Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X and hence find the mean of the distribution.

Hint: To solve this question, we will first find the number of ways to select two numbers from 1, 2, 3, 4, 5 and 6. This will be the sample set. Then we will find the set for which 2 is the larger number and find its probability. We will repeat this process for the remaining numbers, i.e. 3, 4, 5 and 6. After getting probability for various values of X, we can form a probability distribution table. The mean of a probability distribution is given as $E\left( X \right)=\sum{XP\left( X \right)}$. With this relation, we will find the mean of the distribution.

Complete step-by-step answer:
Let S be the event of choosing any two numbers from the first 6 positive integers.
We need to choose 2 items from 6 items and arrange them. This is done in $^{6}{{P}_{2}}$ = 30
Therefore, the number of elements in the event set S is n(S) = 30.
Now, we will consider various event sets for various bigger numbers.
It is given that the bigger number of the two drawn numbers is denoted by X.
Let A be the event of getting X as the bigger number.
If X = 2, then A = {(2, 1), (1, 2)}. Thus, n(A) = 2.
$\Rightarrow $ P(X = 2) = $\dfrac{n\left( A \right)}{n\left( S \right)}$
$\Rightarrow $ P(X = 2) = $\dfrac{2}{30}$
If X = 3, then A = {(1, 3), (3, 1), (2, 3), (3, 2)}. Thus, n(A) = 4
$\Rightarrow $ P(X = 3) = $\dfrac{4}{30}$
If X = 4, then A = {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3)}. Thus, n(A) = 6
$\Rightarrow $ P(X = 4) = $\dfrac{6}{30}$
If X = 5, then A = {(1,5), (5,1), (2,5), (5,2), (3,5), (5,3), (4,5), (5,4)}. Thus, n(A) = 8
$\Rightarrow $ P(X = 5) = $\dfrac{8}{30}$
If X = 6, then A = {(1,6), (6,1), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5)}. Thus, n(A) = 10
$\Rightarrow $ P(X = 6) = $\dfrac{10}{30}$
Therefore, the probability distribution table is as follows:



The mean of the probability distribution is given by the relation $E\left( X \right)=\sum{XP\left( X \right)}$
\[\begin{align}
  & \Rightarrow E\left( X \right)=2\times \dfrac{2}{30}+3\times \dfrac{4}{30}+4\times \dfrac{6}{30}+5\times \dfrac{8}{30}+6\times \dfrac{10}{30} \\
 & \Rightarrow E\left( X \right)=\dfrac{4}{30}+\dfrac{12}{30}+\dfrac{24}{30}+\dfrac{40}{30}+\dfrac{60}{30} \\
 & \Rightarrow E\left( X \right)=\dfrac{140}{30} \\
 & \Rightarrow E\left( X \right)=4.66 \\
\end{align}\]
Therefore, the mean is 4.66.


Note: The number of ways to arrange n things in r spaces is given by $^{n}{{P}_{r}}$, where $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Although the given problem is from probability topic, concepts of permutations and combinations are used in the problem.