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Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/29/
1
Review for the previous lecture Theorem:
How to calculate probabilities of events (1.2.8, 1.2.9, 1.2.11)
Definition:
! n ,^
n r
Theorem:
How to count.
Number of possible arrangements of size
r
from
n
subjects
Without Replacement
(^ r
n ≤^
With Replacement
Ordered
(^
n n
r −
rn
Unordered
n r
n^
r r +^
^
^
^
Example
: Enumerating Outcomes
Chapter 1 – Probability Theory
1.2.4 Enumerating Outcomes
Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/29/
2
Example:
Suppose there are
(^
n n
students in our class, then what is the probability that at least two students
have the same birthday (ignoring the year of their birth and considering there are only 365 days in a year). Solution:
(^
n
P^
n
^ n
=^
^
^
^
Define the event
A
={at least two students have the same birthday}
then
cA^ ={All students have different birthdays}
of elements in the sample space
S
n
of elements in the event
cA =
! n n ^
^
^
Table:
n^
(^
P A
Example 1.2.19 (Sampling with replacement):
Consider sampling
r^ =
items from
n^
=^
objects, with
replacement. The outcomes in the ordered and unordered sample spaces are these:
Unordered
Ordered
Probability
Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/29/
4
Solution 1 (Counting Rules):
^4
^
^
^
(unordered space).
Solution 2 (Conditioning): 4 / 52
×^
×^
×^
Definition 1.3.2:
If
A
and
B
are events in
S
, and
(^
)^
P B
, then the
conditional probability
of
A
given
B
written as
(^
|^
P A B
is
(^
|^
)^
(^
(^
P A B
P A
B^
P B
=^
. Specifically, if
A
and
B
are disjoint, then
(^
|^
)^
(^
|^
)^
P A B
P B
A
=^
=^
Example 1.3.3:
P
(4 aces in four cards|
i^ aces in
i^
cards)=
P
(4 aces in four cards)/
P
( i
aces in
i^
cards), which
equals to
i^
i ^
^
^
^
^
^
Remark:
We require
(^
)^
P B
for the calculation. We will defer the discussion of this until Chapter 4.
For
(^
)^
P B
, you can verify that the function
P^
B
⋅^
satisfies Kolmogarov’s Axioms. It can be considered
as a probability function defined on
B
The calculation of the conditional probability can be very tricky.
Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/29/
5
Example 1.3.4:
Three prisoners. Three prisoners A, B, and C are on death row. One of them will be randomly
selected and pardoned. One day A asks the warden which of B or C will be executed. The warden says B will beexecuted. What is the probability of A is pardoned when he knows this. Solution:
Define the following events:
A^
={A is pardoned},
B
={B is pardoned},
C
={C is pardoned},
W
={Warden says B will be executed when A asks
him if B or C will be executed}, then
(^
)^
(^
)^
(^
)^
P A
P B
P C
=^
=^
=^
,^
(^
)^
P W
=^
. Because
(^
P W
P
=^
(A is pardoned
and Warden says B will be executed when A asks him if B or C will be executed)+
P
(B is pardoned and Warden
says B will be executed when A asks him if B or C will be executed)+
P
(C is pardoned and Warden says B will be
executed when A asks him if B or C will be executed)= 1/ 6
+^
+^
=^
. Actually,
(^
)^
(^
)^
(^
)^
(^
)^
(^
|^
)^
(^
)^
(^
|^
)^
(^
)^
(^
|^
)^
(^
P W
P W
A^
P W
B^
P W
C^
P W
A P A
P W
B P B
P W
C P C
=^
+^
+^
=^
+^
+^
Also we have
(^
)^
P A
W
=^
, therefore we have
(^
|^
)^
(^
(^
)^
P A W
P A
W
P W
=^
=^
But the conditional probability of
(^
|^
)^
c
P A B
=^
is the probability of A is pardoned when B is executed.
In both situations, A knows that B will be executed. You may be wondering what is the difference between
W
and
cB. The difference is that A only asks the status of B and C. If A asks the warden who will be executed (including A himself) and the warden says B will be executed. Then the conditional probability of A is pardoned is 1/ 2 .Please verify this by yourself. Remark: 1.
(^
|^
)^
(^
|^
)^
(^
(^
)^
(^
)^
(^
)^
(^
|^
)^
(^
)^
(^
|^
)^
(^
)^
(^
|^
)^
(^
P A W
P W
A P A
P W
A^
P W
B^
P W
C^
P W
A P A
P W
B P B
P W
C P C
=^
+^
+^
=^
+^
(^
)^
(^
|^
)^
(^
)^
(^
|^
)^
(^
P A
B^
P A B P B
P B
A P A
=^
=^
1
1
(^
)^
(^
)^
(^
|^
)^
(^
i^
i^
i
i^
i
P B
P B
A^
P B
A P A
∞^
∞
=^
=
=^
Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/29/
7
1
1 (^
)^
(^
j^
j
k^
k
i^
i
j
j P^
A^
P A
=
=
=^
Remark:
If a collection of events
1
2 ,^
A^
A^
are mutually independent, then all the pairs of them are independent.
However, even all of pairs are independent; they may not be mutually independent. Example:
If a collection of events
1
2 ,^
A^
A^
are mutually independent,
(^
) i
i
P A
p =^
(^
1,^
i^
n
=^
), then calculate the
probability of the following events:(1)
n^^1
c i i
B^
A
= =^
(none of
A^ occurs): i
1
(^
)^
n
i
i
P B
p
=
=^
n^1
i i
C^
A
= =^
(at least of
A^ occurs): i
1
(^
)^
n
i
i
P C
p
=
=^
−^
(^
n^
c
i^
j
i^
j^ i
D
A^
A
=^
≠
=^
(exactly one
A^ occurs): i
1
(^
)^
(^
n
i^
j
i^
j^ i
P D
p^
p
=^
=^
Example:
A large machine consists of 50 components. Past experience has shown that a particular component has
a probability of 0.1% to fail. Suppose that the performance of one component will not affect another. Theequipment will work if no more than one component fails. What is the probability that the machine will work? Solution:
Define
A^ = i
{the
i^
th component}, then
(^
)^
i P A
=^
and
(^
A ii
are mutually independent.
Define
D
={the equipment will work} ={exactly one component fails }
{none of component fails}, therefore,
50
49
(^
)^
P D
^
=^
−^
+^
−^
^
^
^
We can get
(^
)^
P D
and
(^
)^
P D
if^
(^
)^
i P A
=^
and
(^
)^
i P A
=^
, respectively.
