Probability

2. Probability

3. • Example: Soccer Game
• You are off to soccer, and love being the Goalkeeper, but that
• depends who is the Coach today:
• with Coach Sam the probability of being Goalkeeper is 0.5
• with Coach Alex the probability of being Goalkeeper is 0.3
• Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6).
• So, what is the probability you will be a Goalkeeper today?
• Let's build the tree diagram. First we show the two possible coaches: Sam or Alex:

4. • The probability of getting Sam is 0.6, so the probability of Alex must
be 0.4 (together the
• probability is 1)
• Now, if you get Sam, there is 0.5 probability of being Goalie (and 0.5
of not being Goalie):
If you get Alex, there is 0.3 probability of being Goalie (and 0.7 not):

5. • (When we take the 0.6 chance of Sam being coach and include the 0.5
chance that Sam will
• let you be Goalkeeper we end up with an 0.3 chance.)
• But we are not done yet! We haven't included Alex as Coach:
• An 0.4 chance of Alex as Coach, followed by an 0.3 chance gives 0.12.
• Now we add the column:
• 0.3 + 0.12 = 0.42 probability of being a Goalkeeper today
• (That is a 42% chance)

6. Conditional Probability

12. Fundamentals for Bayes Theorem
● 0 ≤ P(A) ≤ 1, where P(A) is the probability of an event A.
● P(A) = 0, indicates total uncertainty in an event A.
● P(A) =1, indicates total certainty in an event A.
● Event: Each possible outcome of a variable is called an event.
● Sample space: The collection of all possible events is called sample
space.
● Random variables: Random variables are used to represent the events
and objects in the real world.
● Prior probability: The prior probability of an event is probability
computed before observing new information.
● Posterior Probability: The probability that is calculated after all
evidence or information has taken into account. It is a combination of
prior probability and new information.

13. ● Bayes' theorem is also known as Bayes' rule, Bayes' law,
or Bayesian reasoning, which determines the probability
of an event with uncertain knowledge.
● In probability theory, it relates the conditional
probability and marginal probabilities of two random
events.
● Bayes' theorem was named after the British
mathematician Thomas Bayes. The Bayesian inference
is an application of Bayes' theorem, which is
fundamental to Bayesian statistics.
● It is a way to calculate the value of P(B|A) with the
knowledge of P(A|B).

14. • Bayes' Theorem is a way of finding a probability when we know certain other
probabilities. The formula is:
• P(A|B) = P(A) P(B|A)/P(B)
• Which tells us: how often A happens given that B happens, written P(A|B),
• When we know: how often B happens given that A happens, written P(B|A)
• and how likely A is on its own, written P(A)
• and how likely B is on its own, written P(B)

15.  Bayes's theorem is expressed mathematically by the following equation that is
given below.
 𝐏 𝐗 ∣ 𝐘 =
𝐏 𝐘∣𝐗 𝐏 𝐗
𝐏 𝐘
 Where X and Y are the events and P (Y) ≠ 0
 P(X/Y) is a conditional probability that describes the occurrence of event X is
given that Y is true.
 P(Y/X) is a conditional probability that describes the occurrence of event Y is
given that X is true.
 P(X) and P(Y) are the probabilities of observing X and Y independently of each
other. This is known as the marginal probability.
 Let us say P(Fire) means how often there is fire, and P(Smoke) means how often we see
smoke, then:
 P(Fire|Smoke) means how often there is fire when we can see smoke
 P(Smoke|Fire) means how often we can see smoke when there is fire
 So the formula kind of tells us "forwards" P(Fire|Smoke) when we know "backwards"
P(Smoke|Fire)

16.  Example:
 dangerous fires are rare (1%)
 but smoke is fairly common (10%) due to barbecues,
 and 90% of dangerous fires make smoke
 We can then discover the probability of dangerous Fire when there is Smoke:
 P(Fire|Smoke) = P(Fire) P(Smoke|Fire)/P(Smoke)
 = 1% x 90%10%
 = 9%

18. ● Bayes' theorem can be derived using product rule and conditional probability of event A
with known event B:
● As from product rule we can write: P(A ⋀ B)= P(A|B) P(B) or
● Similarly, the probability of event B with known event A: P(A ⋀ B)= P(B|A) P(A)
● According to above statements the equation for Bayes’ rule or Bayes’ theorem is as
below:
 𝐏 𝐀 ∣ 𝐁 =
𝐏 𝐁∣𝐀 𝐏 𝐀
𝐏 𝐁
● P(A|B) is known as posterior, which we need to calculate, and it will be read as
Probability of hypothesis A when we have occurred an evidence B.
● P(B|A) is called the likelihood, in which we consider that hypothesis is true, then we
calculate the probability of evidence.
● P(A) is called the prior probability, probability of hypothesis before considering the
evidence
● P(B) is called marginal probability, pure probability of an evidence.
● In the equation (a), in general, we can write P (B) = P(A)*P(B|Ai), hence the Bayes' rule
can be written as:
 𝐏 𝐀𝐢 ∣ 𝐁 =
𝐏 𝐀𝐢 𝐏 𝐁∣𝐀𝐢
𝐢=𝟏
𝐊 𝐏 𝐀𝐢 ∗𝐏 𝐁∣𝐀𝐢
 Where A1, A2, A3, ..............., An is a set of mutually exclusive and exhaustive events.

19.  Example: In a particular pain clinic, 10% of patients are prescribed narcotic pain
killers. Overall, 5% of the clinic's patients are addicted to narcotics(including
pain killers and illegal substances). Out of all the people restricted pain pills, 8%
are addicts. If a patient is an addict, what is the probability that they will be
prescribed pain pills?
 A: Being prescribed pain pills 10%
 B: Being an addict 5%
 B|A: Out of all 8% addict for narcotics.
 P(A|B)=P(B|A)P(A)/P(B)
 P(A|B)=(0.08 * 0.1)/0.05 =0.16
 That means
 If a patient is an addict, then there are 16% chances to prescribe pain pills.

21. ● According to the given statement the below are statistics:
● The prior probability of having cancer among population is P(Cancer) Which is
P(Cancer)=0.08
● The prior probability of not having cancer among population is P(‫ך‬Cancer) Which is
P(‫ך‬Cancer)=1-0.08=0.992
● The cancer test results are accurate up to 98%. It is indicated as P(+|Cancer)=0.98 and
2% of results are negative such as: P(‫|ך‬Cancer)=0.2(1-0.98)
● Probability of positive result for no cancer is P(+|‫ך‬Cancer)=1-0.97=0.3
● Probability of negative result for no cancer is P(-|‫ך‬Cancer)=0.97

22. Bayes’ Rule
Let S1 , S2 , S3 ,..., Sk be mutually exclusive and
exhaustive events with prior probabilities P(S1),
P(S2),…,P(Sk). If an event A occurs, the posterior
probability of Si, given that A occurred is
,...k
,
i
S
A
P
S
P
S
A
P
S
P
A
S
P
i
i
i
i
i 2
1
for
)
|
(
)
(
)
|
(
)
(
)
|
( 


)
|
(
)
(
)
|
(
)
(
)
(
)
(
)
|
(
)
|
(
)
(
)
(
)
(
)
(
)
|
(
Proof
i
i
i
i
i
i
i
i
i
i
i
i
S
A
P
S
P
S
A
P
S
P
A
P
AS
P
A
S
P
S
A
P
S
P
AS
P
S
P
AS
P
S
A
P








23. We know:
P(F) =
P(M) =
P(H|F) =
P(H|M) =
Example
From a previous example, we know that 49% of the
population are female. Of the female patients, 8% are
high risk for heart attack, while 12% of the male patients
are high risk. A single person is selected at random and
found to be high risk. What is the probability that it is a
male? Define H: high risk F: female M: male
61
.
)
08
(.
49
.
)
12
(.
51
.
)
12
(.
51
.
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(





F
H
P
F
P
M
H
P
M
P
M
H
P
M
P
H
M
P
.12
.08
.51
.49

24. Example
Suppose a rare disease infects one out of every
1000 people in a population. And suppose that
there is a good, but not perfect, test for this
disease: if a person has the disease, the test
comes back positive 99% of the time. On the other
hand, the test also produces some false positives:
2% of uninfected people are also test positive.
And someone just tested positive. What are his
chances of having this disease?

25. We know:
P(A) = .001 P(Ac) =.999
P(B|A) = .99 P(B|Ac) =.02
Example
Define A: has the disease B: test positive
0472
.
02
.
999
.
99
.
001
.
99
.
001
.
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(







 c
A
B
P
c
A
P
A
B
P
A
P
A
B
P
A
P
B
A
P
We want to know P(A|B)=?

26. Example
A survey of job satisfaction2 of teachers was
taken, giving the following results
2 “Psychology of the Scientist: Work Related Attitudes of U.S. Scientists”
(Psychological Reports (1991): 443 – 450).
Satisfied Unsatisfied Total
College 74 43 117
High School 224 171 395
Elementary 126 140 266
Total 424 354 778
Job Satisfaction
L
E
V
E
L

27. Example
If all the cells are divided by the total number
surveyed, 778, the resulting table is a table of
empirically derived probabilities.
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction

28. Example
For convenience, let C stand for the event that the
teacher teaches college, S stand for the teacher being
satisfied and so on. Let’s look at some probabilities
and what they mean.
is the proportion of teachers who are college teachers.
P(C) 0.150

is the proportion of teachers who are satisfied with
their job.
P(S) 0.545

is the proportion of teachers who are college teachers
and who are satisfied with their job.
P(C S) 0.095

Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction

29. Example
is the proportion of teachers who are college
teachers given they are satisfied. Restated:
This is the proportion of satisfied that are
college teachers.
P(C S)
P(C | S)
P(S)
0.095
0.175
0.545

 
is the proportion of teachers who are satisfied
given they are college teachers. Restated:
This is the proportion of college teachers that
are satisfied.
P(S C)
P(S | C)
P(C)
P(C S) 0.095
P(C) 0.150
0.632

 

Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction

30. Example
P(C S) 0.095
P(C) 0.150 and P(C | S) 0.175
P(S) 0.545
   
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C|S)  P(C) so C and S are dependent events.
Are C and S independent events?

31. Example
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.658
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C) = 0.150, P(S) = 0.545 and
P(CS) = 0.095, so
P(CS) = P(C)+P(S) - P(CS)
= 0.150 + 0.545 - 0.095
= 0.600
P(CS)?

32. Tom and Dick are going to take
a driver's test at the nearest DMV office.
Tom estimates that his chances to pass the
test are 70% and Dick estimates his as
80%. Tom and Dick take their tests
independently.
Define D = {Dick passes the driving test}
T = {Tom passes the driving test}
T and D are independent.
P (T) = 0.7, P (D) = 0.8
Example

33. What is the probability that at most one of the two friends will pass the test?
Example
P(At most one person pass)
= P(Dc  Tc) + P(Dc  T) + P(D  Tc)
= (1 - 0.8) (1 – 0.7) + (0.7) (1 – 0.8) + (0.8) (1 –
0.7)
= .44
P(At most one person pass)
= 1-P(both pass) = 1- 0.8 x 0.7 = .44

34. What is the probability that at least one of the two friends will pass the test?
Example
P(At least one person pass)
= P(D  T)
= 0.8 + 0.7 - 0.8 x 0.7
= .94
P(At least one person pass)
= 1-P(neither passes) = 1- (1-0.8) x (1-0.7) =
.94

35. Suppose we know that only one of the two friends passed the test. What is
the probability that it was Dick?
Example
P(D | exactly one person passed)
= P(D  exactly one person passed) / P(exactly
one person passed)
= P(D  Tc) / (P(D  Tc) + P(Dc  T) )
= 0.8 x (1-0.7)/(0.8 x (1-0.7)+(1-.8) x 0.7)
= .63