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Let $T = \{1, 2, 3, ... 30\}$ and let 
$x_1, x_2, x_3$ be members of $T$ with $x_1 < x_2 < x_3$

If $y_1$ and $y_2$ are selected randomly from $T \backslash \{x_1,x_2,x_3\}$ what is the expected value of $y_1 + y_2$?

We can assume $y_1 < y_2$.  The number of equally likely outcomes for $(y_1,y_2)$ is $^{27}C_2$. The sum of the possible values for $y_1+y_2$ is

$S =\large \sum_{(y_1,y_2) \in T \backslash \{x_1,x_2,x_3\}} (y_1+y_2)$

By exclusion / inclusion,

$S = \large \sum_{y_1=1}^{29} \sum_{y_2=y_1+1}^{30} (y_1+y_2)$

$-\large \sum_{y_1=1}^{x_1-1} (y_1+y_2)|_{y_2=x_1}-\large \sum_{y_1=1}^{x_2-1} (y_1+y_2)|_{y_2=x_2}-\large \sum_{y_1=1}^{x_3-1} (y_1+y_2)|_{y_2=x_3}$

$-\large \sum_{y_2=x_1+1}^{30} (y_1+y_2)|_{y_1=x_1}-\large \sum_{y_2=x_2+1}^{30} (y_1+y_2)|_{y_1=x_2}-\large \sum_{y_2=x_3+1}^{30} (y_1+y_2)|_{y_1=x_3}$

$+(x_1+x_2)+(x_1+x_3)+(x_2+x_3)$

which simplifies to

$S = \frac12 \cdot 29 \cdot 30 \cdot 31- 3 \cdot \frac12 \cdot 30 \cdot 31 - 26(x_1+x_2+x_3) = 12090 -26(x_1+x_2+x_3)$

The expected value of $(y_1+y_2)$ is $\large \frac {S} {^{27}C_2}$

For the expected value to be greater than $x_1+x_2$,

$\large \frac {S} {^{27}C_2} > x_1+x_2$

$12090-26(x_1+x_2+x_3) > {^{27}C_2}( x_1+x_2)$

$12090 > 377x_1+377x_2+26x_3$

$930 > 29x_1+29x_2+2x_3$

Tue, 04 May 2021 18:18 GMT