Question

Two positive numbers are selected at random (without replacement) from the first five positive integers. The larger of the two numbers obtained is denoted by X. Find the mean and variance of X.

Answer 1

Hint : Here, in this given question we have to find the mean and variance of X. In order to solve this question, we will first have to find the number of ways to select two numbers from the first five positive integers i.e., $1,2,3,4,5$ . Next, we will find the set for which $2$ is the larger number and find its probability. We will repeat this step for the remaining numbers i.e., $3,4,5$ . After getting all the probabilities, we will make a probability distribution table. Then find the mean and variance of X.
Formula used:
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Mean = $\sum {X \times P\left( X \right)}$
Variance = $\sum {{X^2} \times P\left( X \right)} - {\left[ {\sum {X \times P\left( X \right)} } \right]^2}$

Complete step by step solution:
The first five positive integers are $1,2,3,4,5$
Now, we have to select two numbers from five positive integers. This is done by using formula,
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} \\\ \Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = \dfrac{{120}}{6} = 20 \;$
Here, we are given that X denotes the larger of two numbers. We observe that X can take the values of $2,3,4,5$ .
$P\left( {X = 2} \right)$ = Probability that larger number is $2$ .
$P\left( {X = 2} \right)$ = Probability of getting $1$ in first selection and $2$ in second selection or getting $2$ in first selection and $1$ in second selection.
$\Rightarrow P\left( {X = 2} \right) = \dfrac{1}{5} \times \dfrac{1}{4} + \dfrac{1}{5} \times \dfrac{1}{4} = \dfrac{2}{{20}}$
$P\left( {X = 3} \right)$ = Probability that the larger of two numbers is $3$ .
$\Rightarrow P\left( {X = 3} \right) = \dfrac{2}{5} \times \dfrac{1}{4} + \dfrac{1}{5} \times \dfrac{2}{4} = \dfrac{4}{{20}}$
$P\left( {X = 4} \right)$ = Probability that the larger of two numbers is $4$ .
$\Rightarrow P\left( {X = 4} \right) = \dfrac{3}{5} \times \dfrac{1}{4} + \dfrac{1}{5} \times \dfrac{3}{4} = \dfrac{6}{{20}} \\\ \Rightarrow P\left( {X = 5} \right) = \dfrac{4}{5} \times \dfrac{1}{4} + \dfrac{1}{5} \times \dfrac{4}{4} = \dfrac{8}{{20}} \;$
The probability distribution of X is:

X	$2$	$3$	$4$	$5$
P(X)	$\dfrac{2}{{20}}$	$\dfrac{4}{{20}}$	$\dfrac{6}{{20}}$	$\dfrac{8}{{20}}$

Mean $\left[ {E\left( X \right)} \right]$
$= 2 \times \dfrac{2}{{20}} + 3 \times \dfrac{4}{{20}} + 4 \times \dfrac{6}{{20}} + 5 \times \dfrac{8}{{20}} \\\ = \dfrac{4}{{20}} + \dfrac{{12}}{{20}} + \dfrac{{24}}{{20}} + \dfrac{{40}}{{20}} \\\ = \dfrac{{80}}{{20}} \\\ = 4 \;$
$E\left( {{X^2}} \right) = {2^2} \times \dfrac{2}{{20}} + {3^2} \times \dfrac{4}{{20}} + {4^2} \times \dfrac{6}{{20}} + {5^2} \times \dfrac{8}{{20}} \\\ = \dfrac{8}{{20}} + \dfrac{{36}}{{20}} + \dfrac{{96}}{{20}} + \dfrac{{200}}{{20}} \\\ = \dfrac{{340}}{{20}} \\\ = 17 \;$
Variance of X = $E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} = 17 - {4^2} = 1$
Therefore, $E\left( X \right) = 4$ and $Var\left( X \right) = 1$
Hence, the mean of X is $4$ and variance of X is $1$ .

Note : The given question was not tough. The most important part in these types of questions is to find out the number of ways to select numbers. The given question deals with the concept of probability, but the concept of permutation and combination should also be thoroughly clear.