Probability Theory
Example
Let
A = rain today
B = sunny today
C = snowed yesterday
D = it is April
E = cloudy today
F = we are in Seattle
G = we are in Barbados
H = Jeff passes MTH 351
I = Nick passes MTH 351
J = Everyone passes
K = My favorite color is blue
L = it is January
Discuss the probabilities of conditional events
involving A through L.
Probability Theory
Two concepts:
The occurrence of one event may change the
probability of another event happening.
Probability Theory
Two concepts:
The occurrence of one event may change the
probability of another event happening.
e.g.,
P(B) vs. P(B | C)
Conditional
Probability
Probability Theory
Two concepts:
The occurrence of one event may change the
probability of another event happening.
e.g.,
P(B) vs. P(B | C)
Conditional
Probability
Some events have no effect on the probability of other
events.
Probability Theory
Two concepts:
The occurrence of one event may change the
probability of another event happening.
e.g.,
P(B) vs. P(B | C)
Conditional
Probability
Some events have no effect on the probability of other
events.
e.g.,
P(H | K) same as just P(H)
Independent Events
Independence
Formal Definition:
A and B are independent events if
P(A  B) = P(A)·P(B)
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 1: Using what we’ve already learned…
3 ingredients to probability:
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 1: Using what we’ve already learned…
3 ingredients to probability:
- experiment
- state space
- probabilities
[Do it Yourself]
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 1: Using what we’ve already learned…
3 ingredients to probability:
- experiment
- state space
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 1: Using what we’ve already learned…
3 ingredients to probability:
- experiment
- state space
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 1: Using what we’ve already learned…
3 ingredients to probability:
- experiment
- state space
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
S = { 11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 25, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
61, 62, 63, 64, 65, 66 }
3 ingredients to probability:
- experiment
- state space
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
S = { 11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 25, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
61, 62, 63, 64, 65, 66 }
3 ingredients to probability:
- experiment
Easy with outcomes
- state space
that are equally likely
- probabilities
e.g.: cards
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) =
|AB|
|S|
S = { 11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 25, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
61, 62, 63, 64, 65, 66 }
3 ingredients to probability:
- experiment
Easy with outcomes
- state space
that are equally likely
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
S = { 11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 25, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
|AB| |AB|
=
P(A  B) =
61, 62, 63, 64, 65, 66 }
36
|S|
3 ingredients to probability:
- experiment
- state space
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
S = { 11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 25, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
|AB| |AB|
=
P(A  B) =
61, 62, 63, 64, 65, 66 }
36
|S|
3 ingredients to probability:
- experiment
- state space
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) =
|AB|
|S|
=
S = { 11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 25, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
1
61, 62, 63, 64, 65, 66 }
36
3 ingredients to probability:
- experiment
- state space
- probabilities
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 2: Using Independence
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
How does the red die
affect the green die?
B = green die is a 5
P(A  B) ?
Method 2: Using Independence
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
How does the red die
affect the green die?
B = green die is a 5
P(A  B) ?
IT DOESN’T!!!
Method 2: Using Independence
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 2: Using Independence
 P(A  B) = P(A)·P(B) =
Probability Theory
Example
Roll a red die and a green die.
Let A = red die is a 3
B = green die is a 5
P(A  B) ?
Method 2: Using Independence
 P(A  B) = P(A)·P(B) =
1
6
·
1
6
=
1
36
Conditional Probability
2nd concept
Conditional Probability
2nd concept
P(A|B)
• “Probability of A given B”
• the probability of event A occurring when it is known
that event B has already occurred.
Conditional Probability
Example
Deal 2 cards
A = 1st card is red
B = 2nd card is red
Find P(both cards are red)
Conditional Probability
Example
Deal 2 cards
A = 1st card is red
B = 2nd card is red
Find P(both cards are red)
#1: write in terms of mathematical terms
Conditional Probability
Example
Deal 2 cards
A = 1st card is red
B = 2nd card is red
Find P(A  B)
Conditional Probability
Example
Deal 2 cards
A = 1st card is red
B = 2nd card is red
Find P(A  B)
#2: Identify relationships or equations that may be
useful in solving the given problem.
Conditional Probability
Example
Deal 2 cards
A = 1st card is red
B = 2nd card is red
Find P(A  B) = P(A) · P(B|A)
Prob. B given A has occurred
Conditional Probability
Example
Deal 2 cards
A = 1st card is red
B = 2nd card is red
P(A  B) = P(A) · P(B|A) =
Conditional Probability
Example
Deal 2 cards
A = 1st card is red
B = 2nd card is red
P(A  B) = P(A) · P(B|A) =
1
2
·
25
51
=
25
102
Conditional Probability
Calculation:
P(B | A) =
for P(B) > 0.
P(A  B)
P(A)
Conditional Probability
Calculation:
P(B | A) =
P(A  B)
P(A)
for P(B) > 0.
re-written:
P(A | B) = …
Conditional Probability
Calculation:
P(B | A) =
P(A  B)
P(A)
for P(B) > 0.
re-written:
P(A | B) =
P(A  B)
P(B)
Conditional Probability
Calculation:
P(B | A) =
P(A  B)
P(A)
for P(B) > 0.
re-written:
P(A  B) = …
Conditional Probability
Calculation:
P(B | A) =
P(A  B)
P(A)
for P(B) > 0.
re-written:
P(A  B) = P(B | A) · P(A)
OR
P(A | B) · P(B)
Conditional Probability
Create examples to demonstrate what happens
in these two situations:
• Suppose A & B are mutually exclusive.
P(A | B) =
Conditional Probability
Create examples to demonstrate what happens
in these two situations:
• Suppose A & B are mutually exclusive.
P(A | B) =
P(A  B)
P(B)
=
0
P(B)
= 0
Conditional Probability
Create examples to demonstrate what happens
in these two situations:
• Suppose A & B are mutually exclusive.
P(A | B) =
P(A  B)
• Suppose B  A.
P(A | B) =
P(B)
=
0
P(B)
= 0
Conditional Probability
Create examples to demonstrate what happens
in these two situations:
• Suppose A & B are mutually exclusive.
P(A | B) =
P(A  B)
P(B)
=
0
P(B)
= 0
• Suppose B  A.
P(A | B) =
P(A  B)
P(B)
=
P(B)
P(B)
= 1
Probability Theory
Independence and Conditional Probability
Two events A and B are said to be independent events if
P(A  B) = ?
Probability Theory
Independence and Conditional Probability
Two events A and B are said to be independent events if
P(A  B) = P(A)·P(B)
Probability Theory
Independence and Conditional Probability
Two events A and B are said to be independent events if
P(A  B) = P(A)·P(B)
P(B | A) = ?
P(A | B) = ?
[First answer using intuition]
Probability Theory
Independence and Conditional Probability
Two events A and B are said to be independent events if
P(A  B) = P(A)·P(B)
P(B | A) = P(B)
P(A | B) = P(A)
Why does this make sense mathematically?
Probability Theory
Independence and Conditional Probability
Two events A and B are said to be independent events if
P(A  B) = P(A)·P(B)
P(B | A) = P(B)
P(A | B) = P(A)
Any one of these three conditions implies the other two
(i.e., they’re equivalent)
1.4.2
Do it yourself…
Consider the roll of a fair die.
Let A = the event that a prime number is obtained
B = the event that a five is obtained
C = the event that a six is obtained
Calculate P(B | A), P(C | A), and P(A | B).
Conditional Probability
General problem solving strategies:
• Read the problem as many times as necessary in
order to get a good idea of what’s going on.
• Next, concentrate on setting up the problem properly
- phrase the question mathematically
- what information am I given?
- what am I asked to find?
• Identify relationships or equations that may be
useful in solving the given problem.
1.4.6
A car repair is either on time or late and either
satisfactory or unsatisfactory. If a repair is
made on time, then there is a probability of
0.85 that it is satisfactory. There is a probability
of 0.77 that a repair will be made on time.
What is the probability that a repair is made on
time and is satisfactory?
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
S=…
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
S = {MM MF FM FF} or {MM FM FF} ?
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
S = {MM MF FM FF} or {MM FM FF}
It shouldn’t matter!
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
S = {MM MF FM FF} or {MM FM FF}
It shouldn’t matter! But what’s the difference?
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
S = {MM MF FM FF} or {MM FM FF}
all equally likely
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
S = {MM MF FM FF} or {MM FM FF}
can happen two ways
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
Back to the problem…
P(B | A) =
S = {MM MF FM FF}
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
P(A  B)
P(A)
S = {MM MF FM FF}
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
P(A  B)
P(A)
=
P({MM})
P({MM, MF, FM})
S = {MM MF FM FF}
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
P(B | A) =
P(A  B)
P(A)
=
P({MM})
P({MM, MF, FM})
S = {MM MF FM FF}
1
=
3
4
4
=
1
3
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
Alternatively…
P(B | A) =
P(A  B)
P(A)
S = {MM FM FF}
=
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
Alternatively…
P(B | A) =
P(A  B)
P(A)
S = {MM FM FF}
=
P({MM})
P({MM, FM})
Conditional Probability
Example:
Suppose a couple has two children.
A = at least one male
B = both male
Alternatively…
P(B | A) =
P(A  B)
=
P(A)
S = {MM FM FF}
1
4
2
1
4
4
P({MM})
P({MM, FM})
1
=
3
4
4
=
1
3