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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\(_i\))\(^2\)
=\(P(not 4 and not 4)\)
\(p(x=0\))=P(one 4)
variance X=\(\large\frac{22}{100}\)-(\(\large\frac{22}{100}\))\(^2\)


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

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