Question

Three numbers are selected at random (without replacement) from the first six positive integers. Let $X$denote the largest of the three numbers obtained. Find the probability distribution of$X$. Also, find the mean and variance of the distribution.

Answer 1

Hint: Here to solve this question we need to find out the mean and variance of the given probability distribution. The mean of $X$=$\sum xP(x)$. The variance of $X\begin{array}{rl}& =\sum x^{2}P(x)-{(\sum xp(x))}^2\\ & \end{array}$. The total number of ways of selecting three numbers at random (without replacement) is ${}^6{C}_3=20$.

Complete step by step answer:
The set of first six positive integers is 1, 2, 3, 4, 5, 6.
 The total number of ways of selecting three numbers at random (without replacement) is ${}^6{C}_3=20$.
Here in this question we have that the$X$denote the largest of the three numbers obtained.
So, here the probability of 3 being the largest is possible for 1 set that is {1, 2, 3} =$P(X=3)={\displaystyle \frac{1}{20}}$.
Here the probability of 4 be the largest is possible for 3 sets that is {1, 2, 4} and {1,3,4} and (2,3,4) =$P(X=4)={\displaystyle \frac{3}{20}}$.
Here the probability of 5 be the largest is possible for 6 sets that is {1, 2, 5} and {1,3,5} and {1, 4, 5} and {2,3,5} and {2,4,5} and {3,4,5} =$P(X=5)={\displaystyle \frac{6}{20}}$.
Here the probability of 6 be the largest is possible for 10 sets that is {1, 2, 6} and {1,3,6} and {1, 4, 6} and {1,5,6} and {2,3,6} and {2,4,6} and {2,5,6} and {3,4,6} and {3,5,6} and {4,5,6} =$P(X=6)={\displaystyle \frac{10}{20}}$.
The mean of $X$=$\sum xP(x)$
$\Rightarrow 3\left({\displaystyle \frac{1}{20}}\right)+4\left({\displaystyle \frac{3}{20}}\right)+5\left({\displaystyle \frac{6}{20}}\right)+6\left({\displaystyle \frac{10}{20}}\right)$
$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{3}{20}}+{\displaystyle \frac{12}{20}}+{\displaystyle \frac{30}{20}}+{\displaystyle \frac{60}{20}}\\ & \Rightarrow {\displaystyle \frac{3+12+30+60}{20}}\end{array}$
$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{105}{20}}\\ & \Rightarrow 5.25\end{array}$
The mean of $X$= 5.25
  The variance of $X\begin{array}{rl}& =\sum x^{2}P(x)-{(\sum xp(x))}^2\\ & \end{array}$
             $\begin{array}{rl}& \Rightarrow 9\left({\displaystyle \frac{1}{20}}\right)+16\left({\displaystyle \frac{3}{20}}\right)+25\left({\displaystyle \frac{6}{20}}\right)+36\left({\displaystyle \frac{10}{20}}\right)-{\left({\displaystyle \frac{105}{20}}\right)}^2\\ & \Rightarrow {\displaystyle \frac{9+48+150+360}{20}}-{\left({\displaystyle \frac{105}{20}}\right)}^2\\ & \Rightarrow {\displaystyle \frac{567}{20}}-\left({\displaystyle \frac{11025}{400}}\right)\\ & \Rightarrow {\displaystyle \frac{11340-11025}{400}}\\ & \Rightarrow {\displaystyle \frac{315}{400}}\\ & \Rightarrow 0.7875\end{array}$
  The variance of $X$=0.7875 = 0.79(approx.)
The probability distribution is

P(x)	${\displaystyle \frac{1}{20}}$	${\displaystyle \frac{3}{20}}$	${\displaystyle \frac{6}{20}}$	${\displaystyle \frac{10}{20}}$
$\sum xP(x)$	${\displaystyle \frac{3}{20}}$	${\displaystyle \frac{12}{20}}$	${\displaystyle \frac{30}{20}}$	${\displaystyle \frac{60}{20}}$
$\sum x^{2}p(x)$	${\displaystyle \frac{9}{20}}$	${\displaystyle \frac{48}{20}}$	${\displaystyle \frac{150}{20}}$	${\displaystyle \frac{360}{20}}$

Note: Here we should take care that the variance of $X\begin{array}{rl}& =\sum x^{2}P(x)-{(\sum xp(x))}^2\\ & \end{array}$ not$\sum x^{2}p(x)$. If we miss conceptually consider this it will completely lead us to a different and wrong answer.