# Poisson Question

• Dec 11th 2009, 07:30 PM
Janu42
Poisson Question
The number X of new cases of food poisoning reported in a random month in a cafeteria has a Poisson distribution with mean 0.5. The cafeteria is required by law to request a state inspection of its kitchen whenever 2 or more cases of food poisoning occur in a month.
a) What is the probability that the cafeteria will have to request an inspection this month?
b) The number of new cases of food poisoning in any given month is independent of the number of cases of food poisoning in all other months of the year. Let Z be the number of months (out of a twelve month year) that the cafeteria must be inspected by the state. Find the probability that the cafeteria will be inspected more than once in a year.
• Dec 11th 2009, 08:33 PM
Danneedshelp
Quote:

Originally Posted by Janu42
The number X of new cases of food poisoning reported in a random month in a cafeteria has a Poisson distribution with mean 0.5. The cafeteria is required by law to request a state inspection of its kitchen whenever 2 or more cases of food poisoning occur in a month.
a) What is the probability that the cafeteria will have to request an inspection this month?
b) The number of new cases of food poisoning in any given month is independent of the number of cases of food poisoning in all other months of the year. Let Z be the number of months (out of a twelve month year) that the cafeteria must be inspected by the state. Find the probability that the cafeteria will be inspected more than once in a year.

a) Well, we know $\displaystyle \lambda=0.5$, the average number food poisoning cases over a month period. We want to know the probability of two or more cases, since we are asked to find the probability of a cafeteria inspection. So, you want to find

$\displaystyle P(X\geq\\2)=\sum_{x=2}^{\infty}\frac{\lambda^{x}\\ e^{-\lambda}}{x!}$ or equivalently $\displaystyle 1-\sum_{x=0}^{1}\frac{\lambda^{x}\\e^{-\lambda}}{x!}$, with $\displaystyle \lambda=0.5$.
• Dec 11th 2009, 08:58 PM
matheagle
I assume you want the first sum to start at 2.

(b) is Binomial, n=12 and p is the probability from part (a).

You want $\displaystyle P(Z> 1)=1-[P(Z=0)+P(Z=1)]$
• Dec 11th 2009, 10:00 PM
matheagle
I just noticed the exponent of e is off, you need $\displaystyle -\lambda$
• Dec 11th 2009, 10:51 PM
Danneedshelp
Quote:

Originally Posted by matheagle
I just noticed the exponent of e is off, you need $\displaystyle -\lambda$

Wow, good save. Sorry about that.
• Dec 11th 2009, 11:31 PM
matheagle
NP, that's what my mathEAGLE eyes r4.
• Dec 12th 2009, 09:32 PM
Janu42
Quote:

Originally Posted by matheagle
I assume you want the first sum to start at 2.

(b) is Binomial, n=12 and p is the probability from part (a).

You want $\displaystyle P(Z> 1)=1-[P(Z=0)+P(Z=1)]$

So p from part (a) is the probability it gets inspected in a random month. n is obviously 12 for months of the year, so I use the simple binomial formula right? To get P ( Z = 0) and P (Z = 1)?
• Dec 13th 2009, 03:52 AM
mr fantastic
Quote:

Originally Posted by Janu42
So p from part (a) is the probability it gets inspected in a random month. n is obviously 12 for months of the year, so I use the simple binomial formula right? To get P ( Z = 0) and P (Z = 1)?

What you have done is stated exactly what is in the quote. matheagle is crystal clear on what yuo have to do.