The PMF of a random varlable $$ X $$ taking value in $$ N \cup\{0\} $$ is given by $$ P_{I}(i)=e^{-\lambda} \frac{\lambda^{i}}{n}, i=0,1,2, \ldots $$ where $$ \lambda $$ is a positive real number. Given that $$ \lambda=2 $$, find $$ P(X=0) $$ $$ P(X2) $$

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The PMF of a random varlable $$ X $$ taking value in $$ N \cup\{0\} $$ is given by $$ P_{I}(i)=e^{-\lambda} \frac{\lambda^{i}}{n}, i=0,1,2, \ldots $$ where $$ \lambda $$ is a positive real number. Given that $$ \lambda=2 $$, find $$ P(X=0) $$ $$ P(X>2) $$

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To find the probabilities $$ P(X=0) $$ and $$ P(X>2) $$ for the given probability mass function (PMF) of the random variable $$ X $$, we start with the provided PMF:

$$
P_X(i) = e^{-\lambda} \frac{\lambda^i}{n}, \quad i = 0, 1, 2, \ldots
$$

where $$ \lambda = 2 $$ and $$ n $$ is a normalization constant.

### Step 1: Find the normalization constant $$ n $$

To find $$ n $$, we need to ensure that the total probability sums to 1:

$$
\sum_{i=0}^{\infty} P_X(i) = 1
$$

Substituting the PMF into the sum:

$$
\sum_{i=0}^{\infty} e^{-\lambda} \frac{\lambda^i}{n} = e^{-\lambda} \frac{1}{n} \sum_{i=0}^{\infty} \lambda^i
$$

The series $$ \sum_{i=0}^{\infty} \lambda^i $$ diverges unless we normalize it properly. The correct form for a PMF that resembles a Poisson distribution is:

$$
P_X(i) = e^{-\lambda} \frac{\lambda^i}{i!}
$$

Thus, we can identify $$ n = i! $$ in the denominator.

### Step 2: Calculate $$ P(X=0) $$

Now, substituting $$ \lambda = 2 $$ into the PMF for $$ i = 0 $$:

$$
P(X=0) = e^{-2} \frac{2^0}{0!} = e^{-2} \cdot 1 = e^{-2}
$$

### Step 3: Calculate $$ P(X>2) $$

To find $$ P(X>2) $$, we can use the complement rule:

$$
P(X>2) = 1 - P(X \leq 2) = 1 - (P(X=0) + P(X=1) + P(X=2))
$$

Calculating each term:

1. **For $$ P(X=1) $$**:

$$
P(X=1) = e^{-2} \frac{2^1}{1!} = e^{-2} \cdot 2
$$

2. **For $$ P(X=2) $$**:

$$
P(X=2) = e^{-2} \frac{2^2}{2!} = e^{-2} \cdot \frac{4}{2} = e^{-2} \cdot 2
$$

Now, summing these probabilities:

$$
P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2}
$$

Thus,

$$
P(X>2) = 1 - P(X \leq 2) = 1 - 5e^{-2}
$$

### Final Answers

1. $$ P(X=0) = e^{-2} $$
2. $$ P(X>2) = 1 - 5e^{-2} $$
$$ P(X=0) = e^{-2} $$ $$ P(X>2) = 1 - 5e^{-2} $$

The PMF of a random varlable taking value in is given by

where is a positive real number.
Given that , find

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The PMF of a random varlable taking value in is given by where is a positive real number. Given that , find