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 $X\begin{align}
  & =\sum{{{x}^{2}}P(x)}-{{\left( \sum{xp(x)} \right)}^{2}} \\
 &
\end{align}$. 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} =$P(X=3)=\dfrac{1}{20}$.
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) =$P(X=4)=\dfrac{3}{20}$.
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} =$P(X=5)=\dfrac{6}{20}$.
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} =$P(X=6)=\dfrac{10}{20}$.
The mean of $X$=$\sum xP(x)$
$\Rightarrow 3\left( \dfrac{1}{20} \right)+4\left( \dfrac{3}{20} \right)+5\left( \dfrac{6}{20} \right)+6\left( \dfrac{10}{20} \right)$
$\begin{align}
  & \Rightarrow \dfrac{3}{20}+\dfrac{12}{20}+\dfrac{30}{20}+\dfrac{60}{20} \\
 & \Rightarrow \dfrac{3+12+30+60}{20}
\end{align}$
$\begin{align}
  & \Rightarrow \dfrac{105}{20} \\
 & \Rightarrow 5.25
\end{align}$
The mean of $X$= 5.25
  The variance of $X\begin{align}
  & =\sum{{{x}^{2}}P(x)}-{{\left( \sum{xp(x)} \right)}^{2}} \\
 &
\end{align}$
             $\begin{align}
  & \Rightarrow 9\left( \dfrac{1}{20} \right)+16\left( \dfrac{3}{20} \right)+25\left( \dfrac{6}{20} \right)+36\left( \dfrac{10}{20} \right)-{{\left( \dfrac{105}{20} \right)}^{2}} \\
 & \Rightarrow \dfrac{9+48+150+360}{20}-{{\left( \dfrac{105}{20} \right)}^{2}} \\
 & \Rightarrow \dfrac{567}{20}-\left( \dfrac{11025}{400} \right) \\
 & \Rightarrow \dfrac{11340-11025}{400} \\
 & \Rightarrow \dfrac{315}{400} \\
 & \Rightarrow 0.7875
\end{align}$
  The variance of $X$=0.7875 = 0.79(approx.)
The probability distribution is

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

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