請教3條PROBABILITY既問題

輕描。淡寫

2011-08-03 7:44 am

1) It is known that the number of breakdowns of a printer in a year follows a Poisson distribution with mean 4. Suppose the printer manufacturer now offers a yearly maintenance plan to customers as follows:*The first 3 maintenances in the year are free of charge*The maintenance fee is $1000 each time after the first maintenances.If a customer joins the above maintenance plan at the beginning of a year, finda) The probability that the customer does not need to pay any maintenance fees in the year[ans: 0.4335]b) The expected maintenance fee in the year.[ans: $1348]呢題問B PART

2) Every weekend, there are 5 electronic appliances on sale in a department store. Suppose the number of customers buying on-sale appliances follows a Poisson distribution with mean 5. Find the probability that all the 5 electronic appliances will be sold this weekend.[ans: 0.5595]

3) A company has 10 parking spaces for its staff. Suppose 14 of the staff own private cars and each of them independently has a probability of 0.7 of driving his or her car to work on any weekday. Find the probability that the parking spaces are full on a weekday.[ans: 0.5842]

更新1:

請注意﹗﹗﹗ 第十條唔使答了，改為下面呢一題︰ A car rental company finds that the weekly demand for cars follows a Poisson distribution with mean 6.4

更新2:

問題為︰find the minimum number of cars that the company should keep each week so that the company will have a chance of at least 90% of meeting the demand in the coming week. [ANS: 10]

更新3:

第三條唔使答先岩!!!

回答 (1)

用戶10012

2011-08-04 3:35 pm

✔ 最佳答案

1b. Since no maintenance fee for the first 3 services, fee starts from the 4th service.
For 1 new service after the free period, fee = $1000 x P(X = 4) = $1000 x e^(-4)4^4/4!
For 2 new services after the free period, fee = ($1000 + $1000) x Probability of 2 services after the free period = $2000 x P(X = 5) = $2000 x e^(-4)4^5/5! ans so on.
So expected maintenance fee = $1000e^(-4)4^4/4! + $2000e^(-4)4^5/5! + $3000e^(-4)4^6/6! + .........
= $1000e^(-4)[4^4/4! + 2(4^5)/5! + 3(4^6)/6! + ..................]
Sum of the sequence can be calculated to be = 19 + e^4 ( calculation is quite complicated). So expected maintenance fee = $1000e^(-4)[19 + e^4] = $1348.
2.
P(All 5 items are sold out) = P(At least 5 customers) = P(X > 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) - P(X = 4) = 1 - e^(-5)[1 + 5 + 25/2 + 125/6 + 625/24] = 1 - e^(-5)(65.375) = 0.5595.
3.
P(Keeping n cars) = P(X < n) = e^(-6.4)[1 + 6.4 + 6.4^2/2 + 6.4^3/3! + 6.4^4/4! + ..... + 6.4^n/n!)
For n = 9, P(X < 9) = e^(-6.4)[ 1 + 6.4 + .... + 6.4^9/9!] which is calculated to be less than 90%.
For n = 10, P(X < 10) = P(X < 9) + P( X = 10) which is calculated to be greater than 90%.
So the minimum no. of cars = 10.

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