Lecture 4
description
Transcript of Lecture 4
Lecture 4: Conditional Probability
1
Lecture 4
Conditional Probability, Total Probability Rule
Instructor: Kaveh Zamani
Course material mainly developed by previous instructors:Profs. Mokhtarian and Kendall, Ms. Reardon
Lecture 4: Conditional Probability
2
Reminder
Previous Lecture:
- Axioms of probability• P(S) =1• 0P(A) 1•A1 and A2 with A1 ∩ A2 = ∅P(A1 ∪ A2 ) = P(A1) + P(A2)- Probability of multiplication- Mutually exclusive events
Next session:
Reading: section 2.7 text book HW #2: posted on SmartSite problems:
Lecture 4: Conditional Probability
3
Definition
Sometimes probabilities need to be
reevaluated as additional information
becomes available
The probability of an event B under the
knowledge that the outcome will be in event
A is denoted as: P(B|A)
This is called the conditional probability of
B given A
Lecture 4: Conditional Probability
4
Example 1,2
A: event of rainy day in May B: event of day colder than 40 °F in May
P(A|B) > P(A)Chance of rain in a cold weather is higher than the
average chance of rain! A: event of certain heart disease in people older than 60 B: event of abdominal obesity in people older than 60 P(A)= 10% Probability of the heart diseaseP(B)= 30% Probability of abdominal obesityP(B’)=1-P(B) Probability of being slim P(A|B)= 20% probability of the heart disease given that the person is fat P(A|B) > P(A), P(A|B’)< P(A)
Lecture 4: Conditional Probability
5
Example 3: Welding (Page 41)
Automatic welding devices have error rates of
1/1000Errors are rare, but when they occur, because of wearing of the device, they tend to occur in groups that affect many consecutive welds
If a single weld is performed, we might assume
the probability of an error as 1/1000
However, if the previous welding was wrong,
because of the wearing, we might believe that the
probability that the next welding is wrong is greater
than 1/1000, [P(Wi+1|Wi)>1/1000]
Lecture 4: Conditional Probability
6
Example 4: Manufacturing (Page 42)
D : a part of a steel column is defectiveF : a part of a steel column has a surface defect, P(F) = 0.10, P(D|F) = 0.25 and P(D|F’) = 0.05
Lecture 4: Conditional Probability
7
Conditional Probability
• The conditional probability of an event B given an event A, denoted as P(B|A), is P(B|A) = P(B ∩ A) / P(A) for P(A)>0• Therefore, P(B|A) can be interpreted as the relative frequency of event B among the trials that produce an outcome in event A• It is like scaling down to a smaller sample space
Lecture 4: Conditional Probability
8
Example 4
• D = a part is defective• F = a part has a surface flaw [ P(F) = 0.10 ]• P(D|F) = 0.25 and P(D|F’) = 0.05• P(D|F) = P(D ∩ F) / P(F) = 0.025 / 0.1 = 0.25
Lecture 4: Conditional Probability
9
Example 5: (2-78, Page 45)
•100 samples of a cast aluminum part are summarized as:
P(A) = 82/100 = 0.82P(B) = 90/100 = 0.90P(B|A)= 80/82 = 0.9756P(A|C)= 2/10 = 0.2P(A|B)= 80/90 = 0.889
Lecture 4: Conditional Probability
10
Exercise 2-85 (Page 46)
A batch of 350 steel bars contains 8 that are defective, 2 are
selected, at random, without replacement from the batch
What is the probability that … :
1) both are defective?
P(D1 ∩ D2) = P(D1|D2) P(D2) = P(D1) P(D2) = 8/350 ⋅ 7/349 =
0.000458
2) the second one selected is defective given that the first one
was defective?
P(D2|D1) = P(D2 ∩ D1) / P(D1) = 0.000458/(8/350) = 0.020057
= (7/349)
3) both are acceptable?
P(D1’ ∩ D2’) = P(D1’|D2’) P(D2’) = P(D1’) P(D2’)= 342/350 ⋅
341/349 = 0.954744
11
Exercise 2-85 (Cont.)
Lecture 4: Conditional Probability
Reminder: De Morgan’s rule P(A’)=1-P(A)
The probability that both are acceptable can also be
found as follows:
P(D2 ∩ D1)=P[(D2 ∪ D1)’]=1-P(D2 ∪ D1)
=1-[P(D1)+P(D2)-P(P(D2 ∩ D1)] =
1-[8/350+8/350-0.000458]=0.954744
We are looking at the event (D1 or D2), so P(D2)
depends on the outcome of the first selection, which can
be either defective or acceptable
Therefore, we can use the TOTAL PROBABILITY RULE, to
obtain:
P(D2)=P(D2|D1)P(D1)+P(D2|D’1)P(D’1)
=7/349.8/350+8/349.342/350=8/350
Lecture 4: Conditional Probability
12
Multiplication Rule
When the probability of the intersection is needed:
P(A ∩ B) = P(B|A) P(A) = P(A|B) P(B)
Example: a concrete batch passes compressive tests
with
P(A) = 0.90; a second concrete batch is known to pass the
tests if the first already does, with P(B|A) = 0.95
What is the probability P(A ∩ B) that both pass the tests?
Ans: P(A ∩ B) = P(B|A) P(A) = 0.95⋅0.90 = 0.855
Note: it is also true that P(A ∩ B) = P(A|B) P(B), but the
Information provided in the question does not match this
second formulation
Lecture 4: Conditional Probability
13
Total Probability Rule
Sometimes, the probability of an event can be recovered by summing up a series of conditional probabilities. Everyday life example: If a student is undergrad there is 60% chance he/she passes ECI-114, and if he/she is a grad student there is 70% chance he/she passes Eci-114. P(A): Student is undergrad (55%) P(A’): Student is grad (45%) P(B): he/she passes ECI-114P(B)= P(B ∩ A) + P(B ∩ A’)= P(B|A) P(A) + P(B|A’) P(A’) = 60/100.55/100+70/100.45/100 = 33/100+31.5/100=64.5/100
Lecture 4: Conditional Probability
14
Total Probability Rule
For two events we have:P(B) = P(B ∩ A) + P(B ∩ A’) = P(B|A) P(A) + P(B|A’) P(A’)
Lecture 4: Conditional Probability
15
Total Probability Rule for Multiple Events
For several mutually exclusive and exhaustive events we have:P(B) = P(B ∩ E1) + P(B ∩ E2) + ...+ P(B ∩ Ek) = P(B|E1)P(E1) + P(B|E2) P(E2) + … + P(B|Ek) P(Ek)
Exhaustive Events: E1 ∪ E2 ∪ … ∪ Ek = S Mutually Exclusive Event: can NOT happen at the same time P(Ei ∩ Ej )=0 (i ≠ j)
Lecture 4: Conditional Probability
16
Example 1 (Page 48)
A member fails when subjected to various stress levels, with the following probability
Probability of Failure
Level of stress
Probability of stress level
0.1 High 0.2
0.005 Not high 0.8
Let F: Failure and H: member has high stress level What is the probability of failure of the member?Ans. P(F) = P(F|H) P(H) + P(F|H’) P(H’) = 0.10 ⋅ 0.20 + 0.005 ⋅ 0.80 = 0.024
Sum is 1
Lecture 4: Conditional Probability
17
Example 2 (Page 48)
A water treatment unit fails when subjected to various contamination levels, with the following probabilityProbability
of FailureContaminati
on LevelProbability
of Level
0.1 High 0.2
0.01 Medium 0.3
0.001 Low 0.5
Let H, M, L = member has high/medium/low contamination level
Ans. P(F) = P(F|H) P(H) + P(F|M) P(M) + P(F|L) P(L) = 0.10 ⋅ 0.20 + 0.01 ⋅ 0.30 + 0.001 ⋅ 0.50 = 0.0235
Lecture 4: Conditional Probability
18
Example 2 (Cont.)
Tree diagram for the same example can be used too:
Lecture 4: Conditional Probability
19
Exercise 2-96 (Page 50)
Building failures are due to either natural actions N (87%) or M man-made causes (13%) Natural actions include: earthquakes E (56%), wind W(27%), and snow S (17%) Man-made causes include: construction errors C (73%) ordesign errors D (27%)1)What is the probability of failure due to
construction errors?Ans. P(F) = P(C|M) P(M) = 0.73 ⋅ 0.13 = 0.09492) What is the probability of failure due to wind or
snow?Ans. P(F) = [P(W|N) + P(S|N)] P(N) = [0.27 + 0.17] ⋅ 0.87 = 0.3828
Lecture 4: Conditional Probability
20
Monday before class
Problems HW set 2 2-51 2-62 2-68 2-69 2-75 2-88 2-95 ReadingBayes’ Theorem (Pages 55-59)