independence and law of total probability lecture 7huang251/slides7.pdfindependence and law of total...

Independence andLaw of TotalProbability

Quiz 1

Independence

Law of Total Probability

7.1

Lecture 7Independence and Law of TotalProbabilityText: A Course in Probability by Weiss 4.3 ∼ 4.4

STAT 225 Introduction to Probability ModelsFebruary 2, 2014

Whitney HuangPurdue University


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Independence


7.2

Agenda

1 Quiz 1

2 Independence

3 Law of Total Probability


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Independence


7.3

Quiz 1, Problem 1

Students at Brobbins University were asked to state whetherthey like chocolate ice cream (yes or no) and whether they likevanilla ice cream (yes or no). The partial results of the poll arebelow.

32% like neither18% like vanilla only15% like both chocolate and vanilla

1 Draw a Venn diagram describing the ice cream flavorpreferences at Brobbins University? (5 points)

2 What percent of the students stated that they did not likevanilla at all? (2 points)

3 What percent of those students liking chocolate, did notalso like vanilla ? (2 points)


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Independence


7.4

Quiz 1, Problem 1 solution

Solution.

1

2 By complement rule, we have 1− .18− .15 = .67 = 67%

3 Conditional probability!P(V c |C) = P(V c∩C)

P(C) = .35.50 = 0.70 = 70%


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Independence


7.5

Quiz 1, Problem 2

In a reality television show race there are 12 participants. Inone leg of the race the top 3 finishers are immune fromelimination and will move on to the next leg. The remainingparticipants will have to undergo further challenges to be ableto move on to the next leg.

1 How many ways can three contestants move to the nextround without having to complete further challenges? (3points)

2 If the first place finisher receives $10000, the secondplace finisher receives $5000 and third place gets $2500,how many ways can the prize money be awarded to thecontestants? (3 points)


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Independence


7.6

Quiz 1, Problem 2 solution

Solution.

1(12

3

)= 12!

9!×3! = 12×11×103×2×1 = 1320

6 = 2202 12P3 = 12!

9! = 12× 11× 10 = 1320


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Independence


7.7

Motivating Example

Example

You toss a fair coin and it comes up “Heads" three times. Whatis the chance that the next toss will also be a “Head"?

Solution.

The chance is simply 12 just like ANY toss of the coin⇒What it

did in the past will not affect the current toss!


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Independence


7.8

Independence

Independent events

Let A and B be events of a sample space withP(A) > 0, P(B) > 0. We say that event B is independent ofevent A if the occurrence of event A does not affect theprobability that event B occurs.

P(B|A) = P(B)

If we know A and B are independent then the generalmultiplication rule will reduce to a special form

The special multiplication rule

Two events A and B, are said to be independent events if

P(A ∩ B) = P(A)× P(B)


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Independence


7.9

Example 20

The independence of the events A and B implies that thefollowing are independent:

1 Ac and B2 A and Bc

3 Ac and Bc


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Independence


7.11

Independence for more than two events

Consider more than two events A1,A2, · · · ,Ak , we have twotypes of independence:

Pairwise independent

P(Ai ∩ Aj ) = P(Ai )× P(Aj ) i , j ∈ 1,2, · · · , k i 6= j

Mutually independent

P(Ak1 ∩ Ak2 ∩ · · · ∩ Akn ) = P(Ak1 )× P(Ak2 )× · · · × P(Akn )

ki , kj ∈ 1,2, · · · , k ki 6= kj 1 ≤ n ≤ k


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Independence


7.12

Example 21

Chris and his roommates each have a car. Julia’s MercedesSLK works with probability .98, Alex’s Mercielago Diablo workswith probability .91, and Chris’ 1987 GMC Jimmy works withprobability .24. Assume all cars work independently of onanother. What is the probability that at least 1 car works?


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Independence


7.13

Example 21: Julia’s SLK


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Independence


7.14

Example 21: Alex’s Diablo


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7.15

Example 21: Chris’ 1987 Jimmy


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7.16

Example 21

Solution.

Let S denotes Julia’s SLK works, M denotes Alex’s MercielagoDiablo works, and J denotes Chris’ 1987 GMC Jimmy worksP(S ∪M ∪ J) = 1− P(Sc ∩Mc ∩ Jc) =1− P(Sc)× P(Mc)× P(Jc) = 1− .02× .09× .76 = .9986


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Independence


7.17

Law of total probability

Recall the law of partitions

Let A1,A2, · · · ,Ak form a partition of Ω. Then, for all events B,

P(B) =k∑

i=1

P(Ai ∩ B)

By applying the general multiplication rule, we have the law oftotal probability

P(B) =k∑

i=1

P(B|Ai )× P(Ai )


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Independence


7.18

Example 22

Acme Consumer Goods sells three brands of computers: Mac,Dell, and HP. 30% of the machines they sell are Mac, 50% areDell, and 20% are HP. Based on past experience Acmeexecutives know that the purchasers of Mac machines willneed service repairs with probability .2, Dell machines withprobability .15, and HP machines with probability .25. Find theprobability a customer will need service repairs on thecomputer they purchased from Acme.

Solution.

Apply the law of total probability. Let R denotes a customer willneed service repairs on the computer, D denotes Dell, Mdenotes Mac, and H denotes HP.

P(R) = P(R|D)× P(D) + P(R|M)× P(M) + P(R|H)× P(H)

= .5× .15 + .3× .2 + .2× .25 = .185


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Independence


7.19

Example 23

Let us assume that a specific disease is only present in 5 out ofevery 1,000 people. Suppose that the test for the disease isaccurate 99% of the time a person has the disease and 95% ofthe time that a person lacks the disease. Find the probabilitythat a random person will test positive for this disease.

Solution.

Let D denotes disease, and + denote test positive.Given: P(D) = .005⇒ P(Dc) = .995, P(+|D) =.99, P(−|Dc) = .95⇒ P(+|Dc) = .05P(+) = P(+|D)× P(D) + P(+|Dc)× P(Dc) =.99× .005 + .05× .995 = .0547


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Independence


7.20

Summary

In this lecture, we learnedThe concept of IndependenceSpecial multiplication rule and law of total probability

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