Question

Two numbers are selected at random (without replacement) from the first six positive integers. Let $X$ denote the larger of the two numbers. Find $E\left( X \right)$.

Answer 1

Here, we need to find the expected value of $X$. First, we will find the sample space and the number of favourable outcomes for different possible values of $X$. Then, we will find the probabilities for different possible values of $X$. Finally, we will use the formula for expected value to find the value of $E\left( X \right)$.
Formula Used: We will use the following formulas:

1. The probability of an event, $P\left( E \right) =$ Number of favourable outcomes $\div$ Number of total outcomes.
2. The expected value of a random variable $X$ is given by the formula $E\left( X \right) = \sum {xP\left( {X = x} \right)}$.

Complete step by step solution:
The first positive integers are 1, 2, 3, 4, 5, 6.
Let $\left( {p,q} \right)$ denote the two numbers selected, where $p$ is the first number selected, and $q$ is the second number selected.
We will find the sample space and probabilities using multiple cases.
Case 1: The larger of the two numbers is 2.
The number lesser than 2 is 1.
If the larger of the two numbers is 2, then the other number has to be 1.
Therefore, we get the possible cases $\left( {2,1} \right)$ and $\left( {1,2} \right)$.
Thus, there are 2 favourable outcomes for $X = 2$.
Case 2: The larger of the two numbers is 3.
The numbers lesser than 3 are 1 or 2.
If the larger of the two numbers is 3, then the other number has to be 1 or 2.
Therefore, we get the possible cases $\left( {3,1} \right)$, $\left( {1,3} \right)$, $\left( {3,2} \right)$, and $\left( {2,3} \right)$.
Thus, there are 4 favourable outcomes for $X = 3$.
Case 3: The larger of the two numbers is 4.
The numbers lesser than 4 are 1, 2, or 3.
If the larger of the two numbers is 4, then the other number has to be 1, 2, or 3.
Therefore, we get the possible cases $\left( {4,1} \right)$, $\left( {1,4} \right)$, $\left( {4,2} \right)$, $\left( {2,4} \right)$, $\left( {4,3} \right)$, and $\left( {3,4} \right)$.
Thus, there are 6 favourable outcomes for $X = 4$.
Case 4: The larger of the two numbers is 5.
The numbers lesser than 5 are 1, 2, 3, or 4.
If the larger of the two numbers is 5, then the other number has to be 1, 2, 3, or 4.
Therefore, we get the possible cases $\left( {5,1} \right)$, $\left( {1,5} \right)$, $\left( {5,2} \right)$, $\left( {2,5} \right)$, $\left( {5,3} \right)$, $\left( {3,5} \right)$, $\left( {5,4} \right)$, and $\left( {4,5} \right)$.
Thus, there are 8 favourable outcomes for $X = 5$.
Case 5: The larger of the two numbers is 6.
The numbers lesser than 6 are 1, 2, 3, 4, or 5.
If the larger of the two numbers is 6, then the other number has to be 1, 2, 3, 4, or 5.
Therefore, we get the possible cases $\left( {6,1} \right)$, $\left( {1,6} \right)$, $\left( {6,2} \right)$, $\left( {2,6} \right)$, $\left( {6,3} \right)$, $\left( {3,6} \right)$, $\left( {6,4} \right)$, $\left( {4,6} \right)$, $\left( {6,5} \right)$, and $\left( {5,6} \right)$.
Thus, there are 10 favourable outcomes for $X = 6$.
From the five cases, we can observe that there are 30 possible outcomes.
We will calculate the probabilities of each value of the random variable $X$.
The probability of an event,$P\left( E \right) =$Number of favourable outcomes $\div$ Number of total outcomes.
The number of total outcomes is 30.
Therefore, we get
$P\left( {X = 2} \right) = \dfrac{2}{{30}} = \dfrac{1}{{15}}$
$P\left( {X = 3} \right) = \dfrac{4}{{30}} = \dfrac{2}{{15}}$
$P\left( {X = 4} \right) = \dfrac{6}{{30}} = \dfrac{3}{{15}}$
$P\left( {X = 5} \right) = \dfrac{8}{{30}} = \dfrac{4}{{15}}$
$P\left( {X = 6} \right) = \dfrac{{10}}{{30}} = \dfrac{5}{{15}}$
Arranging the values of the random variable $X$ and the probabilities in a table, we get

$X$	2	3	4	5	6
$P\left( {X = x} \right)$	$\dfrac{1}{{15}}$	$\dfrac{2}{{15}}$	$\dfrac{3}{{15}}$	$\dfrac{4}{{15}}$	$\dfrac{5}{{15}}$

Now, we will find the expected value of the random variable $X$.
The expected value of a random variable $X$ is given by the formula $E\left( X \right) = \sum {xP\left( {X = x} \right)}$.
Therefore, we get
$\Rightarrow E\left( X \right) = 2 \times P\left( {X = 2} \right) + 3 \times P\left( {X = 3} \right) + 4 \times P\left( {X = 4} \right) + 5 \times P\left( {X = 5} \right) + 6 \times P\left( {X = 6} \right)$
Substituting the values using the table, we get
$\Rightarrow E\left( X \right) = 2 \times \dfrac{1}{{15}} + 3 \times \dfrac{2}{{15}} + 4 \times \dfrac{3}{{15}} + 5 \times \dfrac{4}{{15}} + 6 \times \dfrac{5}{{15}}$
Multiplying the terms in the expression, we get
$\Rightarrow E\left( X \right) = \dfrac{2}{{15}} + \dfrac{6}{{15}} + \dfrac{{12}}{{15}} + \dfrac{{20}}{{15}} + \dfrac{{30}}{{15}}$
Adding the terms in the expression, we get
$\begin{array}{l} \Rightarrow E\left( X \right) = \dfrac{{2 + 6 + 12 + 20 + 30}}{{15}}\\\ \Rightarrow E\left( X \right) = \dfrac{{70}}{{15}}\end{array}$
Simplifying the expression, we get
$\Rightarrow E\left( X \right) = \dfrac{{14}}{3}$
$\therefore$ We get the value of $E\left( X \right)$ as $\dfrac{{14}}{3}$.

Note:
We used the term “positive integer” in the solution. An integer is a number that is not a fraction and can be either positive or negative. For example: $- 95$ and 95 are integers.
We know that the sum of all probabilities is always equal to 1. We can verify our probabilities by adding them. If the sum of probabilities is 1, then our probabilities should be correct. From the table, we can observe that $\dfrac{1}{{15}} + \dfrac{2}{{15}} + \dfrac{3}{{15}} + \dfrac{4}{{15}} + \dfrac{5}{{15}} = \dfrac{{15}}{{15}} = 1$.