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probability rate problem

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
Serious railway accidents occur in a Poisson process of rate R per year. It is known that on average, in 9 years out of 10 there are no serious railway accidents.

A.what is the value of R?
I got 0.1 but not sure how exactly got it, and is 0.1 correct?

B.what is the probability that exactly 2 serious accidents will occur in any 2-year period?
I used this formula

(e^-R*R^k)/k!=(e^0.2*0.2^2)/2!=0.01637.... so about 1.6%

is this correct? can anyone correct the formular and teach me how to solve this one using proper mathy way.

many thanks in advance.



The last question.

I can't really recall the concept of this matter.

X is a random variable distibuted as Geo(p). Find the distribution function of X?

thank you!!
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,706
\(\displaystyle \frac{0.10^{0}\cdot e^{-0.10}}{0!}\;=\;0.9048\)

That seems a little too big. Don't guess. Find an analytical solution. Go back to teh definition if nothing else presents itself.

You have:
-----1) Poisson
-----2) 0.90 is Pr(0)

That is enough

\(\displaystyle \frac{\lambda^{0}\cdot e^{-\lambda}}{0!}\;=\;0.90\)

Looks like a little more than 0.10. Let's see what you get.

For the second part, if you stated that your parameter is 0.10, why did you start using 0.20? That makes no sense. Also, you had better check the sign on the exponent on 'e'.

MAKE SURE you are starting in the right place.
 

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
I got the parameter which is 0.105360...

so about 0.11 right?

I used 0.2 because of 2-year period=0.1*2

I was confused,,I should have applied it to k part only. right?

also I found the mistake of omitting the minus sign there as you said,,thanks for pointing out.

so we set parameter as 0.11 then it should be

p(x=2)=(e^-0.11*0.11^2)/2!=5.42*10^-3

is this correct? do I need to plug the actual value of parameter? cos when you apply it the answer is quite different

one with actual parameter is 4.99*10^-3 and for parameter 0.11 I get 5.42*10^-3

which way is correct?
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,706
Ah! 2-year period. I read that wrong. Good work. Poisson is generally scalable just as you have done it.

I do prefer a few more decimal places. 2 really is inadequate.
 
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