Question

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

Answer 1

First six positive integers are {1, 2, 3, 4, 5, 6}

No. of ways of selecting 2 numbers from 6 numbers without replacement ${}^6{C}_2=15$

X denotes the larger of the two numbers, so X can take the values 2, 3, 4, 5, 6. Probability distribution of X:

$\begin{array}{cccccc}x& 2& 3& 4& 5& 6\\ P\left(x\right)& \frac{1}{15}& \frac{2}{15}& \frac{3}{15}& \frac{4}{15}& \frac{5}{15}\end{array}$

Computation of Mean and Variance:

$\begin{array}{cccc}Xi& P(x=xi)& piXi& piX{i}^2\\ 2& \frac{1}{15}& \frac{2}{15}& \frac{2}{15}\\ 3& \frac{2}{15}& \frac{6}{15}& \frac{18}{15}\\ 4& \frac{3}{15}& \frac{12}{15}& \frac{48}{15}\\ 5& \frac{4}{15}& \frac{20}{15}& \frac{100}{15}\\ 6& \frac{5}{15}& \frac{30}{15}& \frac{180}{15}\\ & & \sum piXi=\frac{70}{15}=\frac{14}{3}& \sum piX{i}^2=\frac{350}{15}=\frac{70}{3}\end{array}$

$Mean\sum piXi=\frac{70}{15}=4.67$

$Variance=\sum piX{i}^2={(\sum piXi)}^2=\frac{70}{3}-\frac{196}{9}=\frac{210-196}{9}=\frac{14}{9}$