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Lecture 7
Dustin Lueker
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Experiment
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Random (or Chance) Experiment
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Outcome
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Sample Space
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Event
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Simple Event
◦ Any activity from which an outcome, measurement, or other
such result is obtained
◦ An experiment with the property that the outcome cannot
be predicted with certainty
◦ Any possible result of an experiment
◦ Collection of all possible outcomes of an experiment
◦ A specific collection of outcomes
◦ An event consisting of exactly one outcome
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Let A and B denote two events
Complement of A
◦ All the outcomes in the sample space S that do not
belong to the even A
◦ P(Ac)=1-P(A)
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Union of A and B
◦ A∪B
◦ All the outcomes in S that belong to at least one of
A or B
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Intersection of A and B
◦ A∩B
◦ All the outcomes in S that belong to both A and B
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Let A and B be two events in a sample space S
◦ P(A∪B)=P(A)+P(B)-P(A∩B)
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A and B are Disjoint (mutually exclusive)
events if there are no outcomes common to
both A and B
◦ A∩B=Ø
 Ø = empty set or null set
◦ P(A∪B)=P(A)+P(B)
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Can be difficult
Different approaches to assigning probabilities to
events
◦ Subjective
◦ Objective
 Equally likely outcomes (classical approach)
 Relative frequency
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Relies on a person to make a judgment as to
how likely an event will occur
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Events of interest are usually events that cannot be
replicated easily or cannot be modeled with the
equally likely outcomes approach
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As such, these values will most likely vary from
person to person
The only rule for a subjective probability is
that the probability of the event must be a
value in the interval [0,1]
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The equally likely approach usually relies on
symmetry to assign probabilities to events
◦ As such, previous research or experiments are not
needed to determine the probabilities
 Suppose that an experiment has only n outcomes
 The equally likely approach to probability assigns a
probability of 1/n to each of the outcomes
 Further, if an event A is made up of m outcomes then
P(A) = m/n
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Borrows from calculus’ concept of the limit
a
P( A)  lim
n  n
◦ We cannot repeat an experiment infinitely many
times so instead we use a ‘large’ n
 Process
 Repeat an experiment n times
 Record the number of times an event A occurs, denote this
value by a
 Calculate the value of a/n
a
P( A) 
n
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Let A be the event A = {o1, o2, …, ok}, where
o1, o2, …, ok are k different outcomes
P( A)  P(o1 )  P(o2 ) 
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 P(ok )
Suppose the first digit of a license plate is
randomly selected between 0 and 9
◦ What is the probability that the digit 3?
◦ What is the probability that the digit is less than 4?
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P( A  B)
P( A | B) 
, provided P( B)  0
P( B)
◦ Note: P(A|B) is read as “the probability that A
occurs given that B has occurred”
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If events A and B are independent, then the
events have no influence on each other
◦ P(A) is unaffected by whether or not B has occurred
◦ Mathematically, if A is independent of B
 P(A|B)=P(A)
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Multiplication rule for independent events A
and B
◦ P(A∩B)=P(A)P(B)
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Flip a coin twice, what is the probability of
observing two heads?
Flip a coin twice, what is the probability of
observing a head then a tail? A tail then a
head? One head and one tail?
A 78% free throw shooter is fouled while
shooting a three pointer, what is the
probability he makes all 3 free throws? None?
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A random variable X is called a Bernoulli r.v. if
X can only take either the value 0 (failure) or
1 (success)
 Heads/Tails
 Live/Die
 Defective/Nondefective
◦ Probabilities are denoted by
 P(success) = P(1) = p
 P(failure) = P(0) = 1-p = q
◦ Expected value of a Bernoulli r.v. = p
◦ Variance = pq
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Suppose we perform several, we’ll say n, Bernoulli
experiments and they are all independent of each
other (meaning the outcome of one event doesn’t
effect the outcome of another)
◦ Label these n Bernoulli random variables in this manner: X1,
X2,…,Xn
 The probability of success in a single trial is p
 The probability of success doesn’t change from trial to trial
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We will build a new random variable X using all of
these Bernoulli random variables: n
X  X1  X 2 
 Xn   Xi
i 1
◦ What are the possible outcomes of X? What is X counting?
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The probability of observing k successes in n
independent trails is
 n  k nk
P( X  k )    p q , k  0,1,
k 
, n,
◦ Assuming the probability of success is p
◦ Note:
n
n!
  
 k  k!(n  k )!
 Why do we need this?
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For small n, the Binomial coefficient “n
choose k” can be derived without much
mathematics
n
n!
 
 k  k !(n  k )!
Example:
where n !  1 2  3 
 n and 0!  1
 4
4!
4!
1 2  3  4


6
 
 2  2!(4  2)! 2! 2! 1  2 1  2
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Assume Zolton is a 68% free throw shooter
◦ What is the probability of Zolton making 5 out of 6
free throws?
6
P ( X  5)    0.685 (1  0.68) 65
5
 6  0.1454  0.32  0.279
◦ What is the probability of Zolton making 4 out of 6
free throws?
6
4
6 4
P( X  4)    0.68 (1  0.68)
 4
 15  0.2138  0.1024  0.3284
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  n p
  n  p  (1  p)
2
  n  p  (1  p)
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