Question

X is poisson variate with parameter 3. find the probability that x assumes the values i) 3,2,1,0 ii) less than 3 iii) at least 2

Answer 1

Solution: We are given that X follows Poisson distribution with parameter 3.

i) x = 0, 1, 2, 3

Answer: We know that if X follows poisson distribution with parameter λ. Then the probability distribution is:

$P(X=x) = \frac{e^{-\lambda} \lambda^{x}}{x!}$

We are required to find:

$P(x=0) = \frac{e^{-3}3^{0} }{0!}$

                  $=0.0498$

$P(x=1) = \frac{e^{-3}3^{1} }{1!}$

                  $=0.1494$

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

                  $=0.2240$

$P(x=3) = \frac{e^{-3}3^{3} }{3!}$

                  $=0.2240$

ii) less than 3

Answer: We have to find here

$P(x<3) = P(x=0) +P(x=1) +P(x=2)$

          $=0.0498+0.1494+0.2240=0.4232$

iii) at least 2

Answer: We have to find here:

$P(x\geq 2)$

We know that:

$P(x\geq 2) = 1-P(x<2)$

                        $=1- (P(x=0) +P(x =1))$

                        $=1 - (0.0498+0.1494)$

                        $=1-0.1991$

                        $=0.8009$