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Home  >>  CBSE XII  >>  Math  >>  Probability
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A biased die is such that $P(4)=\Large \frac{1}{10}$ and other scores being equally likely.The die is tossed twice.If X is the number of fours seen',find the variance of the random variable X.

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  • In a toss of baised die
  • P(getting 4)=\(\Large\frac{1}{10}P=\;\frac{1}{10}\;q=\;\frac{1}{10}\)
  • Die is tossed 2 ,X defined as no of '4' seen
  • X can take X=\((0\;1\;2)\)
  • varianceX=\(\sum\)P\(_i\)X\(_i\)\(^2\)+(\(\sum\)P\(_i\)X\(_i\))\(^2\)
  •  
\(P(x=0)\)=\(p(no4\))
=\(P(not 4 and not 4)\)
=\(\large\frac{9}{10}\;\times\;\frac{9}{10}\;=\;\frac{81}{100}\)
or\(P(x=0)\)=\(^2\)c\(_1\)(\(\large\frac{9}{10}\))\(^0\)(\(\large\frac{9}{10}\))\(^1\)
\(p(x=0\))=P(one 4)
=\(^2\)c\(_1\)(\(\large\frac{1}{10}\))\(^1\)(\(\large\frac{9}{10}\))\(^1\)
=2x\(\large\frac{9}{100}\)=\(\large\frac{18}{100}\)
=\(P(x=2)\)=\(^2\)c\(_2\)(\(\large\frac{1}{10}\))\(^2\)(\(\large\frac{9}{10}\))\(^0\)
=\(\large\frac{1}{100}\)
\(\sum\)P\(_i\)X\(_i\)=0\(\times\large\frac{81}{100}\)+1\(\times\large\frac{18}{100}\)+2\(\times\large\frac{1}{100}\)
=\(\large\frac{20}{100}\)
(\(\sum\)P\(_i\)X\(_i\))\(^2\)=0\(\times\large\frac{81}{100}\)+1\(^2\)\(\times\large\frac{18}{100}\)+2\(^2\)\(\times\large\frac{1}{100}\)
\(\large\frac{18}{100}\)+\(\large\frac{4}{100}\)=\(\large\frac{22}{100}\)
variance X=\(\large\frac{22}{100}\)-(\(\large\frac{22}{100}\))\(^2\)
=\(0.18\)

 

answered Feb 27, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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