## Poisson Distribution :

1. It has been observed that the average number of traffic accidents requiring medical assistance on the Hollywood Freeway between 7 and 8 AM on Wednesday mornings is 1. What, then, is the chance that there will be a need for exactly 2 ambulances on the Freeway, during that time slot on any given Wednesday morning? The hospital dispatcher needs to know?
2. Solution:
• P(X2) = (e-1 12) / 2!
• = 2.71828-1 / 2
• = 0.368 / 2
• = 0.184

3. A soap salesman sells on the average 4 soaps per week. Use Poisson's law to calculate the probability that in a given week he will sell
4. 1. Some of the soaps.

2. 2or more soaps but less than 6 soaps.

Assuming that there are 6 working days per week, what is the probability that in a given day he will sell one soaps?

Solution:
• P(x;μ) = (e-μ) (μx) / x!
• Some soaps indicate 1 or more soaps can also be written as
• = 1 – no soaps
• = 1 – P(Xo)
• = 1 - (e-4) (40) / 0!
• = 0.9816
• P (2 ≤ X < 6)
• =P(X2) + P(X3)+ P(X4)+ P(X5)
• = (e-4) (42) / 2! * (e-4) (43) / 3! * (e-4) (44) / 4! * (e-4) (45) / 5!
• = 0.147 + 0.195 + 0.195+ 0.156
• = 0.69
• Average number of soaps sold per day = 4/6 = 0.067
• In a given day = (e-0.067) (0.067) / 1!
• = 0.063

5. ) If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.
6. Solution:
• The average number of failures per week is μ = 3/20 = 0.15
• "Not more than one failure" means we need to include the probabilities for 0 failures plus 1 failure
• P(X0) + P(X1) = (e-0.15 * 0.150) / 0! + (e-0.15 * 0.151) / 1!
• = 0.98981

7. Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
8. 1. Find the probability that none passes in a given minute.

2. What is the expected number passing in two minutes?

3. Find the probability that this expected number actually pass through in a given two-minute period.

Solution:
• 1. a) The average number of cars per minute is:μ= 300/60 = 5
• P(X0) = (e-5 50) / 0!
• = 6.7379 * 10-3
• 1. Expected number each 2 minutes =E(X) = 5 × 2 = 10
• 2. Now, withμ = 10, we have:
• P(X10) = (e-10 1010) / 10!
• = 0.125

9. A company makes electric motors. The probability an electric motor is defective is 0.01. What is the probability that a sample of 300 electric motors will contain exactly 5 defective motors?
10. Solution:
• The average number of defectives in 300 motors is μ = 0.01 × 300 = 3
• The probability of getting 5 defectives is:
• P(X5) = (e-3 35) / 5!
• = 0.10082

11. The average number of flats sold by Troy realty is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow?
12. Solution:
• We plug these values into the Poisson formula as follows:
• P(x) = (e) (μx/ x!)
• P(x3) = (2.71828-2) (23) / 3!
• P(x3)= (0.13534) (8) / 6
• P(x3) = 0.180
• Thus, the probability of selling 3 homes tomorrow is 0.180.

13. The average number of Tigers seen on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari?
14. Solution:
• =P(X0) + P(X1)+ P(X2)+ P(X3)
• =[(e-5)(50)/0!]+[(e-5)(51)/1!]+[(e-5)(52)/2!]+[(e-5)(53)/3!]
• = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ]
• = 0.2650