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).
✓
Answer
Two number are selected without replacement from $\{1,2,3,4,5,6\}$.
Let $S =$ sample space
$ \therefore n ( S )={ }^6 C _2$
$=\frac{6 !}{2 ! \times 4 !}$
$=\frac{6 \times 5 \times 4 !}{2 \times 1 \times 4 !}$
$=15 $
Let $X$ denote the larger of the two numbers obtained.
$\therefore$ Possible values of $X$ are $2,3,4,5,6$.
When $X = 2$,
one of the two numbers is 2 and remaining one is smaller than 2 , i.e., 1.
$\therefore$ Remaining number can be selected in 1 way only.
$ \therefore n(X=2)=1$
$\therefore P(X=2)=\frac{1}{15} $
When $X=3$,
one of the two numbers is 3 and remaining one is smaller than 3 , i.e., 1 or 2.
$\therefore$ Remaining number can be selected in ${ }^2 C _1=2$ ways.
$\therefore n ( X =3)=2$
$\therefore P(X=3)=\frac{2}{15}$
Similarly, $P(X=4)=\frac{3}{15}, P(X=5)=\frac{4}{15}, P(X=6)=\frac{5}{15}$
$\therefore E ( X )=$
$\sum_{i=1}^5 x_i \cdot P \left(x_i\right)$
$=2 \times \frac{1}{15}+3 \times \frac{2}{15}+4 \times \frac{3}{15}+5 \times \frac{4}{15}+6 \times \frac{5}{15}$
$=\frac{1}{15}(2+6+12+20+30)$
$=\frac{14}{3}$
$=4.67$
Let $S =$ sample space
$ \therefore n ( S )={ }^6 C _2$
$=\frac{6 !}{2 ! \times 4 !}$
$=\frac{6 \times 5 \times 4 !}{2 \times 1 \times 4 !}$
$=15 $
Let $X$ denote the larger of the two numbers obtained.
$\therefore$ Possible values of $X$ are $2,3,4,5,6$.
When $X = 2$,
one of the two numbers is 2 and remaining one is smaller than 2 , i.e., 1.
$\therefore$ Remaining number can be selected in 1 way only.
$ \therefore n(X=2)=1$
$\therefore P(X=2)=\frac{1}{15} $
When $X=3$,
one of the two numbers is 3 and remaining one is smaller than 3 , i.e., 1 or 2.
$\therefore$ Remaining number can be selected in ${ }^2 C _1=2$ ways.
$\therefore n ( X =3)=2$
$\therefore P(X=3)=\frac{2}{15}$
Similarly, $P(X=4)=\frac{3}{15}, P(X=5)=\frac{4}{15}, P(X=6)=\frac{5}{15}$
$\therefore E ( X )=$
$\sum_{i=1}^5 x_i \cdot P \left(x_i\right)$
$=2 \times \frac{1}{15}+3 \times \frac{2}{15}+4 \times \frac{3}{15}+5 \times \frac{4}{15}+6 \times \frac{5}{15}$
$=\frac{1}{15}(2+6+12+20+30)$
$=\frac{14}{3}$
$=4.67$
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