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

 

answered Mar 17, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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