Chapter 3 Discrete Probability Distributions

Chapter Outline Section 3.1: Discrete Random Variables Section 3.2: Terminologies in Discrete Random Variables

Probability Mass Function (pmf) Cumulative Distribution Function (cdf) Expected Value and Variance (µ and σ2)

Section 3.3: Binomial Distribution Permutations and Combinations ( nPkand nCk ) pmf and cdf of Binomial Distribution

Section 3.4: Poisson Distribution Section 3.5: Poisson Approximation to the Binomial Distribution Section 3.6: Practice Problems

Why Discrete Probability Distribution? Flip a coin 50 times. Observe the number of heads.

What are the probabilities of obtaining 0, 1, 2, ... 50 heads? An average of 2 holes in 100m2 of paper production. A production of 1000m2of paper.

What are the probabilities of obtaining 0, 1, 2,... holes? Soccer matches between teams A and B. Expected scores between teams A and B are 3:2.

What is the probability that the actual score is 1:0 ?

Section 3.1 Discrete Random Variables

I. Random Variable A Random Variable (RV) is a numeric quantity that takes different values with specified probabilities Two types of random variables: Discrete and Continuous

A random variable for which there exists a discrete set of numeric values is a discrete random variable A random variable which can take a continuous range of value is a continuous random variable.

II. Discrete Random VariableDiscrete random variable:

Takes a discrete set of numeric values. Possible values that discrete random variables can take are separated. (e.g. number of persons suffering from

HIV in 2014)Example 1:

Roll a fair die once Possible outcomes would be {1,2,... , 6}, which is a set of discrete value. Let X be the value of the outcome. X is a discrete random variable

Pr(X=1) = Pr(X=2) = … = Pr(X=6) = 1/6

III. Continuous Random VariableContinuous Random Variable: takes any value on a continuum (with an uncountable number of values)

Example 2:

Let the random variable Z be the weight of an individual. Z is a continuous random variable taking any positive real numbers.

Pr(1<Z<1.7)=0.6

N.B.: We will cover Continuous Random Variable in Chapter 4

Section 3.2 Terminologies in Discrete Random Variables

I. Probability Mass Function

A probability mass function (pmf) assigns a probability to each possible value x of the discrete random variable X. Notation: Pr(X=x). The probability mass function (pmf) is also known as a probability distribution.

The probability mass function is an analog to the frequency distribution of a sample: Probability Mass Function: Population Frequency Distribution: Sample

Example 3:Toss two fair coins.Let X be the (discrete) random variable of the # of heads observed.Compute the probability mass function.

Solutions:4 possible outcomes when tossing two coins:

II. Cumulative Distribution Function

Cumulative Distribution Function (cdf) of a random variable X at value x is the probability that X is less than or equal to the value x.

Notation: F(x) = Pr(X ≤ x). Properties: F(x) is an increasing function from 0 to 1. and the graph for F(x) against x should be a step function

increases from 0 to 1Note: The cdf looks like a series of steps, called the step function. With the increase in the number of values, the cdf will get smoother and smoother For a continuous random variable, the cdf is a smooth curve.

Example 3 (Continued):Toss two fair coins.Let X be the number of heads observed.Compute the cumulative distribution function

Solutions:

pmf:

Pr (X=x )={0.25 x=00.5 x=10.25 x=2

The cdf is therefore:F (x)=Pr (X ≤x )=¿ {0.25 x=00.75x=11.00 x=2

¿

III. Expected Value

A random variable with many different values Hard to describe the random variable Summarize the random variable based on some “location” and “spread” parameters.

Expected value of X: is the sum of product of all possible values with their corresponding probabilities:

where x1, …, xn are the possible values X can take.

Note:µ is also known as the mean or the population mean.Expected value is the analog of the arithmetic mean ẍ of a sample, as it represents the “average” value of the random variable. When the experiment is to repeat many many times, the average of all the results would be E(X).Usually estimated by sample mean.

IV. VarianceVariance of X: is the sum of squares of all possible values of xi - µ with their corresponding probabilities:

σ2 is also called the population variance. Usually estimated by sample variance.

Alternative Formula:

where x1, …, xn are the possible values X can take.

Note:Variance of a random variable X is the analog of the sample variance (s2) of a sample.

V. Expected Value and Variance

Example 3 (Continued):X = # of heads from tossing 2 fair coins.

What is the expected value and variance of X? Pr (X=x )={0.25 x=00.5 x=10.25 x=2

Solutions:

E(X )=∑i=1

3

x iPr (X=¿¿x i)=0(0.25)+1(0.5)+2(0.25)=1¿¿

Intuition: On average, we expect 1 head from tossing 2 fair coins.

Alternative formula:

Example:Consider a discrete random variable X with probability distribution given by

a) Show that k=0.1b) Compute E(X), E(X2) and Var(1-4X)c) Sketch the CDF of X.

Section 3.3 Binomial Distribution

I. Introduction

Binomial distribution: Probability distribution of the number of successes in n independent experiments, each yields a probability of success p.

Derivation of Binomial Distribution: 1) Permutations (rearrange objects and values) 2) Factorial 3) Combinations (rearrange objects and values, ordering doesn’t matter) 4) Binomial Distribution

II. PermutationExample 4:Consider a container with 3 balls (Numbered 1,2 and 3).Two balls are drawn at random without replacement.

Possible pairs: (1,2), (1,3), (2,3) , (2,1), (3,1), (3,2)

Does the order selected matter in terms of winning? If the order matters, then there are 6 possible outcomes, => permutations If the order does not matter, then there are 3 possible outcomes, => combinations

Questions:Select k objects be selected out of n objects (0≤k≤n)How many ways can we select, if the objects are selected without replacement?

Answer:There are

If we denote nPk to be the number of permutations of n objects taken k at a time, then

III. Permutations and Factorial

Define the term n factorial (n!) as a product of n terms:n!= n x (n - 1) x (n - 2) x … x 1

Example 5: 5! = (5)(4)(3)(2)(1)=120.

Permutation nPk and Factorial:

IV. Combination Ordering doesn’t matter => Use combination but not permutation

Denote nCk (or (nk), pronounce as “n choose k”) to be the number of combinations of n objects taken k at a time

It represents the number of ways of selecting k objects out of n where the order of selection does not matter.Example 6:Evaluate 3P2 and 3C2

Solutions:

Examples 7:Consider a lottery from a container with 3 balls (numbering 1,2 and 3).(a) What is the # of permutations if two balls are drawn at random without replacement?(b) What is the # of combinations if two balls are drawn at random without replacement?

Solutions:Let n=3 balls and k=2 balls drawn from the container,

V. Setup of the Binomial DistributionExamples 8:Toss a fair coin 10 timesLet X be the total number of heads observed.What is the distribution of X?

Answer: Binomial Distribution!

General Setup of the Binomial Distribution:If we(1) record a fixed number of experiments n, (2) Probability for the event to happen in each experiment is constant at p.

If X is the total number of events to happen, then X follows Binomial Distribution with parameters n and p

VI. Probability Mass Function of Binomial DistributionThe probability mass function (pmf) of the Binomial distribution with parameters n and p is given by

Therefore, # of possible outcomes =(nx )

VII. Binomial DistributionExample 9:Toss a fair coin twice and let X be the number of heads.Find the probability of obtaining x=0,1 and 2 heads.

Solutions:n=2 and p=0.5. Hence

Therefore,

Example 10:What is the probability of obtaining exactly one success in 10 experiments, if the probability of success is 0.1?

Solutions: n=10 and p=0.1. Therefore

The above probability is pretty large: On average, one would expect to obtain only one success out of the 10 experiments.

In fact, X=1 is the mode of the distribution, followed by the X=0 as follows:

VIII. White Blood Cells CountWhite blood cells: Cells of immune system to defend the body against both infectious disease and foreign material

Major types of White Blood Cells Neutrophil: 70% (most abundant) Eosinophil Basophil Monocyte Lymphocyte

High number of neutrophils counts: Possible bacterial infection/acute viral infections.

A sample of blood is taken => # of neutrophils follow Binomial distribution

Example 11:Assume that 70% of white blood cells are neutrophils from a healthy person. If 10 white blood cells are examined, what is the probability mass function of the number of neutrophils?

Solutions:Let X be the number of neutrophils. ThenX ~Binomial(n=10, p=0.70), with pmf

Note: Pr(X=10) very small at 0.028.If 10 neutrophils are observed from the sample => Indication on bacterial infection (Chapter 7)

IX. Binomial Distribution --- Shape of the Binomial Distribution

1. Binomial Distribution is Symmetric when p=0.5Binomial distribution with p=0.5 and various values of n:

Binomial distribution is symmetric when p=0.5. That is,Pr(X=k) = Pr(X=n-k) for k=0, 1, 2,... N

2. When p<0.5, the distribution is right-skewed.

3. When p>0.5, the distribution is left-skewed.

Binomial Distribution with n=4 Binomial Distribution with n=10

X. Properties of the Binomial Distribution

Mean = μ= E(X) = npIntuition: The expected number of successes in n trials is the probability of success in one trial (p) multiplied by n

Variance = σ2 = np(1- p) = npq ,where q = 1- p is the probability of failure.Note: For fixed value of n, σ2 is maximized when p=0.5

Standard Deviation: σ= √np (1−p ) = √npq

VI. Variance of the Binomial Distributionσ2 = np(1-p) = n¿which is a concave function in p and attains its maximum at p=0.5.

Interpretation: 1) When p=0 or 1: σ2 =0 as there is no uncertainty on the outcome (Why?) 2) When p=1/2: the pmf spreads out widely across x=0 to x=n. Hence variability is the largest.

VII: Why Probability Distribution?

Covered in Chapter 1:Suppose that there are 1,000 students in a class. Eachof them flip the same coin 10 times and record the number of heads each students observed. Below is the frequency table of the number of heads observed by each student:

Question: Is the coin a fair one? (That is, is p=Pr(Head)=0.5?)

Covered in Chapter 3:Let X be the number of heads observed by flipping a coin 10times. Then X~Binomial(n=10, p)Below is the probability mass function for p =0.1, 0.3, 0.5, 0.7 and 0.9. We can then compare the histogram on the previous page with the pmf below to make a guess on what is the true p.

Covered in Later Chapter:Chapter 6 Estimation – Based on the data, try to (1) Find the “best” estimate of the unknown probability p, (2) Find the range of probability p that is likely to generate the data.

Example: 90% of chance p is between 0.6 to 0.9. Chapter 7-8 Hypothesis Testing Question: Is the coin fair? Conclusion: We have 95% confidence that the probability of head p is larger than 0.50.

Section 3.4 Poisson Distribution

I. Poisson Distribution

The second most frequently used discrete distribution Poisson distribution is usually associated with rare events.

Example:Of every 10m2 of paper being produced, we expect to observe an average of one surface defect.Let X be the number of surface defects per 100m2 of production. What is the distribution of X?

Answer: Poisson Distribution

II. Probability Mass Function of Poisson Distribution

Assume that events occur independently with each other.

The probability of k events occurring for a Poisson random variable with parameterμ is

Where e≈ 2.71828 is the Euler’s constant, and μ is expected number of events to occur

Notation:

Example 12:Assume # of bacterial colonies is 0.02 per cm2. For an area of 100cm2, find the probability distribution of the number of bacterial colonies.

Solutions:The expected number of bacterial colonies per 100cm2 is given by μ= 0.02(100) = 2 (Colonies)Let X be the number of colonies in 100cm2, then X~Poisson(μ=2)

Probability Mass Function:

Example 12 (Continued):Now, suppose we are interested in a larger area of 200cm2. What’s the probability distribution of the number of bacterial colonies?Solutions: μ=0.02(200) = 4(colonies). Let Y be the number of colonies in 200cm2, then Y~Poisson(μ=4)

Probability Mass Function:

III. Binomial Distribution vs Poisson Distribution

Binomial distribution: Number of trials n is finite, and the number of events cannot be larger than n.

Poisson distribution: Number of trials is essentially infinite and the number of events can be indefinitely large.

However, probability of k events becomes very small as k increases.

Examples of Binomial Distribution: – Flip a coin n times – Number of neutrophils out of n White Blood Cells

Example of Poisson Distribution: – Electrical system: telephone calls arriving in a system. – Astronomy: photons arriving at a telescope. – Biology: the number of mutations on a strand of DNA per unit time. – Management: customers arriving at a counter or call centre. – Civil Engineering: cars arriving at a traffic light. – Finance and Insurance: Number of Losses/Claims occurring in a given period of time.

IV. Shape of the Poisson Distribution Poisson Distribution is a right-skewed distribution Whenμis small (e.g. μ=0.8), the distribution is heavily skewed to the right. As μ increases (e.g. μ =12), the distribution becomes more symmetric, even though it’s still slightly right-skewed

V. Expected Value and Variance of Poisson DistributionSuppose that X~Poisson (μ), that is,

Then the expected value and the variance of X are given by

Section 3.5 Poisson Approximation to the Binomial Distribution

I. IntroductionBinomial distribution: When n is large and p is small, we can think of observing rare events in a large number of trials. In that case, the binomial distribution can be well approximated by the Poisson distribution, withμ=np.

Rule of Thumb: Let X~Binomial(n,p). When n≥20, p<0.1 and np<5, thenX≈Y~ Poisson(μ) whereμ=np.

Example 12 (Continued):Let Y be the number of bacterial colonies per 100cm2, with Y ~Poisson(μ =2.0).

1) You can think of it as a binomial distribution:Suppose that for each of the small area (at 0.01cm2),Pr(observing a bacterial) = p =2/10,000 = 0.0002

2) If we observe n=10,000 of those small area (i.e. 100cm2), let X = total number of bacterial coloniesX ~Binomial(n=10,000 , p=0.0002)

3) Combine 1) and 2), X ~ Binomial(n=10,000, p=0.0002) would be well approximated by Poisson distribution, withY ~ Poisson (μ=2.0)

The chart below shows how good the approximation is, with μ=np=2

Example 14:It is known that 70% of white blood cells is neutrophil for a healthy person.Suppose that there are 6,000 white blood cells in a microliter (μ L=10-6L) of blood, what isthe probability of more than 4,200 neutrophils?

Solution:Let X be the number of neutrophils out of the 6000 white blood cells. Then X~Binomial(n=6,000, p=0.7)Then

Alternatively, we can approximate X by a Poisson distribution Y, where μ =E(Y)=6000(0.7)=4200.Hence,

Note that Poisson approximation is not good in this case, as np=6000(0.7)=4200 >> 5.

However, the two probabilities computed in the previous page are still close to each other because they are the sum of pmf instead of individual pmf.

Good approximation => Pr(X=x) ≈ Pr(Y=x) => Pr(X ≤ x) ≈ Pr(Y ≤ x)

This example: Bad approximation => Pr(X=x) not close to Pr(Y=x) but still Pr(X ≤ x) ≈ Pr(Y ≤ x)

Summary

•Section 3.1: Discrete Random Variables• Section 3.2: Terminologies in Discrete Random Variables – Probability Mass Function (pmf) – Cumulative Distribution Function (cdf) – Expected Value and Variance (µ and σ2)• Section 3.3: Binomial Distribution – Permutations and Combinations ( nPk and nCk) – Pmf and cdf of Binomial Distribution• Section 3.4: Poisson Distribution• Section 3.5: Poisson Approximation to the Binomial Distribution

Practice Question

Question Type 1: Finding Unknown Value in pmf and cdf

Problem 1Consider a discrete random variable X with pmf where a is a constant.

Find the value of a0.2+a+0.1+a+0.2=1=> a = (1-0.5)/2=0.25

Problem 2Consider a discrete random variable X with pmf

where a, b and c are constants. Suppose the cdf is given by

Find the values of a, b, c and d.a=0.1,b= Pr(X=2)=Pr(X≤2)-Pr(X≤1)=0.5-0.1=0.4,d= Pr(X≤3)=Pr(X≤2)+Pr(X=3)=0.5+0.3=0.8c= Pr(X=4)=Pr(X≤4)-Pr(X≤3)=1-d=0.2

Question Type 2: Manipulation of Expected Value and Variance

Problem 1Consider a discrete random variable X with pmf where a is 0.25

Compute E(X) and Var(X):E(X) = 0.2(-1)+0.25(0)+0.1(1)+0.25(2)+0.2(3)=1.0

E(X2) = 0.2(1)+0.25(0)+0.1(1)+0.25(4)+0.2(9)=3.1 => Var(X)=E(X2)-[E(X)]2=2.1

Consider a new random variable Y = 6-2X, Compute Var(Y)Var(Y)=Var(6-2X)=Var(2X)=4Var(X)=8.4

Problem 2Consider a discrete random variable X with pmf where a=0.1, b=0.4, c=0.2

Find E(X) and Var(X).E(X)=1(0.1)+2(0.4)+3(0.3)+4(0.2)=2.6E(X2)= 12(0.1)+22(0.4)+32(0.3)+42(0.2)=7.76 => Var(X)=E(X2)-E(X)2= 7.6-2.62=0.84

Problem 3If we flip a fair coin three times and let X be the number of heads observed.

Compute E(X) and E(X2) and Var(X):E(X)=1(3/8)+2(3/8)+3(1/8)=1.5E(X2)=12(3/8)+22(3/8)+32(1/8)=3Var(X)= E(X2)- E(X)2=3-1.52=0.75Problem 7:Suppose that the probability you will win a certain game is 0.4. If you play the game 20 times,

(a)What is the expected number of games you win?Expected number of game you win= np = 20(0.4)=8

(b) What is the standard deviation of the number of games you win?Variance = (npq)0.5=2.1909

Question Type 3: Finding Probability and Cumulative Probability

Problem 1Consider a discrete random variable X with pmf where a is 0.25

Compute Pr(X<1) and F(0):

Pr(X<1)=Pr(X=-1)+Pr(X=0)=0.2+0.25=0.45F(0)=Pr(X<1)=0.45

Consider a new random variable Y = 6-2X, Calculate Pr (X Y)≧

Pr(X Y)= Pr(X=3)+ Pr(X=2)=0.2+0.25=0.45≧

Problem 2Consider a discrete random variable X with cdf where d=0.8

Find Pr(3X+2 8):≧Pr(3X+2 8)=Pr(X 2)= 1- Pr(X≤1)=0.9≧ ≧

Problem 4Consider a random variable X which takes integer values from 0 to 10. Suppose that the cdf of X is given by

Calculate Pr(X 5) and Pr (2 ≤ X ≤ 8):≧ Pr(X 5)=1-Pr(X≤4)=1-0.6331=0.3669≧ Pr(2≤X≤8)=Pr(X≤8)-Pr(X≤1)=0.9983-0.0464=0.9519

Question Type 4: Constructing a cdf from a statement

Problem 3:If we flip a fair coin three times and let X be the number of heads observedWhat is the cdf of X?

X~Binomial (3, 0.5)

Pr (X=0) = (0.5)3 = 1/8

Pr (X=1) = (31) (0.5)1 (0.5)2 = 3/8

Pr (X=2) = (32) (0.5)2 (0.5)1 = 3/8

Pr (X=3) = (33) (0.5)3 (0.5)0 = 1/8

Hence,

Question Type 5: Calculate Probability or Cumulative Probability by Binomial Distribution 1

Problem 5Suppose there are 4 seeds in a pot, each seed has germination probability of 0.7.(a) What is the probability of exactly one seed to germinate?

Let X be the number of seeds germinated. Then X~Binomial (4,0.7). HencePr(X=1)=4(0.7)(0.3)3 =0.0756.

(b) What is the probability of at least one seed to germinate?

Probability of at least one seed to germinate= Pr(X≥1)=1-Pr(X=0)=1-(0.3)4=0.9919.

Problem 6It is known that the annual incidence rate of blindness per year was 0.67% among age 30-39 male insulin-dependent diabetics (IDDM), and was 0.74% among age 30-39 female insulin-dependent diabetics.

Beware of the Percentage sign!

a) If a group of 50 IDDM age 30-39 men is observed, what is the probability that exactly 2 will go blind over a 1-year?Let X be the number of blind from a group of 50 IDDM age 30-39 male over 1-year period. X~Binomial(50,0.0067). Hence,Pr(X=2)=n(n-1)/2 (p)2(1-p)n-2=50(49)/2(0.0067)2(0.9933)48=0.0398

b) If a group of 50 IDDM age 30-39 women is observed, what is the probability that at least 2 will go blind over a 1-year?Let Y be the number of blind from a group of 50 IDDM age 30-39 female over 1-year period.Then Y~Binomial (50,0.0074). HencePr(Y=0)=(1-p)n=(0.9926)50=0.6898,Pr(Y=1)=n(p)(1-p)n-1=50(0.0074)(0.9926)49=0.2571

Therefore, Pr(Y≥2)=1-Pr(Y=0)-Pr(Y=1)=1-0.6898-0.2571=0.0531.

c) What is the probability that a 30-year-old IDDM male patient will go blind over the next 10 years?Pr(blind over the next 10 years)=1-Pr(not blind over the next 10 years)=1-Pr(not blind in the 1st year)x Pr(not blind in the 2nd year)x…x Pr(not blind in 10th year)=1-(1-0.0067)10=1-0.9350=0.0650

Check: The answer above should be close to the percentage of blind 0.67% per year x 10 year = 6.7%!!

Question Type 5: Calculate Probability or Cumulative Probability by Binomial Distribution 2Problem 7:Suppose that the probability you will win a certain game is 0.4. If you play the game 20 times,

a) What is the probability that you will win 3 or fewer games?X~Binomial(20,.4).Pr(X=0)=0.620=0.0000366,Pr(X=1)=20(0.4)(0.6)19=0.0004875,Pr(X=2)=20(19)/2 (0.4)2(0.6)18=0.003087,Pr(X=3)=20(19)(18)/6 (0.4)3(0.6)17=0.01235=> Pr(X≤3)=0.01596

Problem 8:If we fair flip a coin 100 times and the probability of obtaining 50 heads is 0.0796. What is the probability of obtaining at least 50 heads?

Let X be the number of heads. Note that X is symmetric as the coin is fair. Now,since Pr(X≤49)+Pr(X=50)+Pr(X≥51)=1=> Pr(X≤49) = (1-Pr(X=50))/2= (1-0.0796)/2 = 0.4602

Hence Pr(X≥50)=1- Pr(X≤49)= 0.5398.

Problem 9:If we roll a die 100 times, what is the probability of rolling a “2” exactly 30 times?

Let X be the number of “2”.Then X~Binomial(100,p) where p=Pr(“2”)=1/6.Therefore Pr(X=30)=100C30(1/6)30(5/6)70=0.00038

Assume Tom gets a fair coin, he repeats the following 2n times,1) Toss a coin2) If the coin shows “head”, he moves forward by 1 step, otherwise he moves backward by 1 stepFind the probability that after the 2n repeated trials, he stays in the original position. (The starting position before he tossed the coin).

Question Type 5: Calculate Probability or Cumulative Probability by Binomial Distribution 3There are two individuals, each of them would independently replicates itself or die independently with equal probability at a stage.Let X be the number of individuals after the 1st stage. Let Y be the number of individuals after the 2nd stage.Find the pmf of X and Pr(Y=0)

Question Type 6: Distinguish Binomial Distribution and Poisson DistributionProblem 10:

Assume that the random variables below follow either the binomial distribution or the Poisson distribution.Which of the following are Poisson distributed?

1. Receive an average of 2 phone calls per minute. X = number of phone calls in 10 minutes time.2. Of the 108 normal cells, we expect an average of 4 normal cells converting into cancer cells within a year. Let X = number of cancer cells after 2 years.3. An average of 60 rainy days in a year. Y = number of rainy days over a 5-year period4. Average number of 20 cars arriving a traffic light/minute. Z = # of cars arriving a traffic light in 10 minutes time.

Infinite outcome is actually possible for 1 and 4,while the outcomes for 2 and 3 are maximized by 108 and 5x365(+leap year) days.Ans: 1 and 4 only

Question Type 7: Distinguish Graph of Binomial Distribution and Poisson Distribution

Problem 11:The histogram on the right shows the probability mass functions of 3 discrete distributions:(i) Binomial(10,0.5)(ii) Binomial(10,0.8)(iii) Poisson(2)

Identify the 3 distributions by the colors of the bars.

Answer: (i) – Black, (ii) –Grey, (iii) – White

Note that Binomial distribution is(i) Symmetric when p=0.5, (ii) left-skewed when p>0.5 and (iii) right-skewed when p<0.5.Also, Poisson distribution is right-skewed for any μ

Question Type 8: Calculate Probability for Poisson Distribution

Problem 12:Suppose the number of admissions to the emergency room follows Poisson distribution, but the incidence rate is different among the week.During the weekday (i.e. Mon to Fri), there is an average of 0.7 admission per day.In the weekend (Sat and Sun), the average admission is smaller at 0.5 admission per day.

a) What is the probability of at least one admission on a Wednesday?Pr(admission ≥1 |weekday) = 1- Pr(admission =0 | weekday) = 1-e-0.7=0.5034

b) What is the probability of at least one admission on a Saturday?Pr(admission ≥1 |weekend) = 1- Pr(admission =0 | weekend) = 1-e-0.5=0.3935

c) What is the probability of having at least one admission for an entire week, if the number of admissions for different days over the week is assumed to be independent?

Note that there are 5 weekdays (Monday to Friday) and 2 weekends (Saturday and Sunday) in a week. Since the number of admission on different days are independent,

Pr(zero admission during the week) = [Pr(admission=0|weekday)]5[Pr(admission=0|weekend)]2

= e-0.7(5)-0.5(2) = e-6 = 0.0111 Therefore, Pr(at least one admission during the week) = 1-0.0111=0.9889.

Question Type 9: Calculate Probability for Poisson Distribution involving manipulation of λ

Problem 13:Otitis media is a common disease of the middle ear in early childhood.Assume that the number of episodes of otitis media for a child follows Poisson distribution with λ=1.6 episodes per year.

a) Find the probability of not getting any episode of otitis media in the first 2 years of life.Let X be the number of otitis media in the first two years.Then X~Poisson(2λ)=Poisson(3.2).

Therefore Pr(X=0)=e-3.2=0.0408

b) Find the probability of 3 or more episodes of otitis media in the first 2 years of life.Pr(X≥3)=1-Pr(X=0)-Pr(X=1)-Pr(X=2)=1-e-3.2(1+3.2+(3.22/2))=0.6201

Suppose that twin girls were delivered and assume that the episodes of otitis media of the two girls are independent.c) What is the probability of exactly one girl to have 3 or more episodes in the first 2 years of life?Let Y be the number of girl to have 3 or more episodes in the first 2 years of life, thenY~Binomial(n,p) where n=2 and p=0.6201 obtained in (b). Therefore

Alternatively,[(0.6201 X e-3.2) + (0.6201 X (e-3.2 X 3.2)) + (0.6201 X (e-3.2 X 3.22 / 2!))] X2= (0.025276642+0.080885256+0.12941641) X2= 0.4711566

D) What is the expected number of girls will have 3 or more episodes in the first 2 years of life?The expected number of girls will have 3 or more episodes in the first 2 years of life isE(Y)=np=2(0.6201)=1.2402

0.135308215+0.270670564+0.270697637+0.180465091+0.09022352+0.036082188

Question Type 10: Binomial Distribution v.s. Combination

Suppose that in a particular sheet of 100 postage stamps, 3 are defective. The inspection policy is to look at 5 randomly chosen stamps on a sheet and to release the sheet into circulation if none of those five is defective. Determine the probability that the sheet described here will be allowed to go into circulations wheni) The stamps are chosen with replacement

ii) The stamps are chosen without replacement

A committee of 4 is to be selected from a group consisting of 5 men and 5 women. Let X be the random variable that represents the number of women in the committee. Find the pmf of X

Assuming in a deck of 52 cards, you are randomly picking up 5 out of the 52 cards, what is the probability that you obtain a full house?