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Home  >>  CBSE XII  >>  Math  >>  Probability
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Two probability distributions of the discrete random variable X and Y are given below.

  X 0 1 2 3 P(X) $\frac{1}{5}$ $\frac{2}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ Y 0 1 2 3 P(Y) $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$ $\frac{1}{10}$ Prove that $E(Y^2)=2E(X)$.  
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1 Answer

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Toolbox:
  • \(E(x)\)=\(\sum\)p\(_i\)X\(_i\)
  • \(E(y)^2\)=\(\sum\)p\(_i\)X\(_i\)\(^2\)
\(E(x)\)=\(\sum\)p\(_i\)X\(_i\)=\(\Large0\times\;\frac{1}{5}\;+1\times\;\frac{2}{5}\;+2\times\;\frac{1}{5}\;+3\times\;\frac{1}{5}\)
=\(\Large\frac{3}{10}\;+\;\frac{8}{5}\;+\;\frac{9}{10}\;=\;\frac{28}{10}\;=\;\frac{14}{5}\)
\(E(y)^2\)=\(2\;E(x)\) proved

 

answered Feb 28, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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