TOPIC 3TOPIC 3

Discrete Probability

Distributions

Discrete Probability

Distributions

Start Thinking Start Thinking

• You’re taking a 33 question multiple choice test. Each question has 4 choices. Clueless on 1 question, you decide to guess. What’s the chance you’ll get it right?

• If you guessed on all 33 questions, what would be your grade? Would you pass?

Discrete Probability DistributionsDiscrete Probability Distributions

Discrete Probability

Distributions

Binomial Poisson Hypergeometric

Negative Binomial

Geometric

Bernoulli Random VariablesBernoulli Random Variables

• Simple random variables with parameter p takes values 0 and 1 with

The values of p :

Then,

The expectation of a Bernoulli random variable

And its variance

10 p pXPandpXP 101

pXE

ppXVar 1

Binomial DistributionsBinomial Distributions

Consider an experiment consisting• n Bernoulli trials (identical trials)• that are independent and• that each have a constant probability p of success (desired

outcome)

Then the total number of successes X is a random variable that has a binomial distribution with parameters n and p, which is written

The probability mass function of a B(n,p) random variable is

for x = 0, 1, 2, …, n with the expectation (mean) and the variance

pnBX ,~

xnx ppx

nxXP

1

pnXE ppnXVar 1

Binomial Distribution ExamplesBinomial Distribution Examples

Number of ‘successes’ in a sample of n observations (trials)

Examples:• Number of reds in 15 spins of roulette wheel

• Number of defective items in a batch of 5 items

• Number correct on a 33 question exam

• Number of customers who purchase out of 100 customers who enter store

ExampleExample

Experiment: Toss 1 coin 5 times in a row. Note number of tails.

What’s the probability of exactly 3 tails ?

What’s the probability of less than or equal to 3? (0, 1,2, or 3) Note: This is known as ‘Cumulative Probability Function’

844.03125.0312.01570.003125.0

)3()2()1()0()3(

3125.05.015.0!3!35

!5)3(

5.0

1!!

!

353

xPxPxPxPxP

xP

p

ppxxn

nqp

x

nxP xnxxnx

Symmetric Binomial DistributionsSymmetric Binomial Distributions

n = 5 p = 0.1 n = 5 p = 0.5

For p = 0.5, written B(n, 0.5), the binomial probability distribution is symmetric. The distribution is symmetric about the expected value of n/2 In above case where B(5, 0.5), the expected value is 2.5.

Proportion of SuccessesProportion of Successes

Let the proportion of successes of a binomial distribution is given by

n

XY

and has an expected value and variance of

n

ppYpYE

12and

As the number of trial n increases, the proportion Y tends to become closer and closer to the success (trial) probability p

0.00

0.25

0.50

0.75

1.00

0 25 50 75 100 125

ExercisesExercises

1) You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20 in your last 100 calls. If you call 12 people tonight, what’s the probability of

a) No sales?

b) Exactly 2 sales?

c) At most 2 sales?

d) At least 2 sales?

2) A multiple choice quiz consists of 10 questions, each with 5 possible answers of which only one is correct. A student passes the quiz if seven or more correct answers obtained. What is the probability that a student who guesses blindly at all of the questions will pass the quiz? What is the probability of passing the quiz if, on each question, a student can eliminate three incorrect answers then guesses between the remaining two?

3) A biologist has a culture consisting of 13 cells. In a period of one hour, independent of the other cells, there is a probability of 0.4 that each of these cells splits into two cells. What is the probability that after one hour the biologist has at least 16 cells? What is the expected number of cells after one hour?

Answers to the ExercisesAnswers to the Exercises

1) p = 20/100 = 0.2 and n = 12

a) No sales means P(X = 0)

b) Exactly 2 sales means P(X = 2)

c) At most 2 sales means P(X ≤ 2)

d) At least 2 sales means P(X ≥ 2)

0687.02.012.00

1210 0120

xnx ppx

nXP

283.02.012.02

122 2122

XP

558.0283.0206.00687.02102 XPXPXPXP

725.0206.00687.0110112322 PPPPPXP

206.02.012.01

121 1121

XP

Answers to the ExercisesAnswers to the Exercises

2) a)

b)

00086.000000010.00000041.0000074.0000786.0

109877

00000010.02.012.010

1010

0000041.02.012.09

109

000074.02.012.08

108

000786.02.012.07

107

10,2.051

101010

9109

8108

7107

PPPPXP

XP

XP

XP

XP

np

172.000098.00098.0044.0117.0

109877

10,5.021

PPPPXP

np

Answers to the ExercisesAnswers to the Exercises

3) a) At least 16 cells means that there are at least three cells split

b)

Jadi paling tidak ada 6 sel yang membelah sehingga ekspektasi jumlah sel setelah 1 jam adalah 13 + 6 = 19

942.0045.0011.00013.01

210113433

045.04.014.02

132

011.04.014.01

131

0013.04.014.00

130

?3,13,4.0

2132

1131

0130

PPPPPPXP

XP

XP

XP

XPnp

2.54.013 pnXE

Poisson DistributionsPoisson Distributions

• A random variable X distributed as a Poisson random variable with parameter λ , which is written

• Has a probability mass function

• λ is the expected (mean) number of ‘successes’

• e = 2.71828 (base of natural logarithm or Euler number)

• Then, the mean or expectation and the variance

PX ~

,2,1,0!

xforx

exXP

x

XVarXE

Poisson Distribution ExamplesPoisson Distribution Examples

• Number of events that occur in an interval

• events per unit time / length / area / volume

• Examples

• Number of customers arriving in 20 minutes

• Number of strikes per year in the U.S.

• Number of defects per lot (group) of DVD’s

Poisson Distribution PropertiesPoisson Distribution Properties

• Constant event probability

• Average of 60/hr is1/min for 60 1-minuteintervals

• One event per interval

• Don’t arrive together

• Independent events

• Arrival of 1 person doesnot affect another’sarrival

Poisson Distribution GraphsPoisson Distribution Graphs

.0

.2

.4

.6

.8

0 1 2 3 4 5

X

P(X)

.0

.1

.2

.3

0 2 4 6 8 10X

P(X)

= 0.5

= 6

ExampleExample

Experiment: Customers arrive at a rate of 72 per hour. What is the probability of 4 customers arriving in 3 minutes?

72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

1912.0

!4

6.3

!4

6.36.34

e

x

eXP

arrivalsx

ExercisesExercises

1) You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction?

2) On average there are about 25 imperfections in 100 meters of optical cable. Use the Poisson distribution to estimate the probability that there are no imperfections in 1 meter of cable. What is the probability that there is no more than one imperfection in 1 meter of cable?

3) Assume that the number of network errors experienced in a day on a local area network (LAN) is distributed as a Poisson random variable. The mean number of network errors experienced in a day is 2.4. What is the probability that in any given daya) Zero network errors will occur?b) Exactly one network error will occur?c) Two or more network errors will occur?d) Less than three network errors will occur?

Answers to the ExercisesAnswers to the Exercises

1) 75 words/min = 75 × 60 = 4500 words/hour.If a 4500-word has 6 errors, then for a 255-word

2) 25 imperfections in 100 meters of cableIn 1 meter cable, there is

712.0

!0

34.0

!0

34.02554500

6

034.0

e

x

eXP

error

x

974.0195.0779.0!1

25.0

!0

25.0101

779.0!0

25.0

!0

25.01100

25

125.0025.0

025.0

eeXPXPXP

e

x

eXP

onimperfecti

x

Answers to the ExercisesAnswers to the Exercises

3)

a)

b)

c)

d)

57.0261.0218.0091.02103

691.0218.0091.011012

218.0!1

4.21

091.0!0

4.2

!0

4.2

14.2

04.2

PPPXP

PPXP

eXP

e

x

eXP

errorx

• For the binomial distribution, the sample data are selected with replacement from a finite population or without replacement from an infinite population.

• For the hypergeometric distribution, the sample data are selected without replacement from a finite population.

• The hypergeometric represents the distribution of the number of items of a certain kind in a random sample of size n drawn without replacement from a population of size N that contains r items of this kind

Hypergeometric DistributionsHypergeometric Distributions

• The hypergeometric distribution has a probability mass function given by

• for max (0, n + r – N) ≤ x ≤ min (n, r) , with• An expected value of

• And a variance of

Hypergeometric DistributionsHypergeometric Distributions

n

Nxn

rN

x

r

xXP

N

nrXE

N

r

N

rn

N

nNXVar 1

1

ExampleExample

• A box of N = 10 computer chips contains r = 5 defective computer chips, and n = 3 chips are selected at random without replacement. What is the probability that the sample contains exactly one defective chip?

4170

238910

245

5

331010

2255

1155

3

10

13

510

1

5

1

.

!!!

!!!

!!!

n

N

xn

rN

x

r

xP

• It is sometimes of interest to count the number of trials performed until the first success occurs.

• The number of trials up to and including the first success in a sequence of independent Bernoulli trials with a constant success probability p has a geometric distribution. The p.m.f. is

• The cumulative distribution function is

• The expected value and the variance are

Geometric DistributionsGeometric Distributions

,3,2,11 1 xforppxXP x

xpxXP 11

2

11

p

pXVarand

pXE

• The number of trials up to and including the r th success in a sequence of independent Bernoulli trials with a constant success probability p has a negative binomial distribution. The p.m.f. is

• for x = r, r + 1, r + 2, r + 3, ….• The expected value and the variance are

Negative Binomial DistributionsNegative Binomial Distributions

rrx ppr

xxXP

11

1

2

1

p

prXVarand

p

rXE

ExampleExample

• When a fisherman catches a fish, it is a young one with a probability of 0.23 and it is returned to the water. On the other hand, an adult fish is kept to be eaten later.

a) What is the expected number of fish caught by the fishermen before an adult fish is caught?

b) What is the probability that the fifth fish caught is the first young fish?

• Suppose that the fisherman wants three fish to eat for lunch

c) What is the probability that the first time the fisherman can stop for lunch is immediately after the sixth fish has been caught?

d) If the fisherman catches five fish, what is the probability that there are sufficient fish for lunch?

Answer to the ExampleAnswer to the Example

a) padult fish = 1 – 0.23 = 0.77

b) pyoung fish = 0.23 , x = 5

c) p = 0.77 , x = 6 , r = 3

d)

29.177.0

11 p

XE

081.023.023.0115 151 ppXP x

056.077.077.0113

161

1

16 336

rrx ppr

xXP

9164.077.077.0113

1577.077.01

13

14

77.077.0113

135435

335334

333

PPPXP

• Consider a sequence of n independent trials where each individual trial can have k outcomes that occur with constant probability values p1, …, pk with p1 + … + pk = 1. The random variables X1, … , Xk that count the number of occurrences of each outcome are said to have a multinomial distribution, and their joint probability mass function is

• satisfying

• The random variables X1, … , Xk expected value and the variance are (they are not independent)

Multinomial DistributionsMultinomial Distributions

kxk

x

kkk pp

xx

nxXxXP

1

11

11 !!

!,,

iiii ppnXVarandpnXE 1

nxx k 1

ExampleExample

• Patients being treated with a particular drug run the risk of being allergic to the drug. A patient’s reaction to the drug is characterized by doctors as being hyper allergic, allergic, mildly allergic, or not allergic, and these have probability values of 0.12, 0.28, 0.33, and 0.27. If there are nine patients,

a) What is the probability that two patients are hyper allergic, one patient is allergic, four patients are mildly allergic, and two patients exhibit no reaction?

b) What is the probability that no patients are hyper allergic, one patient is allergic, four patients are mildly allergic, and four patients exhibit no reaction?

c) What is the expected number of hyper allergic reaction?

d) What is the expected number of allergic reaction?

Answer to the ExampleAnswer to the Example

a)

b)

c)

d)

013.027.033.028.012.0!2!4!1!2

!9

2,4,1,2

2412

4321

XXXXP

011.027.033.028.012.0!4!4!1!0

!9

4,4,1,0

4410

4321

XXXXP

08.112.0911 pnXE

52.228.0922 pnXE