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Home  >>  CBSE XII  >>  Math  >>  Probability
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A die is thrown three times.Let X be 'the number of twos seen'.Find the expectation of X.

$\begin{array}{1 1}(A)\;\large\frac{1}{2}\\(B)\;\large\frac{1}{4}\\(C)\;\large\frac{1}{8}\\(D)\;\large\frac{1}{3}\end{array} $

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1 Answer

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Toolbox:
  • Die is thrown 3 times
  • Probability of getting '2'=\(\frac{1}{6}\)in a single through of die
  • P=\(\frac{1}{6}\)q=1-\(\frac{1}{6}\)=\(\frac{5}{6}\)
  • X=defined as number '2'2seen
  • X=(\(0\;1\;2\;3\))
  • E(X)=\(\sum\)P\(_i\)X\(_i\)
  • \(P(x-r\))=\(^n\)C\(_r\) P\(^r\) q\(^{n-r}\)
\(p(x=0)\)=\(^3\)C\(_0\)(\(\frac{1}{6}\))\(^0\)(\(\frac{5}{6}\))\(^3\)=(\(\frac{5}{6}\))\(^3\)
\(p(x=1)\)=\(^3\)C\(_1\)(\(\frac{1}{6}\))\(^1\)(\(\frac{5}{6}\))\(^2\)
\(p(x=2)\)=\(^3\)C\(_2\)(\(\frac{1}{6}\))\(^2\)(\(\frac{5}{6}\))
\(p(x=3)\)=\(^3\)C\(_3\)(\(\frac{1}{6}\))\(^3\)(\(\frac{5}{6}\))\(^0\)=(\(\frac{1}{6}\))\(^3\)
=\(\frac{1}{63}\)[75+30+3]
=\(\frac{108}{219}\)
=\(\frac{1}{2}\)
answered Feb 27, 2013 by poojasapani_1
 

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