Chapter 4. Discrete Probability Distributions Sections 4.7, 4.8: Poisson and Hypergeometric Distributions

Jiaping Wang Department of Mathematical Science 03/04/2013, Monday The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Outline

Poisson: Probability Function Poisson: Mean and Variance Hypergeometric: Probability Function Hypergeometric: Mean and Variance More Examples The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Part 1. Poisson: Probability Function

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Consider the probability of the number of accidents that occur at a particular highway intersection in a period of one week.

Split one week into n subintervals such that P(One accident in a subinterval)=p P(No accident in a subinterval) = 1-p Here we assume p holds for all subintervals and P(More than two accidents in a subinterval)=0 Also assume n



∞, p



0 but np=λ.

So, the probability that x accidents happen in these n subintervals is

𝒏 λ 𝒙 𝟏 − λ 𝒙 𝒏 𝒏 𝒏 − 𝒙

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Probability Function

The Poisson probability function:

P(X=x)=p(x)= λ 𝑥 𝑥!

𝑒 − 𝑥 , x= 0, 1, 2, …., for λ> 0

The distribution function is

F(x)=P(X≤x)= 𝑥 𝑖=0 λ 𝑖 𝑒 − 𝑖 𝑖!

Recall that λ denotes the mean number of occurrences in one time period, if there are t non-overlapped time periods, then the mean would be λt. Poisson distribution is often referred to as the distribution of rare events.

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Example 4.22

During business hours, the number of calls passing through a particular relay system averages five per minute.

1. Find the probability that no call will pass through the relay system during a given minute.

2. Find the probability that no calls will pass through the relay system during a 2-minute period.

3. Find the probability that three calls will pass through the relay system during a 2-minute period.

Answer: 1. λ = 5, so P(X=0)=p(0)=5 0 /0! e -5 = e -5 = 0.007.

2. λt=2x5=10, so P(X=0)=p(0)= 10 0 /0! e -10 = e -10 = 0.00005.

3. λt=2x5=10, so P(X=3)=p3)= 10 3 /3! e -10 = 0.0076.

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Example 4.23

Refer to Example 4.22, find the following probabilities: 1. No more than four calls in the given minute.

2. At least four calls in the given minute.

3. Exactly four calls in the given minute.

Answer: λ = 5, 1. so P(X ≤ 4)=F(4)= 4 𝑥=0 5 𝑥 𝑥!

𝑒 − 5 =0.44.

2. P(X ≥ 4) = 1-P(X ≤ 3) = 1 – F(3) = 1 3 𝑥=0 5 𝑥 𝑥!

𝑒 − 5 = 1 − 0.265 = 0.735

.

3. P(X=4) = P(X ≤ 4) – P(X ≤ 3) = F(4) – F(3) = 0.44 – 0.265 = 0.175.

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Part 2. Poisson:Mean and Variance

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Mean and Variance

𝒆 𝒙 = 𝟏 + 𝒙 + 𝒙 𝟐 + 𝟐!

𝒙 𝟑 𝟑!

+ … Similarly, we can find V(X)= λ. So E(X)= V(X) = λ for Poisson random variable.

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Example 4.24

The manager of an industrial plant is planning to buy a new machine of either type A or type B. For each day’s operation, the number of repairs that machine A requires is a Poisson random variable with mean 0.10t, where t is the time (in hours) of daily operation. The number of daily repairs Y for machine B is a Poisson random variable with mean 0.12t. The daily cost of operating A is C A (t)=20t+40X 2 ; for B, the cost is C B (t)=16t+40Y 2 . Assume that the repairs take negligible time and that each night the machines are to be closed so that they operate like new machines at the start of each day. Which machine minimizes the expected daily cost for the following times of daily operation? 1. 10 hour; 2. 20 hours.

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Answer: E[C A (t)]=20t+40E(X 2 ) = 20t+40[V(X)+E 2 (X)] = 24t+0.4t

2 .

E[C B (t)]=16t+40E(Y 2 ) = 16t+40[V(X)+E 2 (Y)] =20.8t+0.576t

2 .

For Option I: 10 hours. E[C A (t)]= 24(10) + 0.4(10) 2 E[C B (t)]=20.8(10)+0.576(10) 2 = 280.

= 265.6.

which results in the choice of B.

For Option 2: 20 hours. E[C A (t)]= 24(20) + 0.4(20) 2 E[C B (t)]=20.8(20)+0.576(20) 2 = 640.

= 646.4.

which results in the choice of A.

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Part 3. Hypergeometric: Probability Function

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Until now, we learned distributions based on the independent Bernoulli trials. Now considering a dependent case. Example: A lot has 10 computer chips, of which 4 are defective. Now we select two chips randomly without replacement, what is the probability of choosing one defective chip?

Answer: P(X=1)=

𝟒 𝟏 𝟏𝟎−𝟒 𝟏 / 𝟏𝟎 𝟐

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Probability Function

Now we consider a general case: Suppose a lot consists of N items, of which k are of one type (called successes) and N-k are of another type (called failures). Now n items are sampled randomly and sequentially without replacement. Let X denote the number of successes among the n sampled items. So What is P(X=x) for some integer x?

The probability function is: P(X=x) = p(x) = 𝑘 𝑥 𝑁−𝑘 𝑛−𝑥 𝑁 𝑛 , 0 ≤ 𝑥 ≤ 𝑘 ≤ 𝑁, 0 ≤ 𝑥 ≤ 𝑛 ≤ 𝑁 Which is called hypergeometric probability distribution.

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Example 4.25

Two positions are open in a company. Ten men and five women have applied for a job at this company and all are equally qualified for either position. The manager randomly hires two people from the applicant pool to fill the positions. What is the probability that a man and a woman were chosen?

Answer: N=15, k=10 men (as successes), x=1, n=2. So using hypergeometric probability function to have P(X=1) = p(1) = 10 1 15−10 1 15 2 = 10 21

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Part 4. Hypergeometric:Mean and Variance

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Mean and Variance

𝑽 𝑿 = 𝒏 𝑬 𝑿 = 𝒏 𝒌 𝑵 𝒌 𝑵 𝒌 𝟏 − 𝑵 𝑵 − 𝒏 𝑵 − 𝟏

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Example 4.26

In an assembly line production of industrial robots, gearbox assemblies can be installed in 1 minute each, if the holes have been properly drilled in the boxes, and in 10 minutes each if the holes must be installed. There are 20 gearboxes in stock, and 2 of them have been improperly drilled holes. From the 20 gearboxes available, 5 are selected randomly fro installation in the next 5 robots in line. 1. Find the probability that all 5 gearboxes will fit properly.

2. Find the expected value, the variance and the standard deviation of the time it will take to install these five gearboxes.

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Answer: N=20, k=2 (as successes for noncomforming boxes), n=5 1. x=0 means all of the 5 boxes fit properly, so P(X=0)=p(0)= 2 0 20−2 5 20 5 2. Total time T=10X+(5-X) = 9X+5, and E(X)=n(k/N)=5(2/20)=0.5, V(X)=n(k/N)(1-k/N)[(N-n)/(N-1)]=5(2/20)(1-2/20)[(20-5)/(20-1)]=0.355

So E(T)=9E(X)+5=9.5, V(T)=(9) 2 V(X)=28.755, and the standard deviation is [V(T)] 1/2 =(28.755) 1/2 =5.4(minutes).

= 0.55

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Part 5. More Examples

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Additional Example 1

A certain kind of sheet metal has, on the average, five defects per 10 square feet. If we assume a Poisson distribution, what is the probability that a 15-square feet sheet of the metal will have at least six defects?

Answer: λs=5(15/10)=7.5, so P(X≥6)=1-P(X≤5)=1 5 𝑥=0 7.5

𝑥 𝑒 𝑥!

− 7.5

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Additional Example 2

Let X be a Poisson random variable with mean λ.

If P(X = 1|x≤ 1) = 0.8, what is the value of ?

Answer: P(X=1|X ≤ 1) = P(X=1, X ≤ 1)/P(X≤1)=P(X=1)/P(X≤1)= λ/(λ+1)=0.8

 λ=4.

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Additional Example 3

A box contains 10 white and 15 black marbles. Let X denote the number of white marbles in a selection of 10 marbles selected at random and without replacement. Find Var(X)/E(X).

Answer: N=25, k=10, n=10, E(X)=n(k/N)=10(10/25)=4, V(X)=n(k/N)(1-k/N)[(N k)/(N-1)]=4(1-10/25)(15/24)=1.5, so V(X)/E(X)=0.375

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