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Lecture 3
Dustin Lueker

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)

Multiplication rule for independent events A
and B
◦ P(A∩B)=P(A)P(B)
STA 291 Winter 09/10 Lecture 3



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?
STA 291 Winter 09/10 Lecture 3

X is a random variable if the value that X will
assume cannot be predicted with certainty
◦ That’s why its called random

Two types of random variables
◦ Discrete
 Can only assume a finite or countably infinite number
of different values
◦ Continuous
 Can assume all the values in some interval
STA 291 Winter 09/10 Lecture 3

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
STA 291 Winter 09/10 Lecture 3

Suppose we perform several, we’ll say n, Bernoulli
experiments and they are all independent of each
other (meaning the outcome of one even 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

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?
STA 291 Winter 09/10 Lecture 3

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?
STA 291 Winter 09/10 Lecture 3

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 
 4
4!
4!
1 2  3  4


6
 
 2  2!(4  2)! 2! 2! 1  2 1  2
STA 291 Winter 09/10 Lecture 3
 n and 0!  1

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
STA 291 Winter 09/10 Lecture 3
E ( X )    np
Var ( X )    npq
2
SD( X )    npq
STA 291 Winter 09/10 Lecture 3

A list of the possible values of a random
variable X, say (xi) and the probability
associated with each, P(X=xi)
◦ All probabilities must be nonnegative
◦ Probabilities sum to 1
0  P( xi )  1
 P( x )  1
i
STA 291 Winter 09/10 Lecture 3

X
0
1
2
3
4
P(X)
.1
.2
.2
.15
.1
5
6
7
.05 .05 .15
The table above gives the proportion of
employees who use X number of sick days in
a year
◦ An employee is to be selected at random
 Let X = # of days of leave




P(X=2) =
P(X≥4) =
P(X<4) =
P(1≤X≤6) =
STA 291 Winter 09/10 Lecture 3

Expected Value (or mean) of a random
variable X
◦ Mean = E(X) = μ = ΣxiP(X=xi)

Example
X
2
4
6
8
10
12
P(X)
.1
.05
.4
.25
.1
.1
◦ E(X) =
STA 291 Winter 09/10 Lecture 3

Variance
◦ Var(X) = E(X-μ)2 = σ2 = Σ(xi-μ)2P(X=xi)

Example
X
2
4
6
8
10
12
P(X)
.1
.05
.4
.25
.1
.1
◦ Var(X) =
STA 291 Winter 09/10 Lecture 3

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
STA 291 Winter 09/10 Lecture 3

Suppose we perform several, we’ll say n, Bernoulli
experiments and they are all independent of each
other (meaning the outcome of one even 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

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?
STA 291 Winter 09/10 Lecture 3

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?
STA 291 Winter 09/10 Lecture 3

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 
 4
4!
4!
1 2  3  4


6
 
 2  2!(4  2)! 2! 2! 1  2 1  2
STA 291 Winter 09/10 Lecture 3
 n and 0!  1

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
STA 291 Winter 09/10 Lecture 3
E ( X )    np
Var ( X )    npq
2
SD( X )    npq
STA 291 Winter 09/10 Lecture 3