ProblemA newly married couple plans to have four children and would like to have three girls and a boy. What are the chances (probability) their desire will be fulfilled ?

Problem

The probability that a person will die within a month after a heart transplant is 0.18 (18%). What are the probabilities that in three such operations one, two or all three persons will survive ?

BINOMIAL

DISTRIBUTION

AND ITS

APPLICATION

Probability ScaleProbability Scale

Absolute certainty—1.0—( a stone thrown up will come done Absolute certainty—1.0—( a stone thrown up will come done

to earth)to earth)

0.90.9

0.80.8

0.70.7

0.60.6

Equally Likely --- 0.5--- (when a coin is tossed , the Head Equally Likely --- 0.5--- (when a coin is tossed , the Head

will come up)will come up)

0.40.4

0.30.3

0.20.2 0.1 0.1

Absolute Absolute

Impossibility–--- 0.0– ( I will travel to the sun tomorrow)Impossibility–--- 0.0– ( I will travel to the sun tomorrow)

Characteristics of Probabilities

• Probabilities are expressed as fractions between 0.0 and 1.0– e.g., 0.01, 0.05, 0.10, 0.50, 0.80– Probability of a certain event = 1.0– Probability of an impossible event =

0.0

• Application to biomedical research– e.g., ask if results of study or

experiment could be due to chance alone

– e.g., significance level and power– e.g., sensitivity, specificity,

predictive values

Basic Probability Concepts

• Foundation of statistics because of the concept of sampling and the concept of variation or dispersion and how likely an observed difference is due to chance

• Probability statements used frequently in statistics– e.g., we say that we are 90%

sure that an observed treatment effect in a study is real

For example, if a surgeon performs renal For example, if a surgeon performs renal transplants in 150 cases and succeeds in 63 transplants in 150 cases and succeeds in 63 cases then probability of survival after cases then probability of survival after operation in calculated as:operation in calculated as:

No. of survivals after operationsNo. of survivals after operations

p = --------------------------p = -------------------------- Total no. of cases operatedTotal no. of cases operated

6363

= ------------------------ = 0.42= ------------------------ = 0.42

150150

And chances of not surviving i.e., probability of dying cases, And chances of not surviving i.e., probability of dying cases, q or (1-p) is 1-0.42 = 0.58 which can be shown as:q or (1-p) is 1-0.42 = 0.58 which can be shown as:

No. of cases died after the operationNo. of cases died after the operation

q = ------------------------------------------------------------------q = ------------------------------------------------------------------

Total number of cases operatedTotal number of cases operated

8787

= ------------------------------- = 0.58= ------------------------------- = 0.58

150150

Definition of Probabilities

• If some process is repeated a large number of times, n, and if some resulting event with the characteristic of E occurs m times, the relative frequency of occurrence of E, m/n, will be approximately equal to the probability of E: P(E)=m/n

• Also known as relative frequency

Elementary Properties of Probabilities - I

• Probability of an event is a non-negative number– Given some process (or

experiment) with n mutually exclusive outcomes (events), E1, E2, …, En, the probability of any event Ei is assigned a nonnegative number

– P(Ei) 0

Elementary Properties of Probabilities - II

• Sum of the probabilities of mutually exclusive outcomes is equal to 1– Property of exhaustiveness

• refers to the fact that the observer of the process must allow for all possible outcomes

– P(E1) + P(E2) + … + P(En) = 1

Elementary Properties of Probabilities - III

• Probability of occurrence of either of two mutually exclusive events is equal to the sum of their individual probabilities– Given two mutually

exclusive events A and B

– P(A or B) = P(A) + P(B)

Examples:(1) Suppose we toss a die. What

is the probability of 4 coming up?

since there are six mutually exclusive and equally likely outcomes out of which 4 in only one, the probability of 4 coming up is 1/6.

(2) Suppose we toss 2 coins. We can have the following outcomes: both heads, HH; one head and the other tail, TH or HT; and both tails, TT (H=Head; T=Tail).

Suppose we want to know the probability of HH.

HH being one of the four equally likely outcomes, the probability of obtaining HH is ¼.

(3) Suppose we throw 2 dice and we want the probability of a total of 7 points.

A total of 7 can come in 6 ways

(1-6,2-5,3-4,4-3,5-2, or 6-1). So the numerator will be 6. Since we have 6 sides for each die, the total number of ‘equally likely’ ‘mutually exclusive’ outcomes is 6 x 6 =36. So the chance of getting a total of 7 when we throw 2 dice is 6/36 (or 1/6).

Examples: (Addition Law)

When we toss a die, what is the probability of getting 2 or 4 or 6 ?

The prob. of 2 =1/6

The prob. of 4=1/6

The prob. of 6=1/6

Probability of 2 or 4 or 6 is:

1/6 +1/6+1/6 = 3/6 = ½

(Multiplication Law)

In tossing 2 coins,

Prob. of head in one coin =1/2

Prob. of head in another coin=1/2

Thus prob. of head in both coins

=1/2 x 1/2 =1/4

Combinations

• Based on last example, it is clear that we need to calculate more easily the probability of a particular result– If a set consists of n objects, and we

wish to form a subset of x objects from these n objects, without regard to order of the objects in the subset, the result is called a combination

• The number of combinations of n objects taken x at a time is given by

– nCk = n! / (k! ( n-k)!)

– Where k! (factorial) is the product of all numbers from k to 0

• 0! = 1

Permutations

• Similar to combinations– If a set consists of n objects, and we

wish to form a subset of x objects from these n objects, taking into account the order of the objects in the subset, the result is called a permutation

• The number of permutations of n objects taken x at a time is given by

– nPk = n! / ( n-k)!

Probability distributions of discrete variables

• A table, graph, formula, or other device used to specify all possible values of a discrete random variable along with their respective probabilities– P(X=x)

• Tables – value, frequency, probability• Graph – usually bar chart• Formula - Binomial distribution

Theoretical Probability Distributions

-- If we know (reasonably) that data are from a certain distribution, than we know a lot about it

-- Means, standard deviations, other measures of dispersion

– That knowledge makes it easier to make statistical inference; i.e., to test differences

• Many types of distributions

– 1300+ have been documented in the literature

• Three main ones

– Binomial (discrete - 0,1)

– Poisson (discrete counts)

– Normal (continuous)

Binomial Distribution

• Derived from a series of binary outcomes called a Bernoulli trial

• When a random process or experiment, called a trial, can result in only one of two mutually exclusive outcomes, such as dead or alive, sick or well, the trial is called a Bernoulli trial

Bernoulli Process

• A sequence of Bernoulli trials forms a Bernoulli process under the following conditions– Each trial results in one of two

possible, mutually exclusive, outcomes: “success” and “failure”

– Probability of success, p, remains constant from trial to trial. Probability of failure is q = 1-p.

– Trials are independent; that is, success in one trial does not influence the probability of success in a subsequent trial.

Bernoulli Process - Example

• Probability of a certain sequence of binary outcomes (Bernoulli trials) is a function of p and q.

• For example, a particular sequence of 3 “successes” and 2 “failures” can be represented by p*p*p*q*q; = p3q2

• However, if we ask for the probability of 3 “successes” and 2 “failures” in a set of 5 trials, then we need to know how may possible combinations of 3 successes and 2 failures out of all of the possible outcomes there are.

Binomial Distribution

• The binomial probability density function

–f(x) = nCx px qn-x

for x=0,1,2,3…,n

= n! / (k! ( n-k)!) px qn-x

–This is called the binomial distribution

Parameters of Binomial distribution

n (the number of objects) and p (the probability of a ‘success’) are the two unknown quantities which define a binomial distribution. They are called the parameters of the binomial distribution.

The Binomial distribution is applicable to the situation where:

i) the n trails are independent (ie., what occurs in one trail does not affect what will occur in the next trail),

ii) at each trail there is only two possible outcomes (‘success’ or ‘failure’)

iii) the probability of a ‘success’ (p) should be known and is the same for all trails

Mean and Variance of the Binomial distribution

To obtain the frequency distribution for a particular combination of ’ n ‘and ‘p ‘we need to calculate the probabilities

p (X=0), p (X=2), ------, p (X=n)Where X is the random variable representing

the number of ‘successes’ in ‘n’ trails. Where ‘n’ is large, the calculations of these probabilities becomes very tedious and time consuming.

However, we can obtain the mean and variance of ‘X’ as a summary of the distribution.

For a binomial distribution mean ( ) is given by ‘np’ and the variance (2 ) is equal to

‘ np(1-p)’.The mean is the average value of the

random variable that would be expected to occur in the long run and the standard deviation is the expected variation from the mean

Binomial Table

• Normally, we would look up probabilities in the Binomial Table

• Tables of the Binomial probability distribution function– P (X=k)– Find probability that x=4 successes

when n trials = 10 and p of success = 0.3

– Find probability that x4– Find probability that x5

In medical research, an In medical research, an outcome of interest can outcome of interest can often be expressed as the often be expressed as the presence or absence of a presence or absence of a particular disease, sign or particular disease, sign or symptom or as whether the symptom or as whether the patient lived or died, or patient lived or died, or recovered or did not recover. recovered or did not recover. In each case we are dealing In each case we are dealing with an outcome in which with an outcome in which exactly one of two exactly one of two alternatives can occur.alternatives can occur.

Suppose we know that the Suppose we know that the survival rate for a particular survival rate for a particular disease is 20% and we have disease is 20% and we have 10 patients with this disease. 10 patients with this disease. We would use the binomial We would use the binomial distribution to calculate the distribution to calculate the probability of having 3 or probability of having 3 or fewer patients survive. The fewer patients survive. The answer is 0.88, so that we answer is 0.88, so that we have about a 90% chance have about a 90% chance of having 3 or fewer patients of having 3 or fewer patients surviving (or 7 or more surviving (or 7 or more dying)dying)

(d) P( two or fewer) = P(X<=2)(d) P( two or fewer) = P(X<=2)

=1- P(x=3)=1- P(x=3)

=1-0.729=1-0.729

=0.271=0.271

(e)(e) P( two or three) = P(X=2 orP( two or three) = P(X=2 or

X=3)X=3)

=P(X=2) +P(X=3)=P(X=2) +P(X=3)

=0.972=0.972

(f) P( exactly three) = P(X=3)(f) P( exactly three) = P(X=3)

= 0.729= 0.729