Question

For a poisson variate X, if p(0)=p(2) then find p(3).​

Answer 1

Answer:

The value of P(3) is

$\bf\frac{e^{-\sqrt{2}}\,\sqrt{2}}{3}$

Step-by-step explanation:

Concept:

The probability mass fuction of poisson distribution is

$\boxed{\bf\,P(x)=\frac{e^{-\lambda}\,{\lambda}^x}{x!}}$

Given:

$P(0)=P(2)$

$\implies\frac{e^{-\lambda}\,{\lambda}^0}{0!}}=\frac{e^{-\lambda}\,{\lambda}^2}{2!}$

$\implies\frac{e^{-\lambda}.1}{1}}=\frac{e^{-\lambda}\,{\lambda}^2}{2}$

$\implies\,1=\frac{{\lambda}^2}{2}$

$\implies\,{\lambda}^2=2$

$\implies\,\lambda=\sqrt{2}$

Now,

$P(3)=\frac{e^{-\sqrt{2}}\,(\sqrt{2})^3}{3!}$

$P(3)=\frac{e^{-\sqrt{2}}\,2\sqrt{2}}{6}$

$P(3)=\frac{e^{-\sqrt{2}}\,\sqrt{2}}{3}$

Answer 2

p(3)=$\frac{\sqrt2e^{-\sqrt2}}{3}$

Step-by-step explanation:

We are given that X is random variable

p(0)=p(2)

The probability mass function is given by

$f(x,\lambda)=p(x=k)=\frac{\lambda^{k}e^{-\lambda}}{k!}$

Where

$\lambda$=Mean value of  X=V(X)>0

$p(x=k)$=Probability at x=k

$p(x=0)=\frac{\lambda^{0}e^{-\lambda}}{0!}=e^{-\lambda}$

$p(2)=\frac{\lambda^2e^{-\lambda}}{2!}$

$\frac{\lambda^2e^{-\lambda}}{2!}=e^{-\lambda}$

$\lambda^2=2!=2\times 1=2$

$\lambda=\sqrt2$

Substitute the values then we get

$p(3)=p(x=3)=\frac{(\sqrt2)^3e^{-\sqrt2}}{3!}=\frac{e^{-\sqrt2}2\sqrt2}{3\times 2\times 1}=\frac{\sqrt2e^{-\sqrt2}}{3}$

$p(3)=\frac{\sqrt2e^{-\sqrt2}}{3}$

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Explaining of poisson distribution

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