# Passion Distribution

## Passion Distribution

It is limiting case of binomial distribution. Let the number of events n is large (n→∞) and probability of success in each experiment is 0 and np = λ (say)

$$P(X=r)\ (or)\ P(r)=\frac{{{e}^{-\lambda }}{{\lambda }^{r}}}{r!}$$.

Where r = 0, 1, 2, …

And λ = np. Here, λ is know as parameter of Poisson distribution.

P (r + 1) = λ/ (r + 1) P(r) is known as recurrence formula

Note:

$$\sum\limits_{r=0}^{\infty }{P(r)=1}$$.

If λ₁ and λ₂ are parameter of variables X and Y, then parameter of (x + y) will be (λ₁ + λ₂)

In poisson distribution, mean = variance = λ = np

Example: if mean of a poisson distribution of a random variable x is 2, then the value of P (X > 1.5) is

Solution: It is limiting case of binomial distribution. Let the number of events n is large (n → ∞) and probability of success in each experiment is 0 and np = λ (say)

$$P(X=r)\ (or)\ P(r)=\frac{{{e}^{-\lambda }}{{\lambda }^{r}}}{r!}$$.

P (X = r > 15) = P (2) + P (3) + … + ∞

= 1 – P (X = r ≤ 1)

= 1 – P (0) – P (1)

$$=1-\left( \frac{{{e}^{-2}}\times {{2}^{0}}}{0!}+\frac{{{e}^{-2}}\times {{2}^{1}}}{1!} \right)$$.

$$=1-\left( \frac{{{e}^{-2}}\times {{2}^{0}}}{1}+\frac{{{e}^{-2}}\times {{2}^{1}}}{1} \right)$$.

$$=1-\left( {{e}^{-2}}+\frac{{{e}^{-2}}\times 2}{1} \right)$$.

= 1 – e⁻² – 2e⁻²

= 1 – 1/e² – 2/e²

= 1 – 3/e².