Question

These are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let $X$ denote the sum of the numbers on the two drawn cards. Find the mean and variance of $X$ .

Answer 1

Hint: We can see there are 4 cards and out of 4 cards, 2 cards are drawn and we will make sample space of all the possibilities. As we know sample space is the group of all the possibilities. After that, we will sum up the possibilities of the sample space which is denoted by $s$ . As discussed in the question, we know,$X$is the sum of the possibilities. Now, the probability of $X$will be $p\left( X \right)$ and it will be calculated for all values of $X$. After that, we need to calculate the values of $p\left( X \right)\left( X \right)$ and $p\left( X \right){\left( X \right)^2}$ and put them all in the table. Now, we can calculate mean and variance by using formula.
Mean \[ = \bar X = \sum {p\left( X \right)\left( X \right)} \] and variance $Var(X) = \sum {p\left( X \right){{\left( X \right)}^2}} - {\left( {\sum {p\left( X \right)\left( X \right)} } \right)^2}$

Complete step-by-step answer:
According to question,
There are 4 cards numbered 1, 3, 5 and 7.
$X$ Denote the sum of the numbers on the two drawn cards.
Now, sample space can be written as
$s = \left\{ {\left( {1,3} \right),\left( {1,5} \right),\left( {1,7} \right),\left( {3,1} \right),\left( {3,5} \right),\left( {3,7} \right),\left( {5,1} \right),\left( {5,3} \right),\left( {5,7} \right),\left( {7,1} \right),\left( {7,3} \right),\left( {7,5} \right)} \right\}$
Now, the probability distribution of $X$is given below in table.

$X$	$p\left( X \right)$	$p\left( X \right)\left( X \right)$	$P\left( X \right){\left( X \right)^2}$
4	$\dfrac{2}{{12}}$	$\dfrac{8}{{12}}$	$\dfrac{{32}}{{12}}$
6	$\dfrac{2}{{12}}$	$\dfrac{{12}}{{12}}$	$\dfrac{{72}}{{12}}$
8	$\dfrac{4}{{12}}$	$\dfrac{{32}}{{12}}$	$\dfrac{{256}}{{12}}$
10	$\dfrac{2}{{12}}$	$\dfrac{{20}}{{12}}$	$\dfrac{{200}}{{12}}$
12	$\dfrac{2}{{12}}$	$\dfrac{{24}}{{12}}$	$\dfrac{{288}}{{12}}$

We have,
 \[
  \sum {p\left( X \right)\left( X \right)} = \dfrac{{96}}{{12}} \\
  \dfrac{{96}}{{12}} = 8 \\
 \]
Now,
Mean \[ = \bar X = \sum {p\left( X \right)\left( X \right)} = 8\]
So, mean = $8$
Now, substituting the value to find variance,
$
  Var(X) = \sum {p\left( X \right){{\left( X \right)}^2}} - {\left( {\sum {p\left( X \right)\left( X \right)} } \right)^2} \\
   = \dfrac{{848}}{{12}} - {\left( 8 \right)^2} \\
   = \dfrac{{848}}{{12}} - 64 \\
   = \dfrac{{212}}{3} - 64 \\
   = \dfrac{{20}}{3} \\
 $
So, Variance = $\dfrac{{20}}{3}$
Hence, Answer is Mean = $8$ and Variance =$\dfrac{{20}}{3}$.

Note: As we know, high attention is needed to do these types of questions because we have to take a lot of terms in calculation and we cannot make mistakes in the number of terms. The calculation of median is simple but it should also be rechecked twice to avoid any mistakes in these types of questions.