Math Help - Poisson Probabilities

1. Poisson Probabilities

I have the following questions.

The number of episodes per year of obtaining a certain disease follows poisson distribution has $\mu = 1.6$

Find the probability of 3 or more episodes occuring over 2 years.

I have $\mu = 1.6\times 2 = 3.2$

$\displaystyle P(X\geq 3) = 1- (P(X=0)+P(X=1)+P(X=2) )$

$= 1-\left(\frac{e^{3.2}\times 3.2^0}{0!}+\dots \right) = 0.62$

Happy with that.

Find the probability of not getting any episodes of the disease in a certain year

so $\displaystyle P(X=0) = \frac{e^{3.2}\times 3.2^0}{0!}= 0.201897$

Also happy with that.

What is the probability that 2 siblings will both have 3 or more episodes in the first to years?

Is this binomial given p = 0.62 finding P(X=2)? but what would n be?

Or is it still poisson? maybe a different $\mu = 0.62$

Thx!

2. Re: Poisson Probabilities

Originally Posted by Bushy
I have the following questions.

The number of episodes per year of obtaining a certain disease follows poisson distribution has $\mu = 1.6$

Find the probability of 3 or more episodes occuring over 2 years.

I have $\mu = 1.6\times 2 = 3.2$.

$\displaystyle P(X\geq 3) = 1- (P(X=0)+P(X=1)+P(X=2) )$

$= 1-\left(\frac{e^{3.2}\times 3.2^0}{0!}+\dots \right) = 0.62$

Happy with that.

Find the probability of not getting any episodes of the disease in a certain year

so $\displaystyle P(X=0) = \frac{e^{3.2}\times 3.2^0}{0!}= 0.201897$

Also happy with that.

What is the probability that 2 siblings will both have 3 or more episodes in the first to years?

Is this binomial given p = 0.62 finding P(X=2)? but what would n be?

Or is it still poisson? maybe a different $\mu = 0.62$

Thx!
Calculate $p = \Pr(X \geq 3)$

Let Y be the random variable 'Number of children that have an episode in the first two years'.

Then Y ~ Binomial(p = from above, n = 2).

Calculate Pr(Y = 2). The answer will be p^2 = ....