# Consider a Poisson distribution with μ=3. What is P(x≥2)?

Jan 13, 2017

The answer is $= 0.8009$

#### Explanation:

The Poisson distribution is

P(x)=(e^-mu mu^x)/(x!)

$\mu = 3$

3! =3*2*1

$P \left(x \ge 2\right) = 1 - P \left(1\right) - P \left(0\right)$

P(1)=(e^-3 *3^1)/(1!)=3/e^3

$P \left(0\right) = {e}^{-} 3 = \frac{1}{e} ^ 3$

Therefore,

$P \left(x \ge 2\right) = 1 - P \left(1\right) - P \left(0\right) = 1 - 0.0498 - 0.1494 = 0.8009$

Jan 13, 2017

$P \left(x \ge 2\right) = 0.800852$

#### Explanation:

In a Poisson probability distribution, if mean value of success is $\mu$,

the probability of getting $x$ successes is given by

P(x)=(e^(-mu)mu^x)/(x!)

Now $P \left(x \ge 2\right)$ means $1 - P \left(x = 0\right) - P \left(x = 1\right)$

Here $\mu = 3$ and ${e}^{- \mu} = {e}^{- 3} = 0.049787$

and hence, desired probability is

1-(e^(-3)3^0)/(0!)-(e^(-3)3^1)/(1!)

= $1 - {e}^{- 3} \times \left(1 + 3\right)$

= $1 - 0.049787 \times 4$

= $1 - 0.199148$

= $0.800852$