lecture 4

Lecture 4: Conditional Probability

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Lecture 4

Conditional Probability, Total Probability Rule

Instructor: Kaveh Zamani

Course material mainly developed by previous instructors:Profs. Mokhtarian and Kendall, Ms. Reardon


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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:


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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


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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)


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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]


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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


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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


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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


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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


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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

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Exercise 2-85 (Cont.)


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


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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


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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


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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’)


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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)


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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


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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


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Example 2 (Cont.)

Tree diagram for the same example can be used too:


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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


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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)

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