Introduction to Probability

Elementary properties of probability

• Definition: Probability is the likelihood or chance of an event occurring.• Probability of an event is usually referred to as relative frequency• Probability statements used frequently in biostatistics – e.g., we say that we are

90% probably sure that an observed treatment effect in a study is real; the success probability of this surgery is only 10%; the probability of tumor to develop is 50%......

THE RELATIVE FREQUENCY INTERPRETATION OF PROBABILITY

• We are interested in learning about the probability of some event to occur in a certain process based on the number of actual occurrence of that event relative to number of times that process repeated.

• For example, our process could be rolling a dice, and we are interested in the probability in the event that the number on the dice is equal to 6.

• I did this dice experiment 20 times. Each time I recorded the top face number: results are below:

1 3 2 2 4 6 4 5 3 1 3 5 6 4 2 1 2 5 6 2

Experimental Probability Vs Theoretical Probability• Experimental Probability = 3/20

• Theoretical probability = 1/6• The theoretical probability is what you expect to happen, but it isn't

always what actually happens. • When n increases experimental probability approaches theoretical

probability. When n ∞, experimental probability = theoretical probability.

In general, the probability of an event can be approximated by the relative frequency , or proportion of times that the event occurs. PROBABILITY (EVENT) is approximately = # of times event occurs/# of experiments

THE SUBJECTIVE INTERPRETATION OF PROBABILITY • The relative frequency notion of probability is useful when the

process of interest, say tossing a coin, can be repeated many times under similar conditions. But we wish to deal with uncertainty of events from processes that will occur a single time. e.g. probability of a success of a surgery, probability to get A in class.

• A subjective probability reflects a person's opinion about the likelihood of an event. Different between people.

Elementary properties of probability

Given some process (or experiment) with n mutually exclusive outcomes (called events), E1, E2, E3, …, En, the probability of any event Ei is assigned a nonnegative number. That is:

The sum of probabilities of all mutually exclusive outcomes is equal to 1 (exhaustiveness):

0)( iEP

1)(...............)()()( 321 nEPEPEPEP

Probability as a Numerical Measureof the Likelihood of Occurrence

0 1.5

Increasing Likelihood of Occurrence

Probability:

The eventis veryunlikelyto occur.

The occurrenceof the event is

just as likely asit is unlikely.

The eventIs almostcertain

to occur.

Some Basic Relationships of Probability

There are some basic probability relationships thatcan be used to compute the probability of an eventwithout knowledge of all the sample point probabilities.

Complement of an Event Complement of an Event

Intersection of Two Events Intersection of Two Events

Mutually Exclusive Events Mutually Exclusive Events

Union of Two EventsUnion of Two Events

Elementary properties of probability

Example: Tossing a coin: the outcome of head or tail is mutual exclusive events. No way in a certain toss you will have head and tail at the same time. It is either H or T

Mutual Exclusiveness: Two or more events are said to be mutually exclusive if the occurrence of any one of them means the others will not occur (That is, we cannot have 2 events occurring at the same time)

H T

Mutually exclusive events

A B

Non-Mutually exclusive events

Example:

A B

Calculating the probability of an event

• Sample of 75 men and 36 women were selected to study the cocaine addiction as function of gender. The subjects are representative sample of typical adult who were neither in treatment nor in jail.

Frequency of cocaine use by gender among adult cocaine users

Lifetime frequency of cocaine use

Male (M) Female (F) Total

1-19 times (A) 32 7 39

20-99 times (B) 18 20 38

100+ times (c) 25 9 34

Total 75 36 111

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Suppose we pick a person at random from this sample, what is the probability that this person is a male?• 111 subjects are our population. • Male and female are mutually exclusive categories• The likelihood of selecting any one person is equal to the likelihood of selecting any other• The desired probability is the number of subjects with the characteristic of interest (Male)

divided by the total number of subjectsP(M) = Number of males / Total number of subjects = 75/111 = 0.6757

Conditional probability, P(B|A)

• Conditional probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. This probability is written P(B|A)

• Suppose we pick a subject at random from the 111 subjects and find that he is a male (M), what is the probability that this male will be one who has used cocaine 100 times or more during his lifetime?

• What is the probability that a subject has used cocaine 100 times or more given he is a male?

• P(C100ΙM) = 25/75 = 0.33

• Joint probability is defined as the probability of both A and B taking place together (at same time)

• The joint probability is given the symbolic notation: P(A∩B) in which the symbol “∩” is read either as “intersection” or “and”.

• What is the probability that a person picked at random from the 111 subjects will be male and be a person who had used cocaine 100 times or more?

• The statement M∩C100 indicates the joint occurrence of conditions M and C100. P(M∩C100) = 25/111 = 0.2252

Joint probability, P(A∩B)

The Addition Rule, P(AUB) for mutual exclusive events

• In the previous example, if we picked a person at random from the 111 subject sample, what is the probability that this person will be a male (M) or female (F)?

• We state this probability as P(M∪F) were the symbol ∪ is read either “union” or “or”.

• Since the two genders are mutually exclusive and variables:

P(M∪F) = P(M) + P(F) =(75/111)+(36/111 =0.6757+0.3243=1

P(AB) P(A)P(B)A B

Calculating the probability of an event

• Complimentary events:• The probability of an event A is equal to 1 minus the

probability of its compliment which is written as Ā, andP(Ā)=1-P(A)

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Ā

A

1. If the probability of having smokers in this class is 5%, so what is the probability of having non-smokers in the same class?

2. If you toss a dice, what is the probability not to have 2 “face up”?

Biostatistics

Lecture 6Probability Distributions

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Probability Distributions

• The probability distribution of random variable is a graph, table, or formula used to specify all possible values of a random variable along with the respective probabilities.

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graph table

formula

Probability Distributions“General Example”• The following table shows the prevalence of

prescription and nonprescription drugs use in pregnancy among a population of women who were delivered of infants.

• We wish to construct the probability distribution of the variable X = number of prescription and nonprescription drugs used by the study subjects.

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Variable

Variable Probability

Varia

ble

Prob

abili

ty

Variable (# Drugs)

Probability DistributionsCurve

Probability Distributions Table

Note: 0≤P(X=x)≤1; ΣP(X=x)=1

Probability distributions “General Example”

Q1: What is the probability that a randomly selected woman will be one who used exactly three prescription and nonprescription drugs?

P(X=3)=0.0832

Q2: What is the probability that a randomly selected woman used wither one or two drugs?

We use the addition rule for mutually exclusive events

P(1∪2)=P(1)+P(2)=0.3228+0.1895=0.5123

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Probability distributions “General Example”

• Sometimes we may want to work with a cumulative probability distribution of a random variable.

• The cumulative probability distribution is obtained by successively adding the probabilities, P(X=x) given in the last column of the previous table.

• Q3: What is the probability that a woman picked at random will be one who used two or fewer drugs? P=0.8528

• Q4: What is the probability that a randomly selected woman will be one who used fewer than two drugs? P= 0.6633

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Probability distributions “General Example”

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Q5: What is the probability that a randomly selected woman is one who used between three and five drugs, inclusive?

P(x≤5)=0.9872 is the probability that a woman used between zero and five drugs,

inclusive.P(X≤2)= 0.8528 is the probability that a

woman used between zero and two drugs, inclusive.

P(3≤x≤5)=P(x≤5)-P(X≤2)= 0.9872- 0.8528=0.1344

“BE CARFULL HERE!”

Probability Distributions (Types)

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Probability Distribution

Discrete

Binomial Distribution

Poisson Distribution

Continues

Normal Distribution

Continues distribution• A continuous variable is one that can assume any value within a specified

interval of values.

• Consequently, between any two values assumed by a continuous variable, there exist an infinite number of values.

• In Normal distributions, as the number of possible outcomes (observations, n) approaches infinity, and the width of the class intervals approaches zero, the frequency polygon approaches a smooth curve.

•

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Normal distribution• The most important of continuous probability distributions is the normal

distribution (some times referred to as Gaussian distribution, Carl Friedrich Gauss, 1777-1855).

• The parameters of distribution are:

μ: mean and σ: the standard deviation.

• π=3.14159, e=2.71828

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-∞<x<∞

Characteristics of the normal distribution1. is considered the most prominent probability distribution in statistics2. arises as the outcome of the central limit theorem, which states that under

mild conditions the sum of a large number of random variables is distributed approximately normally

3. It is symmetrical about its mean, µ4. The curve on either side of µ is a mirror image of the other side5. The mean, the median, and the mode are all equal6. The total area under the curve above the x-axis is one square unit. This

characteristic follows from the fact that the normal distribution is a probability distribution

Dr Alkilany 2012-x x

μσ

Mode

Characteristics of the normal distribution

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7 . The relative frequency of occurrence of values between any two points (a, b) on the x-axis is equal to the total area bounded by the curve (gray area).

Characteristics of the normal distribution8. If we erect perpendiculars a distance of 1 standard deviation from the

mean in both directions, the area enclosed by these perpendiculars, the x-axis, and the curve will be approximately 68 percent of the total area.

9. If we extend these lateral boundaries a distance of 2 standard deviations on either side of the mean, approximately 95 percent of the area will be enclosed.

10. If we extend these lateral boundaries a distance of 3 standard deviations will cause approximately 99.7 percent of the total area to be enclosed.

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Characteristics of the normal distribution11. The normal distribution is completely determined by the parameters

µ and . Different values of µ, shift the graph of the distribution along the x-axis. Different values of determine the degree of flatness or peakedness of the graph of the distribution.

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3 Normal distributions with different µ values and equal values

3 Normal distributions with equal µ values and different values

1<2<3

Summaryµ: location: shape

Biostatistics

Lecture 9Probability Distributions

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Normal distribution• The most important of continuous probability distributions is the normal

distribution (some times referred to as Gaussian distribution, Carl Friedrich Gauss, 1777-1855).

• The parameters of distribution are:

μ: mean and σ: the standard deviation.

• π=3.14159, e=2.71828

Dr Alkilany 2012

-∞<x<∞

Characteristics of the normal distribution1. is considered the most prominent probability distribution in statistics2. arises as the outcome of the central limit theorem, which states that under

mild conditions the sum of a large number of random variables is distributed approximately normally

3. It is symmetrical about its mean, µ4. The curve on either side of µ is a mirror image of the other side5. The mean, the median, and the mode are all equal6. The total area under the curve above the x-axis is one square unit. This

characteristic follows from the fact that the normal distribution is a probability distribution

Dr Alkilany 2012-x x

μσ

Mode

Characteristics of the normal distribution

Dr Alkilany 2012

7 . The relative frequency of occurrence of values between any two points (a, b) on the x-axis is equal to the total area bounded by the curve (gray area).

Characteristics of the normal distribution8. If we erect perpendiculars a distance of 1 standard deviation from the

mean in both directions, the area enclosed by these perpendiculars, the x-axis, and the curve will be approximately 68 percent of the total area.

9. If we extend these lateral boundaries a distance of 2 standard deviations on either side of the mean, approximately 95 percent of the area will be enclosed.

10. If we extend these lateral boundaries a distance of 3 standard deviations will cause approximately 99.7 percent of the total area to be enclosed.

Dr Alkilany 2012

Characteristics of the normal distribution11. The normal distribution is completely determined by the parameters

µ and . Different values of µ, shift the graph of the distribution along the x-axis. Different values of determine the degree of flatness or peakedness of the graph of the distribution.

Dr Alkilany 2012

3 Normal distributions with different µ values and equal values

3 Normal distributions with equal µ values and different values

1<2<3

Summaryµ: location: shape

The Standard Normal Distribution• A standard normal distribution is a normal distribution which

has a mean (µ)=0 and a standard deviation ()=1.

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Z (standard score, z-score)= (x — µ)/

f (z) 1

2e z 2 / 2

Standardization

Normal Distribution Standard Normal Distribution

-∞<x<∞ -∞<z<∞

What is z-score??• Z-scores are scores converted into the number of standard deviations that the values (x) are from the mean (μ):

• Z-scoring is used to standardize scores in order to provide a way of comparing them (compare score of 3.0 in IELTS with a 580 in the old TOEFL exam??)

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z x

• A Z-score= +2.0 means that the original score (x) was 2 standard deviations above the mean. A Z-score= –3.0 means that the original score was three standard deviations below the mean

Z=+2.0Z=-3.0

(Important note: for z-scores: mean (µ)=0 and a standard deviation ()=1)

Do it @ home (NOT a homework) for values

0, 5, 10, 15, 20 calculate z-score for every value and then calculate the mean and standard deviation for all z-scores

The Standard Normal Distribution• To find the probability that z takes on a value between any two points on the z-

axis (z0 and z1), we must find the area bounded by the perpendiculars erected at these points, the curve and the horizontal z-axis.

• The area can be found by integrating the f(z) function between two values of the variable. In the standard normal distribution, the integral (area) is given by the equation:

1

2e z 2 / 2dz

z0

z1Dr Alkilany 2012

z0 z1

f(z)

The Standard Normal Distribution• There are CUMULATIVE TABLES that provide the results of all such

integrations.

• These tables provide the areas under the curve between - and z (cumulative relative frequency again!!).

• These tables are used to find the probability that a statistic is observed below, above, or between values on the standard normal distribution

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Dr Alkilany 2012

-3.8 +3.8

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P(Z ≤ 1.57) = 0.9418

P(Z ≤ 1.96) = 0.9750

P(Z ≤ 1.0) = 0.8413

P(2.4 ≤ Z ≤ 2.5) = 0.9938-0.9918=0.002

P(Z > 3.0) =1-P(Z≤3)=1-0.9987=0.0013

Example I• As a part of a study of Alzheimer’s disease, data reported were compatible

with the hypothesis that brain weights of victims of the disease are normally distributed. From the reported data we may compute a mean of 1076.80 g and a standard deviation of 105.76 g. Find the probability that a randomly selected patient will have a brain that weighs less than or equal to 800 grams.

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• Given from the question:1. Brain weight mean (µ)=1076.80 g2. Brain weight standard deviation (σ)=105.76 g3. Brain weight measurements are normally distributed

• The question asks for the probability randomly selected patients to have a brain weights less than or equal to 800 g……..> P(x≤800)

What do we need to do first??

Application of Normal Distribution

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62.276.105

80.1076800

z

First, we need to calculate the z-score for x=800 g

Now we can use the standard normal distribution table:What is the probability that randomly selected patients to have a brain weights less than or equal to Z=-2.62……..> P(z≤ -2.62)=0.0044

Example II• Suppose it is known that the Dissolution time (Dt) of a certain

pharmaceutical tablets are approximately normally distributed with a mean of 70 min and a standard deviation of 3 min. What is the probability that a tablet picked at random from this batch will have Dt between 65 and 74 min?

z65 (x )

65 703

1.67

z74 (x )

74 703

1.33

P(65x 74) P( 1.67z1.33) P( z1.33) P( z1.67) 0.9082 0.0475 0.8607

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Example III

• In a batch of 10,000 tablets described in the previous example, how many tablets would you expect to have Dt at least 77min?

• Out of 10,000 tablets, we would expect 10,000*0.0099=99 to have dissolution time at least 77 min

z77 77 703

2.33

P(x 77) P(z2.33) 1 0.99010.0099

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Z=+2.33