For a Poisson variate X,if P(X=2)=3 P(X=3),th | vedclass

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(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda$ is the mean of th

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$1$

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(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda$ is the mean of the distribution. Given the condition $P(X=2) = 3 P(X=3)$. Substituting the formula: $\frac{e^{-\lambda} \lambda^2}{2!} = 3 \cdot \frac{e^{-\lambda} \lambda^3}{3!}$ Dividing both sides by $e^{-\lambda} \lambda^2$ (assuming $\lambda 
eq 0$): $\frac{1}{2} = 3 \cdot \frac{\lambda}{3 \cdot 2 \cdot 1}$ $\frac{1}{2} = \frac{3 \lambda}{6}$ $\frac{1}{2} = \frac{\lambda}{2}$ $\lambda = 1$. Thus,the mean of the Poisson distribution is $1$.

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

Outcome	Probability
$\omega_{1}$	$1/8$
$\omega_{2}$	$2/3$
$\omega_{3}$	$1/3$
$\omega_{4}$	$1/3$
$\omega_{5}$	$-1/4$
$\omega_{6}$	$-1/3$

For a Poisson variate $X$,if $P(X=2)=3 P(X=3)$,then the mean of $X$ is:

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If the probability distribution of a random variable $X$ is given by the following table,then the mean of $X$ is:
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Outcome Probability

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$\omega_{3}$ $1/3$

$\omega_{4}$ $1/3$

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$A$ fair six-faced die is rolled $12$ times. The probability that each face turns up exactly twice is equal to:

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