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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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For a poisson variate X if $P(X=0) = 0.2$ then its variance is

$\begin {array} {1 1} (A)\;\log_ e5 & \quad (B)\;\log_e4 \\ (C)\;\log_e2 & \quad (D)\;0.2 \end {array}$

 

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1 Answer

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$P(X=0) = 0.2$
$ \Rightarrow \large\frac{e^{-\lambda} \times\lambda^0}{0!} $$= 0.2$
$ \Rightarrow e^{-\lambda}=0.2$
$ \Rightarrow -\lambda= \log_e (0.2)$
$ \Rightarrow\lambda = \log _e5$
answered Jan 6, 2014 by thanvigandhi_1
edited Jan 20, 2014 by sreemathi.v
 

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