Question

How can I calculate the mean and variance of the event $Y$ ?
Given that a box contains balls numbered from 1-10. We will pick a ball randomly from the box when the toss of a regular coin results in a head. Let $Y$ denote the total of all the tossing’s in the given experiment. What is the mean and variance of $Y$ ?

Answer 1

Hint: Here in this question we have been asked to calculate the mean and variance of $Y$ . From the basic concepts we know that the mean is the average value in a collection of numbers and variance is defined as the average of the squares of the differences between the individual (observed) and the expected value.

Complete step-by-step solution:
Now considering from the question we have been asked to calculate the mean and variance of $Y$ . When it is given that a box contains ball numbered from 1-10. We will pick a ball randomly from the box when the toss of a regular coin results in head. Let $Y$ denote the total of all the tossing’s in the given experiment.
From the basic concepts we know that the mean is the average value in a collection of numbers and variance is defined as the average of the squares of the differences between the individual (observed) and the expected value.
As it is given that $Y$ is the number of tosses for a person until he gets a head. Then we can say that probability of event $Y$ when the number of tosses $n$ then we will have $P\left( Y=n \right)={{\left( \dfrac{1}{2} \right)}^{n}}$ as a coin has a head and a tail and both has equal probability.
The probability of the event $Y$ will be given as
$\begin{align}
  & E\left( Y \right)=\dfrac{1}{2}.1+\dfrac{1}{4}.2+\dfrac{1}{8}.3........ \\
 & \Rightarrow E\left( Y \right)=\sum\limits_{n=1}^{\infty }{n.{{\left( \dfrac{1}{2} \right)}^{n}}} \\
\end{align}$ .
We can consider this as an Arithmetic-geometric progression. We know that the formula will be given as ${{S}_{\infty }}=\dfrac{a}{1-r}+\dfrac{dr}{{{\left( 1-r \right)}^{2}}}$. By using this formula in this case we will have $\begin{align}
  & \Rightarrow E\left( Y \right)=\dfrac{1}{1-\dfrac{1}{2}}+\dfrac{1\left( \dfrac{1}{2} \right)}{{{\left( 1-\left( \dfrac{1}{2} \right) \right)}^{2}}} \\
 & \Rightarrow E\left( Y \right)=4 \\
\end{align}$ .
Hence we can conclude that the mean of the event $Y$ is 4.
Now we need to calculate the variance of the event $Y$ . For doing that we will use the concept given as $V\left( Y \right)=E{{\left( Y \right)}^{2}}-E\left( {{Y}^{2}} \right)$ .
We can say that
$\begin{align}
  & E\left( {{Y}^{2}} \right)=\dfrac{1}{2}{{.1}^{2}}+\dfrac{1}{4}{{.2}^{2}}+\dfrac{1}{8}{{.3}^{2}}........ \\
 & \Rightarrow E\left( {{Y}^{2}} \right)=\sum\limits_{n=1}^{\infty }{{{n}^{2}}.{{\left( \dfrac{1}{2} \right)}^{n}}} \\
 & \Rightarrow E\left( {{Y}^{2}} \right)={{\left( \dfrac{1}{2} \right)}^{2}}\sum\limits_{n=1}^{\infty }{n.\left( n-1 \right).{{\left( \dfrac{1}{2} \right)}^{n-2}}}+\left( \dfrac{1}{2} \right)\sum\limits_{n=1}^{\infty }{n.{{\left( \dfrac{1}{2} \right)}^{n-1}}} \\
\end{align}$

Let us assume $p=\dfrac{1}{2}$ then we will have
 $\begin{align}
  & \Rightarrow E\left( {{Y}^{2}} \right)={{p}^{2}}\sum\limits_{n=1}^{\infty }{n.\left( n-1 \right).{{p}^{n-2}}}+p\sum\limits_{n=1}^{\infty }{n.{{p}^{n-1}}} \\
 & \Rightarrow {{p}^{2}}\left[ \dfrac{{{d}^{2}}}{d{{p}^{2}}}\left( \dfrac{1}{1-p} \right) \right]+p\left[ \dfrac{d}{dp}\left( \dfrac{1}{1-p} \right) \right] \\
 & \Rightarrow {{p}^{2}}\left( \dfrac{2{{p}^{2}}}{{{\left( 1-p \right)}^{3}}} \right)+p\left( \dfrac{p}{{{\left( 1-p \right)}^{2}}} \right) \\
 & \Rightarrow \dfrac{2{{p}^{4}}+{{p}^{2}}\left( 1-p \right)}{{{\left( 1-p \right)}^{3}}} \\
\end{align}$
By substituting $P=\dfrac{1}{2}$ we will have $E\left( {{Y}^{2}} \right)=2$ .
Hence we can say that variance will be given as
 $\begin{align}
  & V\left( Y \right)=E{{\left( Y \right)}^{2}}-E\left( {{Y}^{2}} \right) \\
 & \Rightarrow V\left( Y \right)={{\left( 4 \right)}^{2}}-2 \\
 & \Rightarrow V\left( Y \right)=16-2 \\
 & \Rightarrow V\left( Y \right)=14 \\
\end{align}$
Hence we can conclude that the mean and variance of $Y$ are 4 and 14 respectively.

Note: While answering questions of this type, we should be sure with our concepts that we are going to apply in between the steps during the process of answering. In questions of this type we should clearly understand the given question any misunderstanding can lead us to end up having a wrong conclusion.