Question

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

• A. $\frac{14}{3}$
• B. $\frac{13}{3}$
• C. $\frac{12}{3}$
• D. $\frac{11}{3}$

Answer 1

Answer: (A)

$\frac{14}{3}$

E(X) is nothing but mean of X

$\to$ first sic natural numbers are

$1,2,3,4,5,6$

$\to$We can select two positive no.s in

$6\times 5=30$ ways

$\to 2$ numbers are selected at random among which X denotes the larger of the two no.s

$\to$But as X is largest among the two ,

it can be $2,3,4,5$ or $6$

$\therefore P(X=2)=P$ (larger no. is $2$) $(1,2),(2,1)$

$=\frac{2}{30}$

$\to P(X=3)=P$ (Larger no. is $3$)

$=\{(1,3),(3,1),(2,3),(3,2)\}$

$=\frac{4}{30}$

$\to P(X=4)=P$ (larger no. is $4$)

$=\{(1,4),(4,1),(2,4),(4,2),(3,4),(4,3)\}$

$=\frac{6}{30}$

$\to P(X=5)=P$ (Larger no. is $5$)

$=\{(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4)\}$

$=\frac{8}{30}$

$\to P(X=6)=P$ (Larger no. is $6$)

$=\{(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6),(6,4),(5,6),(6,50)\}$

$=\frac{10}{30}$

$\therefore E\left(X\right)=$ mean $=\sum P\left(X_i\right).X_i$

$=2\left(\frac{2}{30}\right)+3\left(\frac{4}{30}\right)+4\left(\frac{6}{30}\right)+5\left(\frac{8}{30}\right)+6\left(\frac{10}{30}\right)$

$=\frac{140}{30}=\frac{14}{3}$