Probability Theory

Modelling random phenomena

Permutationsthe number of ways that you can order n objects is:

n! = n(n-1)(n-2)(n-3)…(3)(2)(1)

Definition: 0! = 1

the number of ways that you can choose k objects from n objects in a specific order:

)1()1()!(

!

knnn

knnPkn

Combinationsthe number of ways that you can choose k objects from n objects (order irrelevant) is:

!!( )!n k

n nCk k n k

( 1) ( 1)( 1) (1)

n n n kk k

are called Binomial Coefficients

)!(!!

knkn

kn

Ckn

Reason:The Binomial Theorem

nyx022

211

10

0 yxCyxCyxCyxC nnn

nn

nn

nn

022110

210yx

nn

yxn

yxn

yxn nnnn

2 2 22x y x xy y 2 2 2

1, 2, 10 1 2

3 3 2 2 33 3x y x x y xy y 3 3 3 3

1, 3, 3, 10 1 2 3

4 4 3 2 2 3 44 6 4x y x x y x y xy y

4 4 4 4 41, 4, 6, 4, 1

0 1 2 3 4

Binomial Coefficients can also be calculated using Pascal’s triangle

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 1

Random Variables

Probability distributions

Definition:A random variable X is a number whose value is determined by the outcome of a random experiment (random phenomena)

Examples1. A die is rolled and X = number of spots

showing on the upper face.2. Two dice are rolled and X = Total number

of spots showing on the two upper faces.3. A coin is tossed n = 100 times and

X = number of times the coin toss resulted in a head.

4. A person is selected at random from a population and

X = weight of that individual.

5. A sample of n = 100 individuals are selected at random from a population (i.e. all samples of n = 100 have the same probability of being selected) .

X = the average weight of the 100 individuals.

In all of these examples X fits the definition of a random variable, namely:– a number whose value is determined by the

outcome of a random experiment (random phenomena)

Random variables are either• Discrete

– Integer valued – The set of possible values for X are integers

• Continuous– The set of possible values for X are all real

numbers – Range over a continuum.

Discrete Random VariablesDiscrete Random Variable: A random variable usually assuming an integer value.

• a discrete random variable assumes values that are isolated points along the real line. That is neighbouring values are not “possible values” for a discrete random variable

Note: Usually associated with counting• The number of times a head occurs in 10 tosses of a coin• The number of auto accidents occurring on a weekend• The size of a family

Examples• Discrete

– A die is rolled and X = number of spots showing on the upper face.

– Two dice are rolled and X = Total number of spots showing on the two upper faces.

– A coin is tossed n = 100 times and X = number of times the coin toss resulted in a head.

Continuous Random Variables

Continuous Random Variable: A quantitative random variable that can vary over a continuum

• A continuous random variable can assume any value along a line interval, including every possible value between any two points on the line

Note: Usually associated with a measurement• Blood Pressure• Weight gain• Height

Examples• Continuous

– A person is selected at random from a population and X = weight of that individual.

– A sample of n = 100 individuals are selected at random from a population (i.e. all samples of n = 100 have the same probability of being selected) . X = the average weight of the 100 individuals.

Probability distribution of a Random Variable

The probability distribution of a discrete random variable is describe by its :

probability function p(x).p(x) = the probability that X takes on the value x.

Probability Distribution & Function Probability Distribution: A mathematical description of how probabilities are distributed with each of the possible values of a random variable.

Notes: The probability distribution allows one to determine probabilities

of events related to the values of a random variable. The probability distribution may be presented in the form of a

table, chart, formula.

Probability Function: A rule that assigns probabilities to the values of the random variable

Comments:Every probability function must satisfy:

1)(0 xp

1. The probability assigned to each value of the random variable must be between 0 and 1, inclusive:

x

xp

1)(

2. The sum of the probabilities assigned to all the values of the random variable must equal 1:

b

ax

xpbXaP )(3.

)()1()( bpapap

Examples• Discrete

– A die is rolled and X = number of spots showing on the upper face.

– Two dice are rolled and X = Total number of spots showing on the two upper faces.

x 1 2 3 4 5 6p(x) 1/6 1/6 1/6 1/6 1/6 1/6

x 2 3 4 5 6 7 8 9 10 11 12p(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

GraphsTo plot a graph of p(x), draw bars of height p(x) above each value of x.Rolling a die

01 2 3 4 5 6

Rolling two dice

0

x 0 1 2 3p(x) 6/14 4/14 3/14 1/14

ExampleIn baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below:

P X( )the random variable equals 2 p ( ) 23

14

Note: This chart implies the only values x takes on are 0, 1, 2, and 3. If the random variable X is observed repeatedly the probabilities,

p(x), represents the proportion times the value x appears in that sequence.

2least at is variablerandom the XP 32 pp 144

141

143

A Bar GraphNo. of persons on base when a home run is hit0.429

0.286

0.214

0.071

0.000

0.100

0.200

0.300

0.400

0.500

0 1 2 3# on base

p(x)

Note: In all of the examples1. 0 p(x) 1

2.

3.

x

xp 1

b

ax

xpbXaP )(

An Important discrete distributionThe Binomial distributionSuppose we have an experiment with two outcomes – Success(S) and Failure(F).Let p denote the probability of S (Success).In this case q=1-p denotes the probability of Failure(F).Now suppose this experiment is repeated n times independently.

Let X denote the number of successes occuring in the n repititions.Then X is a random variable.It’s possible values are

0, 1, 2, 3, 4, … , (n – 2), (n – 1), nand p(x) for any of the above values of x is given by:

xnxxnx qpxn

ppxn

xp

1

X is said to have the Binomial distribution with parameters n and p.

Summary:X is said to have the Binomial distribution

with parameters n and p.1. X is the number of successes occurring in

the n repetitions of a Success-Failure Experiment.

2. The probability of success is p.3.

xnx ppxn

xp

1

Examples:1. A coin is tossed n = 5 times. X is the

number of heads occuring in the 5 tosses of the coin. In this case p = ½ and

3215

215

21

21

555

xxxxp xx

x 0 1 2 3 4 5

p(x) 321

325

325

321

3210

3210

Another Example2. An eye surgeon performs corrective eye

surgery n = 10 times. The probability of success is p = 0.92 (92%). Let X denote the number of successful operations in the 10 cases.

1010.92 .08x xp x

x

0,1,2,3, ,8,9,10x

1010.92 .08x xp x

x

0,1,2,3, ,8,9,10x

x 0 1 2 3 4 5p(x) 0.00000 0.00000 0.00000 0.00000 0.00004 0.00054

x 6 7 8 9 10p(x) 0.00522 0.03427 0.14781 0.37773 0.43439

-

0.10

0.20

0.30

0.40

0.50

0 1 2 3 4 5 6 7 8 9 10

Continuous Random Variables

Continuous Random VariablesThe probability distribution of a continuous random variable is described by its :

probability density curve f(x).

i.e. a curve which has the following properties :1.      f(x) is always positive.2.      The total are under the curve f(x) is one.3.      The area under the curve f(x) between a

and b is the probability that X lies between the two values.

An Important continuous distributionThe Normal distribution

The normal distribution (with mean and standard deviation ) is a continuous distribution with a probability density curve that is:

1. Bell shaped2. Centered at the value of the mean 3. The spread is determined by the value of the

standard deviation . The points of inflection on the bell curve occur at - and + .

The graph of the Normal distribution

-

0.005

0.010

0.015

0.020

0.025

0.030

0 20 40 60 80 100 120

Points of Inflection

Mean and Variance of aDiscrete Probability Distribution

• Describe the center and spread of a probability distribution

• The mean (denoted by greek letter (mu)), measures the centre of the distribution.

• The variance (2) and the standard deviation () measure the spread of the distribution.

is the greek letter for s.

Mean of a Discrete Random Variable• The mean, , of a discrete random variable x is found by

multiplying each possible value of x by its own probability and then adding all the products together:

Notes: The mean is a weighted average of the values of X.

x

xxp

kk xpxxpxxpx 2211

The mean is the long-run average value of the random variable.

The mean is centre of gravity of the probability distribution of the random variable

-

0.1

0.2

0.3

1 2 3 4 5 6 7 8 9 10 11

2

Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x )2, by its own probability and then adding all the products together:

Standard Deviation of a Discrete Random Variable: The positive square root of the variance:

x

xpx 22

2

2

xx

xxpxpx

22 x

xpx

ExampleThe number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.

x p (x ) xp(x) x 2 x 2 p(x)0 0.429 0.000 0 0.0001 0.286 0.286 1 0.2862 0.214 0.429 4 0.8573 0.071 0.214 9 0.643

Total 1.000 0.929 1.786

)(xp )(xxp )(2 xpx

• Computing the mean:

Note: • 0.929 is the long-run average value of the random variable • 0.929 is the centre of gravity value of the probability distribution

of the random variable

929.0x

xxp

• Computing the variance:

x

xpx 22

2

2

xx

xxpxpx

923.0929.786.1 2

• Computing the standard deviation:

2

961.0923.0

The Binomial distribution1. We have an experiment with two outcomes

– Success(S) and Failure(F).

2. Let p denote the probability of S (Success).

3. In this case q=1-p denotes the probability of Failure(F).

4. This experiment is repeated n times independently.

5. X denote the number of successes occuring in the n repititions.

The possible values of X are0, 1, 2, 3, 4, … , (n – 2), (n – 1), n

and p(x) for any of the above values of x is given by:

xnxxnx qpxn

ppxn

xp

1

X is said to have the Binomial distribution with parameters n and p.

Summary:X is said to have the Binomial distribution with

parameters n and p.1. X is the number of successes occurring in the n

repetitions of a Success-Failure Experiment.2. The probability of success is p.3. The probability function

xnx ppxn

xp

1

Example:1. A coin is tossed n = 5 times. X is the

number of heads occurring in the 5 tosses of the coin. In this case p = ½ and

3215

215

21

21

555

xxxxp xx

x 0 1 2 3 4 5

p(x) 321

325

325

321

3210

3210

0.0

0.1

0.2

0.3

0.4

1 2 3 4 5 6

number of heads

p(x

)

Computing the summary parameters for the distribution – , 2,

x p (x ) xp(x) x 2 x 2 p(x)0 0.03125 0.000 0 0.0001 0.15625 0.156 1 0.1562 0.31250 0.625 4 1.2503 0.31250 0.938 9 2.8134 0.15625 0.625 16 2.5005 0.03125 0.156 25 0.781

Total 1.000 2.500 7.500

)(xp )(xxp )(2 xpx

• Computing the mean: 5.2

x

xxp

• Computing the variance:

x

xpx 22

2

2

xx

xxpxpx

25.15.25.7 2

• Computing the standard deviation:2

118.125.1

Example:• A surgeon performs a difficult

operation n = 10 times. • X is the number of times that the operation is

a success.

• The success rate for the operation is 80%. In this case p = 0.80 and

• X has a Binomial distribution with n = 10 and p = 0.80.

xx

xxp

1020.080.0

10

x 0 1 2 3 4 5p (x ) 0.0000 0.0000 0.0001 0.0008 0.0055 0.0264

x 6 7 8 9 10p (x ) 0.0881 0.2013 0.3020 0.2684 0.1074

Computing p(x) for x = 1, 2, 3, … , 10

The Graph

-

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8 9 10

Number of successes, x

p(x

)

Computing the summary parameters for the distribution – , 2,

)(xxp )(2 xpx

x p (x ) xp(x) x 2 x 2 p(x)0 0.0000 0.000 0 0.0001 0.0000 0.000 1 0.0002 0.0001 0.000 4 0.0003 0.0008 0.002 9 0.0074 0.0055 0.022 16 0.0885 0.0264 0.132 25 0.6616 0.0881 0.528 36 3.1717 0.2013 1.409 49 9.8658 0.3020 2.416 64 19.3279 0.2684 2.416 81 21.743

10 0.1074 1.074 100 10.737Total 1.000 8.000 65.600

• Computing the mean: 0.8

x

xxp

• Computing the variance:

x

xpx 22

2

2

xx

xxpxpx

60.10.86.65 2

• Computing the standard deviation:2

118.125.1