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# There are 5 cards numbered 1 to 5,one number on one card.Two cards are drawn at random without replacement.Let X denote the sum of the numbers on two cards drawn.Find the mean and variance of X.

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• $if\;5\;cards\;are\;numbered\;1or5\;and\;two\;cards\;are\;drawn\;without\;replacement\;$
• $X\;defined\;as\;sum\;of\;numbers\;on\;two\;cards$
• $X=(3\;,4\;,5\;,6\;,7\;,8\;,9\;,)$
• $s=sample\;space\;cointains\;of\;20\;events\;without\;replacement[5\;,4]\;cases$
• $variance\;=\;\sum\;P_iX_i^2\;-\;(\sum\;P_iX_i)^2$
$\large\;p(X=3)=p[(1\;,2)or(2\;,1)]$
=$\Large\frac{2}{20}\;=\frac{1}{10}$
$\large\;p(X=5)=p((1\;,4)or(4\;,1)or(3\;,2)or(2\;,3))$
=$\Large\frac{4}{20}\;=\frac{2}{10}$
$\large\;p(X=4)=p[(1\;,3)or(3\;,1)]$
=$\Large\frac{2}{20}\;=\frac{1}{10}$
$\large\;p(X=6)=p[(1\;,5)or(5\;,1)or(4\;,2)or(2\;,4)]$
=$\Large\frac{4}{20}\;=\frac{2}{10}$
$\large\;p(X=7)=p[(3\;,4)or(4\;,3)(5\;,2)or(2\;,5)]$
=$\Large\frac{4}{20}\;=\frac{2}{10}$
$\large\;p(X=8)=p[(5\;,3)or(3\;,5)]$
=$\Large\frac{2}{20}\;=\frac{1}{10}$
$\large\;p(X=9)=p[(4\;,5)or(5\;.4)]$
=$\Large\frac{2}{20}\;=\frac{1}{10}$
$\large\sum\;P_iX_i=\frac{1}{10}[3+4+10+12+14+8+9]$
$E(X)=\Large\frac{60}{10}=6$
$\Large\sum\;P_iX_i^2=\frac{1}{10}\;x\;3^2+\;\frac{1}{10}\;x\;4^2\;+\;\frac{2}{10}\;x\;5^2\;+\;\frac{2}{10}\;x\;6^2\;+\;\frac{2}{10}\;x\;7^2\;+\frac{1}{10}x\;8^2\;+\;\frac{1}{10}x\;9^2\;$+
=$\Large\frac{1}{10}[9+16+50+72+98+64+81]$
=$\Large\frac{390}{10}=39$
$variance=39-6^2=39-36$
=$3$

answered Mar 17, 2013