Question

Two numbers are selected at random (without replacement) the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E\left(X\right)$.

Answer 1

The two positive integers can be selected from the first six positive integers without replacement in $6\times 5=30$ ways.
$X$ represent the larger of the two numbers obtained. Therefore, $X$ can take the value of 2, 3, 4, 5, or 6.
For $X=2$, the possible observations are $(1,2)$ and $(2,1)$.
$\therefore P(X=2)=\frac{2}{30}=\frac{1}{15}$
For $X=3$, the possible observations are $(1,3),(2,3),(3,1),and(3,2)$.
$\therefore P(X=3)=\frac{4}{30}=\frac{2}{15}$
For $X=4$, the possible observations are $(1,4),(2,4),(3,4),(4,3),(4,2)$, and $(4,1)$.
$\therefore P(X=4)=\frac{6}{30}=\frac{1}{5}$
For $X=5$, the possible observations are $(1,5),(2,5),(3,5),(4,5),(5,4),(5,3),(5,2)$, and $(5,1)$.
$\therefore P(X=5)=\frac{8}{30}=\frac{4}{15}$
For $X=6$, the possible observations are $(1,6),(2,6),(3,6),(4,6),(5,6),(6,4),(6,3),(6,2)$, and $(6,1)$.
$\therefore P(X=6)=\frac{10}{30}=\frac{1}{3}$.
Therefore, the required probability distribution is as follows.
Then, $E\left(X\right)=\sum X_{i}P\left(X_i\right)$
$=2\cdot \frac{1}{15}+3\cdot \frac{2}{15}+4\cdot \frac{1}{5}+5\cdot \frac{4}{15}+6\cdot \frac{1}{3}$
$=\frac{2}{15}+\frac{2}{5}+\frac{4}{5}+\frac{4}{3}+2$
$=\frac{70}{15}$
$=\frac{14}{3}$