Question

A variate X follows Poisson distribution with
parameter 3. If e 3 = 0.0489, then probability
of exactly 1 success p(1) is
O a. 0.9004
O b. 0.1494
O c. 0.0894
O d. 0.0489​

Answer 1

Appropriate Question:

A variate X follows Poisson distribution with parameter 3. If $\sf \: e^{- 3}$ = 0.0498, then probability of exactly 1 success, p(1) is

O a. 0.9004

O b. 0.1494

O c. 0.0894

O d. 0.0489

Answer:

$\boxed{\bf \: P(1) = 0.1494 \: }$

Step-by-step explanation:

Given that, a variate X follows Poisson distribution with

parameter 3.

Let m denotes the mean of Poisson distribution.

$\implies\sf \: m = 3$

We know, Probability of success, where r is any random variable associated with the event, having mean m is given by

$\sf \: \boxed{\sf \: P(r) = \dfrac{ {e}^{ - m} \: {m}^{r} }{r!} \: }$

So, on substituting the values, we get

$\sf \: P(1) = \dfrac{ {e}^{ - 3} \times {3}^{1} }{1!} \:$

$\sf \: P(1) = 3 {e}^{ - 3} \:$

$\sf \: P(1) = 3 \times 0.0498 \:$

$\sf \: P(1) = 0.1494 \:$

Hence,

$\implies\sf \: \boxed{\bf \: P(1) = 0.1494 \: }$

$\rule{190pt}{2pt}$

Additional Information:

In a Poisson distribution, mean and variance, both are same and is given by

$\boxed{\sf \: Mean = Variance = np \: }$

where,

n is number of trials

p is probability of success.