar+a{{r}^{2}}+...........$ is $\dfrac{a}{1-r}$.
So, we get the value of $\dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+...........$ as,
$\begin{align}
  & \Rightarrow \dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+...........=\dfrac{\dfrac{2}{3}}{1-\dfrac{2}{3}} \\
 & \Rightarrow \dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+...........=\dfrac{\dfrac{2}{3}}{\dfrac{1}{3}} \\
 & \Rightarrow \dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+...........=2 \\
\end{align}$
Substituting this value in equation (1) we get,
$\begin{align}
  & \Rightarrow C\left( \dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( \dfrac{2}{3} \right)}^{3}}+............ \right)=1 \\
 & \Rightarrow C\left( 2 \right)=1 \\
 & \Rightarrow C=\dfrac{1}{2} \\
\end{align}$
Hence answer is Option C.

Note:
There is a possibility of one making a mistake while solving this question by using the formula for infinite G.P as $a+ar+a{{r}^{2}}+...........$ is $\dfrac{1}{1-r}$. In that case the answer will be $\dfrac{1}{3}$. But here we need to remember that sum of infinite GP is equal to $\dfrac{1}{1-r}$ if the G.P is $1+r+{{r}^{2}}+...........$, that is if the first term is 1. In our problem the first term is not equal to 1, so we need to use the formula $\dfrac{a}{1-r}$.