- 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?
*P(X*_{2}) = (e^{-1}1^{2}) / 2!- = 2.71828
^{-1}/ 2 - = 0.368 / 2
*= 0.184***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***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(X
_{2}) + P(X_{3})+ P(X_{4})+ P(X_{5}) - = (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***) 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.***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(X
_{0}) + P(X_{1}) = (e^{-0.15}* 0.150) / 0! + (e^{-0.15}* 0.151) / 1! *= 0.98981***Vehicles pass through a junction on a busy road at an average rate of 300 per hour.***1. a) The average number of cars per minute is:μ= 300/60 = 5*- P(X
_{0}) = (e^{-5}5^{0}) / 0! - = 6.7379 * 10
^{-3}

- 1. Expected number each 2 minutes =E(X) = 5 × 2 = 10
- 2. Now, withμ = 10, we have:
- P(X
_{10}) = (e^{-10}10^{10}) / 10! *= 0.125***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?***The average number of defectives in 300 motors is μ = 0.01 × 300 = 3*- The probability of getting 5 defectives is:
- P(X
_{5}) = (e^{-3}3^{5}) / 5! *= 0.10082***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?***We plug these values into the Poisson formula as follows:*- P(x) = (e
^{-μ}) (μ^{x}/ x!) - P(x
_{3}) = (2.71828^{-2}) (2^{3}) / 3! - P(x
_{3})= (0.13534) (8) / 6 - P(x
_{3}) = 0.180 *Thus, the probability of selling 3 homes tomorrow is 0.180.***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?***=P(X*_{0}) + P(X_{1})+ P(X_{2})+ P(X_{3})- =[(e
^{-5})(5^{0})/0!]+[(e^{-5})(5^{1})/1!]+[(e^{-5})(5^{2})/2!]+[(e^{-5})(5^{3})/3!] - = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ]
*= 0.2650*

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?**

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.