Question

A random variable $X$ takes values $0,1,2,3....$ with probability $P(X=x)=k(x+1){\left({\displaystyle \frac{1}{5}}\right)}^x$, where $k$ is a constant, then $P(X=0)$ is equal to
$1)7/25$
$2)\text{ 18}/25$
$3)\text{ 13}/25$
$4)\text{ 19}/25$
$5)\text{ 16}/25$

Answer 1

Hint: In this question we have been given with a probability function for which the random variable $X$ takes values from $0,1,2,3$ upto infinity. Based on the given probability function we have to find the value of $P(X=0)$. We will solve this question by first finding the value of $k$. We will also use the formula of the sum of series which is $a+(a+d)r+(a+d)r^2+...={\displaystyle \frac{a}{1-r}}+{\displaystyle \frac{dr}{{(1-r)}^2}}$. We will then substitute $X=0$ and get the required probability.

Complete step-by-step solution:
We have the function given to us as:
$\Rightarrow P(X=x)=k(x+1){\left({\displaystyle \frac{1}{5}}\right)}^x$
Now we know that the sum of all the probabilities of an event is $1$ therefore, we can write:
$\Rightarrow \sum _{x=0}^{\mathrm{\infty }}P(X=x)=1$
Now from the question we have been given that $P(X=x)=k(x+1){\left({\displaystyle \frac{1}{5}}\right)}^x$ therefore, on substituting, we get:
$\Rightarrow \sum _{x=0}^{\mathrm{\infty }}k(x+1){\left({\displaystyle \frac{1}{5}}\right)}^x=1$
Now since $k$ is a constant, we can take it out and write the expression as:
$\Rightarrow k\sum _{x=0}^{\mathrm{\infty }}(x+1){\left({\displaystyle \frac{1}{5}}\right)}^x=1$
Now on expanding the sum by substituting the values, we get:
$\Rightarrow k[(0+1){\left({\displaystyle \frac{1}{5}}\right)}^0+(1+1){\left({\displaystyle \frac{1}{5}}\right)}^1+(2+1){\left({\displaystyle \frac{1}{5}}\right)}^2+....]=1$
On simplifying, we get:
$\Rightarrow k[1+(1+1){\left({\displaystyle \frac{1}{5}}\right)}^1+(2+1){\left({\displaystyle \frac{1}{5}}\right)}^2+....]=1$
Now we know the formula $a+(a+d)r+(a+d)r^2+...={\displaystyle \frac{a}{1-r}}+{\displaystyle \frac{dr}{{(1-r)}^2}}$therefore, on substituting $a=1$, $d=1$ and $r={\displaystyle \frac{1}{5}}$, we get:
$\Rightarrow k[{\displaystyle \frac{1}{1-{\displaystyle \frac{1}{5}}}}+{\displaystyle \frac{1\times {\displaystyle \frac{1}{5}}}{{(1-{\displaystyle \frac{1}{5}})}^2}}]=1$
On taking the lowest common multiple, we get:
$$\Rightarrow k\left[{\displaystyle \frac{1-{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{5}}}{{(1-{\displaystyle \frac{1}{5}})}^2}}\right]=1$$
on using the expansion of ${(a-b)}^2$, we get:
$$\Rightarrow k\left[{\displaystyle \frac{1-{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{5}}}{1+{\displaystyle \frac{1}{25}}-{\displaystyle \frac{2}{5}}}}\right]=1$$
On simplifying, we get:
$$\Rightarrow k\left[{\displaystyle \frac{1}{{\displaystyle \frac{16}{25}}}}\right]=1$$
On rearranging the terms, we get:
$$\Rightarrow k\left[{\displaystyle \frac{25}{16}}\right]=1$$
On transferring the terms, we get:
$$\Rightarrow k={\displaystyle \frac{16}{25}}$$
Therefore, we get the probability function as:
$P(X=x)=\left({\displaystyle \frac{16}{25}}\right)(x+1){\left({\displaystyle \frac{1}{5}}\right)}^x$
Now we have to find $P(X=0)$ therefore, on substituting $x=0$, we get:
$P(X=0)=\left({\displaystyle \frac{16}{25}}\right)(0+1){\left({\displaystyle \frac{1}{5}}\right)}^0$
On simplifying, we get:
$P(X=0)={\displaystyle \frac{16}{25}}$, which is the required solution.
Therefore, the correct answer is option $\left(5\right)$.

Note: It is to be noted that the general principle applied in this question is the sum of all the probabilities of an event is $1$. It is to be remembered that the total probability can never exceed $1$ neither can it be negative. The various series formulas should be remembered to convert an infinite series to a finite sum.