How do I find the probability when
Poisson random variable = 4 
What is probability for P(X=5) ,  P(X=0)  ,  P(X>1)

20% people use motorway 1 and 10% take motorway 2. People who take motorway 1, 
40% also take motorway 2.

what percentage of people take motorway 1 and motorway 2.
What percentage of people do not take either motorway.

I would appreciate your help to solve the above problems.
Elshana

Hi elshana!!

Problem 1:
Poisson random variable = 4  
What is probability for P(X=5) ,  P(X=0)  ,  P(X>1) 

For Poisson distribution with expected value (Poisson parameter) h we
have that:
          e^-h * h^k  
P(X=k) = ------------
              k!


For this problem we have that h = 4, then:

P(X=5) = e^-4 * 4^5 / 5! =
       = 0.01831563 * 1024 / 120 = 
       = 0.15629338

P(X=0) = e^-4 * 4^0 / 0! =
       = 0.01831563 * 1 / 1 =
       = 0.01831563

P(X>1) = 1 - P(X=<1) = 1 - [P(X=0) + P(X=1)];

P(X=1) = e^-4 * 4^1 / 1! = 0.01831563 * 4 / 1 = 0.07326252;

then:
P(X>1) = 1 - [0.01831563 + 0.07326252] = 
       = 1 - 0.09157815 =
       = 0.90842185

For reference see:
Poisson distribution From Wikipedia:
http://www.wikipedia.org/wiki/Poisson_distribution

 -----------------------------------------------------

Problem 2:
20% people use motorway 1 and 10% take motorway 2. People who take
motorway 1 (M1), 40% also take motorway 2 (M2).
 
what percentage of people take motorway 1 and motorway 2. 

This is a Conditional probability problem and we must use the
following formula:
P(M1 and M2) = P(M1) . P(M2/M1) =
             = 0.20 . 0.40 = 
             = 0.08 = 8%

A little explanation:
The % of people who take M1 is 20%, from this 20% of the total of
people only the 40% also take M2, then the % of people taking M1 and
M2 is the 40% of the 20% of the total:
0.40 . 0.20 . 1 = 0.08.


What percentage of people do not take either motorway.

P(no M1 and no M2) = 100% - P(M1 or M2) =
                   = 100% - [P(M1) + P(M2) - P(M1 and M2)] =
                   = 100% - [20% + 10 % - 8%] =
                   = 100% - 22% =
                   = 78%

For reference with Probability and Conditional Probability visit the
following pages:
"Conditional Probability":
http://www.fmi.uni-sofia.bg/vesta/Virtual_Labs/prob/prob5.html

"Stats: Probability Rules":
http://www.richland.cc.il.us/james/lecture/m170/ch05-rul.html


I hope this helps you, if you need a clarification, please post a
request for it and I will glad to reply to your request.

Best regards.
livioflores-ga

thank you for your input 
Elshana