Transcript BINOMIAL distribution and applications.ppt

Problem
A 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 Scale
Absolute certainty—1.0—( a stone thrown up will come done
to earth)
0.9
0.8
0.7
0.6
Equally Likely ---
Absolute
Impossibility–---
0.5--- (when a coin is tossed , the Head
will come up)
0.4
0.3
0.2
0.1
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
transplants in 150 cases and succeeds in 63
cases then probability of survival after
operation in calculated as:
No. of survivals after operations
p =
-------------------------Total no. of cases operated
63
=
------------------------ = 0.42
150
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:
No. of cases died after the operation
q
= -----------------------------------------------------------------Total number of cases operated
87
= ------------------------------- = 0.58
150
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 x4
– Find probability that x5
In medical research, an
outcome of interest can
often be expressed as the
presence or absence of a
particular disease, sign or
symptom or as whether the
patient lived or died, or
recovered or did not recover.
In each case we are dealing
with an outcome in which
exactly one of two
alternatives can occur.
Suppose we know that the
survival rate for a particular
disease is 20% and we have
10 patients with this disease.
We would use the binomial
distribution to calculate the
probability of having 3 or
fewer patients survive. The
answer is 0.88, so that we
have about a 90% chance
of having 3 or fewer patients
surviving (or 7 or more
dying)
(d) P( two or fewer) = P(X<=2)
=1- P(x=3)
=1-0.729
=0.271
(e) P( two or three) = P(X=2 or
X=3)
=P(X=2) +P(X=3)
=0.972
(f) P( exactly three) = P(X=3)
= 0.729