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Home  >>  CBSE XII  >>  Math  >>  Probability
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The random variable X can take only the values 0,1,2.Given that P(x=0)=P(X=1)=P and that E($X^2$)=E[X],find the value of p.

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  • X={\(0\;1\;2\)}
  • \(P(X=0)+P(X=1)+P(X=2)\)=1
  • \(P+P+P(X=2)\)=1
  • \(p(X=2)\)=1-2p
  • Also\(E(X)\)=\(P(X=0)\times0+P(X=1)\times1+P(X=2)\times2\)
  • \(E(X)^2\)=\(P(X=0)\times0^{2}+P(X=1)\times1^{2}+P(X=2)\times2^{2}\)
E(\(x^2)\)=\(O\times\;P+1^{2}\times\;P+2^{2}(1-2p)\)=\(P+4-8p\)=\(4-7p\)
E(\(x)\)=\(O\times\;P+1\times\)P+2(1-2p)=\(P+2-4p\)=\(2-3p\)
sinceE(\(X^2\))=E(\(x\))
=\(2-3p=4-7p\)
=\(4p=2\)
\(p=\Large\frac{1}{2}\)

 

answered Mar 1, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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