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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Probability

If X is a Poisson variate such that $ P(X=1)=P(x=2)$ the $P(x=4)$ is equal to

$\begin {array} {1 1} (A)\;\large\frac{1}{2e^2} & \quad (B)\;\large\frac{1}{3e^2} \\ (C)\;\large\frac{2}{3e^2} & \quad (D)\;\large\frac{1}{e^2} \end {array}$

 

1 Answer

Given that $ P(x=1)=P(x=2)$
$ = \large\frac{e^{-\lambda}\lambda'}{1!} = \large\frac{e^{-\lambda}\lambda^2}{2!}$
$ \lambda = 2$
$ P(x=4) = \large\frac{e^{-2}(2)^4}{4!}$
$ = \large\frac{e^{-2} \times \not{16}2}{\not{24}3}$
$ = \large\frac{2}{3e^2}$
Ans : (C)
answered Jan 16, 2014 by thanvigandhi_1
 

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