# Poisson Problem

• Jan 6th 2009, 11:16 PM
zorro
Poisson Problem
If the number of accidents occuring in an industrial plant during aa day is given by a Poisson random variable with parameter 3.
Find
i) probability that no accident occurs on a day
ii)the expected number of accidents per day.
• Jan 7th 2009, 05:53 AM
mr fantastic
Quote:

Originally Posted by zorro
If the number of accidents occuring in an industrial plant during aa day is given by a Poisson random variable with parameter 3.
Find
i) probability that no accident occurs on a day
ii)the expected number of accidents per day.

These questions are totally trivial and suggest a complete lack of effort on your part in trying to understand this material. Go to your class notes and/or textbook and read the section on the Poisson distribution.
• Jan 7th 2009, 09:31 PM
zorro
Then is this right
probability of an accident happening in a day is
$\displaystyle p(x;\lambda) \ = \ \frac{\lambda^x \ e^{- \lambda}}{x!} \ = \ \frac{3^x \ e^{-3}}{x!} = \frac{3^x \ 0.050}{x!}$

probability of accident happening is
$\displaystyle \left ( 1 - \frac{3^x \ 0.050}{x!} \right)$

Expected no. of accidents per day is
$\displaystyle E(X) = \sum x . p(x;\lambda) = \sum x \ . \ \frac{3^x \ . \ 0.050}{x!}$
• Jan 7th 2009, 10:51 PM
mr fantastic
Quote:

Originally Posted by zorro
probability of an accident happening in a day is
$\displaystyle p(x;\lambda) \ = \ \frac{\lambda^x \ e^{- \lambda}}{x!} \ = \ \frac{3^x \ e^{-3}}{x!} = \frac{3^x \ 0.050}{x!}$

probability of accident happening is
$\displaystyle \left ( 1 - \frac{3^x \ 0.050}{x!} \right)$

Expected no. of accidents per day is
$\displaystyle E(X) = \sum x . p(x;\lambda) = \sum x \ . \ \frac{3^x \ . \ 0.050}{x!}$

$\displaystyle \Pr(\text{accident}) = 1 - \Pr(\text{no accident}) = 1 - \Pr(X = 0) = 1 - e^{-3} = \, ....$

$\displaystyle E(X) = \lambda = 3$.

You need to thoroughly review this topic - the questions, especially the second question, are extremely straightforward.
• Jan 8th 2009, 12:04 AM
zorro
hw did u get E[x]
Quote:

Originally Posted by mr fantastic
$\displaystyle \Pr(\text{accident}) = 1 - \Pr(\text{no accident}) = 1 - \Pr(X = 0) = 1 - e^{-3} = \, ....$

$\displaystyle E(X) = \lambda = 3$.

You need to thoroughly review this topic - the questions, especially the second question, are extremely straightforward.

Isnt

$\displaystyle E[x] \ =\ \sum x .\ [1-p(x;\lambda)]= \ 0 \ . \ [1-p(x;\lambda) ]=0$
• Jan 8th 2009, 01:29 AM
mr fantastic
Quote:

Originally Posted by zorro
Isnt

$\displaystyle E[x] \ =\ \sum x .\ [1-p(x;\lambda)]= \ 0 \ . \ [1-p(x;\lambda) ]=0$

No.

$\displaystyle E[X] = \sum x \cdot p(x; \, \lambda)$.

And the result of this calculation (which can be used without proof I would have thought unless the calculation is specifically asked for: http://www.mathhelpforum.com/math-he...tributiom.html) is $\displaystyle E[X] = \lambda$.