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A die is tossed twice.A 'success' is getting an even number on a toss.Find the variance of the number of successes.

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• $In\;a\;single\;toss\;of\;die\;getting\;even\;no\;=\;\Large\frac{3}{6}=\frac{1}{2}$
• $\large\;p(S)=\frac{1}{2}\;p(F)=\frac{1}{2}$
• $in\;a\;die\;tossed\;twice\;we\;define\;X=\;no\;of\;sucesses$
• $X=(0\;1\;2\;3\;)$
• $Mean\;=\sum\;P_i\;X_i$
• $\sigma\;=\sqrt{variance}=\sqrt{\sum\;P_iX^2_i\;-\;(\sum\;P_iX_i)^2}$
$P(X=0)=p(no\;success)$=$p(S\;S)$
$\Large\frac{3}{2}\times\frac{1}{2}$
$P(X=1)=p(1S\;,1F)or\;p(1F\;,1S)$
$\Large\;2(\frac{1}{2}\times\frac{1}{2})$
$P(X=2)=p(both\;failurs)$=$p(F\;,F)$
$\Large\frac{1}{2}\times\frac{1}{2}$
$E(X)=\sum\;P_iX_i$
=$\Large\;0\times\frac{1}{4}\;+\;1\times\;\frac{1}{2}\;+\;2\;\times\;\frac{1}{4}$
=$\Large\;1$
$\sum\;P_iX_i^2=\Large\;0\;\times\;\frac{1}{4}\;+\;1^2\;\times\;\frac{1}{2}\;+\;2^2\times\frac{1}{4}$
=$\Large\frac{1}{2}+1=\frac{3}{2}$
$var\;=\Large\frac{3}{2}-\;1^2$
=$\Large\frac{1}{2}$

edited Jun 4, 2013