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UNIT 2 THEOREMS PROBABILITYStructure

2.1 IntroductionObjectives

2.2 Some Elementary Theorems2 3 General Addition Rule2.4 Conditional Probability and Independence

2 4 1 Conditional Probability2 4 2 Independent Events and Multiplication Rule2 4 3 Theorem of Total Probability and Bayes Theorem

2 5 Summary

2 1 INTRODUCTIONYou have already learnt about probability axioms and ways to evaluate probabilityof events in some simple cases.In this unit, we discuss ways to evaluate the probability of combination of events.For this, we derive the addition rule which deals with the probability of union oftwo events and the multiplication rule which deals with thc probability ofintersection of two events. Two important concepts namely ConditionalProbability and independence of events, are introduced and Bayes theorem, whichdeals with conditional probability is presented.ObjectivesAfter reading this unit, you should be able to

evaluate the probability of certain combination of events involving union,intersection and complementation,evaluate conditional probability,check independence of two or more events, andapply Bayes theorem to find the probability that the effectY'Awascaused by the event B.

2 2 SOME ELEMENT RY THEOREMSRecall the axiomatic definition of probability which you have read in Section 1.4.Using these axioms of probability, it is possible to derive many results which arevery useful in applications. We present some of these results in this section.Theorem 1

If 4 is the empty set thenP 4 ) 0

ProofFor any eventA, A A U 9 Also and re mutually exclusive as

l 4 4 Hence by Axiom 3 of probability axioms,

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robabilityConceptswhich implies that P ( ) 0

Theorem 0 for all i

In the below given figure (Figure 2.5) the eventsB1 Bz, Bs, B4,85 and Bsdefine a partition of the sample spaceS.

IF i i 5It should be clear from the definition that events Bi , B2, ...Bk define a partition ofthe sample spaceS if the random experimentE results in one and only one of the

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events Bi For example if the random experiment E represents the tossing of adie, then = { 1,2,3,4,5,6 } and for this sample spaceBi = [ 1,3 ,6 and B2 - { 2,4,5 } definea partition of S.On the other handB1 - {1,4} ,B2 = (5 ,6 }, B3 2 ,3 ,5 }donotde f iheapar t i t iono f~because B2 l l B3 p. We now present the Theorem of total probabilityw(Theorem 5):

heoremetB1, B2, Bkbe a partition of the sample space S and letA be some event

with respect toS.Then

Proof:

Note thatA A n .Buts = U Bi, and hencei - 1A - A n ( B l U B 2 U ...UBk),i.e.

I = U ( A n B i )i - 1

(The following Venn diagram (Figure 2.6) illustrates this fact for n = 6.)Now for j (A n Bi ) and (A l l Bj ) are mutually exclusive asBi i Bj = .Therefore from the probability axioms, equation (2.19) gives

Theorem ofProbability

But by the definition of conditional probabilityP A fl Bi ) = Bi ) A/Bi )(recollect the multiplication rule) and hence equation (2.20) implies

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ProbabfitycoIke~~k 3 P ( A ) 2 P ( B i ) P ( A/B i ),i -1

which proves the theorem of total probability.xample 9

An assembly plant receives its voltage regulators from two different suppliers75 from supplier B1 and 25 from supplier B2. Suppose that 95 of thevoltage regulators supplied by B1 and 80 of those supplied by B2 performaccording to specifications. What is the probability that a randomly selectedvoltage regulator from the assembly plant will perform according tospecification

Solution :LetA be the event that a randomly selected voltage regulator from theassembly plant performs according to specifications. Note that B1 ,B2 definepartition of the sample space (as a regulator has certainly to be supplied byeither Bi or B2, but not from both), the Theorem of total probability becomesapplicable. The relevant probabilities can be computed as follows:

Hence by the theorem of total probability,P ( A ) = P ( B 1 ) P ( A / B i ) + P ( B 2 ) P (A /B 2)

xample 10A building contractor requires a roll of roofing felt. There are 3 suppliers,B1 B2, B3 in the area and the probabilities (based on his previous experienceand the location of the supplier) that the contractor will instruct his van man visit a particular supplier are 0.6,0.2 and 0.2 respectively. Each supplier stocroofing felt produced by two manufacturers,X and Y Both types of roofingfelt sell for the same price and both satisfy the current building regulations.The stock situation at each of the supplier is:Supplier

BB2B

No. of X rolls No. of Y rolls10 3030 2030 10

Assuming that the van man will be told by his employer which supplier tovisit, which roll type ( or Y ) is the van man most likely to return with.

SolutionLet P ( X ) be the probability that the van man will return with roll typeX ,P(Y ) eing defined similarly. From the given data,

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Hencep ( X ) = P ( B 1 ) P ( X / B l ) + P ( B 2 ) P ( X / B 2 ) + P ( B 3) P ( X/ B3 )

In a similar manner, P Y ) 0.58, Hence it is more likely that the van manwill return with a roll of type YNext we proceed to establish the Bayes Theorem (also called Bayes Rule orBayes formula) which helps to evaluate the probability that the effectmAwascaused by the event Bi i.e. P Bi/A ) where events Bi, B2, Bk representsthe partition of the sample spaceS.

Theorem6Let Bi, B2, ...Bk be a partition of the sample spaceS and letA be some eventwith respect to S Then for k

P m fThe proof follows immediately from the definition of conditional probabilityand the theorem of total probability (Theorem 5) because

xampleWith reference to the data given in Example 9, suppose that a voltage regulatorfrom the assembly plant is chosen at random and found to perform accordingto specifications. Find the probability that it has been supplied by the supplierB2.

SolutionLetA, Bi, B2 be the events a s defined in the solution of Example 9. So wehave

P(B1) 0 .75, P(B2) = 0.25

and we have to determineP BdA ) This is precisely the situation whenBayes rule is applicable and hence using Theorem 6,

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robability conceps mmple 12In a bolt factory, machines Bi , B and B3 manufacture 25 35 and 4 0 per centof the total output respectively. Of their outpu ts 5,4 and 2 per cent, respectivelyare defective bolts. bolt is chosen at random and found to be defective. Whatis the probability that the bolt come from m achine B3

SolutionLet us introduce the following events:

{the item is defective)B1 = {the bolt came from machine BiB2 = {the bolt came from machine B2B3 {the bolt cam e from m achine Bg

From the given data,P Bl ) = 0.25, B2 ) = 0.35, B3 ) = 0.40,( A / B 1 ) = 0.05, P ( A / B a ) 0.04, and ( A / B 3 ) = 0.02. The

required probability is

RemarkThe theorem of total probability Theorem 5 can be visualized by con structinga tree diagram like that of Figure 2.7, where the probability of the finaloutcom e is given by the sum of the products of the probabilities correspondingto each individual branch.

m 7:TrccDiqpmTh e tree diagram, so named because of its appearance, can often be useful indecision making. Each branch illustrate the path that will be taken whenever aparticular decision is made. Figure 2.7 illustrates the tree diagram w ith

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reference to the theorem of total probability only. In general there may bemore than one branch from i 1 2 k resulting in different eventswhich may h ave further branches. Th e tree diagram for Example 1 s shownin Figure 2.8and sh ows all the possible combinations of events that could beinvolved in finishing up with an or Y type of roofing felt.

ThtomnsofProbability

igure 2 8 T m iagram for Exnmple 1

Consider a lot consisting of 2 defective and 8 non-defective items. If twoitems are chosen at random without replacement find the probability that thesecond chosen item is defective.

There are four boxes numbered 1 2 3 nd 4 Box 1 contains 2 red and 8whiteballs Box contains 5 red and 5white balls Box 3 ontains 6 red and 4 whiteballs and Box 4 contains 9 red and 1white balb. A box is chosen at randomand then a ball is chosen a t random from the box. find the probability that thechosen ball is red. Use tree diagram to evaluate the required probability.

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Consider the d t of Example 10 and suppose that the van man returned withan X type of roofing felt. Determine the probability that he obtained it fromsupplierB2

2 5 SUMM RYWe briefly sum up what has been done in this unit.1) Using the axioms of probability, certain results have been derived forevaluating the probability of combination of events. Specifically we have

proved the followingi P ( ) 0i i ) ~ ( x ) 1 - P ( A )

Suppose that a vacuum tube may come from any one of three manufacturerswith probabilitiesPI 0.25, P = 0.50 and P3 = 0.25 The probabilitiesthat the tube will function during a specified period of time equal 0.1,0.2 and0.4 respectively for the three manufacturers. Compute the probabiliy that arandomly chosen tube will work for the specified period of time.

ElWith reference to the data of E 13 above, suppose that the randomly chosentube has been found to work for the specified period of time. Find theprobability that it has been manufactured by the first manufacturer.

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iv) A C B -P B - A ) = P B ) - P A )V ) A C B -P A ) s P B ) .A general addition rule has been given for finding the probability ofoccurrence of at least one of the given n eventsAl,A~,..An. This rule statesthat

P A 1 UA2U An) Z P A ~ ) -C P A i n A j ) + ......i -1 i c j - 2

The concept of conditional probability has been introduced and P A/B hasbeen defined as

Using the definition of conditional probability, the multiplication rule forfinding the probability of joint occurrence of two events, has been stated asP A n B ) = P A ) P B / A) = P B ) P A /B ).The concept of mutually independent events has been introduced. Two eventsA and B are said to be independent i P A n B ) = P A ) P B ) ThreeeventsA ,B, C are s id to be mutually independent if every two of these areinde pe nde n ta ndP AnBnC) = P A ) P B ) P C ) .The partition of the two sample space is defined and for a given partitionBi , B2, ...Bk of andA any event, theorem of total probability and Bayestheorem have been proved. These are stated as

andP B r ) P A / B r )P B r / A ) respectively

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ProbabilityConccplP

E2LetA be the event that he will ask to have tyres checked and B be the eventthat he will ask to have his oil checked. From the given data,P A ) 0 .1 2, P B ) = 0.29 and P A n B ) 0.07.

i Required Probability P AU B )

ii Required Probability = P AU B )

EP A U B U C ) = P A ) + P B ) + P C ) - P A n B ) - P B n C )- P A n C ) + P A n B n C )

Here everything is known exceptP AnB nC . ButA n B n C )c A n B ) a n d h e n c e P A n B n C )s P A n B ) = 0.

This implies thatP AnB C = 0Therefore

E We know thatP A U B ) = P A ) + P B ) - P A n B )s P A ) + P B ) ,s P A B ) 0. Let us denoteB U C by D and therefore

P A U B U C ) P A U D ) P A ) + P D )P A ) + P B U C )

s P A ) + P B ) + P C )

LetA be the event that the die shows an even number andB be the event thatthe cord is from a red suit. Then P A ) = and P B ) Also, underthe given situation,A and B are independent and so

ii The required probability is given by P AU B )which equalsP A ) + P B ) - P A n B ) .e.

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E6Let Ei be the event that the th relay is closed ( = 1,2,3,4) and E be the eventthat there is current betweenL andR Then

E = ( E i n E 2 ) U ( E g n E q ): P ( E ) = P ( ( E i n E 2 ) U ( E g n E 4 ) ) , a s (E i n E2 ) a nd ( Eg n E q)are mutually exclusive

= ~ ~ l f l ~ ~ ) + P E g n E 4 ) - P ~ l n E 2 n E s n E 4 )= P ( E 1 ) P ( E 2 ) + P ( E 3 ) P ( E 4 ) - P ( E 1 ) P ( E 2 ) P ( E 3 ) P ( E 4 )( as Ej s are mutually independent )= p 2 + p 2 - p 4

2= 2 P P .E

As the events are independent, the required probability is given byP ( late on Monday ) P ( on time on Tuesday ) P ( on time onWednesday) P ( on time on Thursday ) P ( late on Friday )

= ( 0.60 ) ( 0.40 ) ( 0.40 ) ( 0.40 ) ( 0.60 )

ES ILetA be the event that the lorry delivers sand and B be the event that it ispetrol operated. Then the required probability is

= p ( A n B )= P ( A ) P ( ) ( as A and B are independent refer to theproblem given atE 9 )

E9~ ( A n i i ) P A 1 - P ( A U B )

= 1 - ( P ( A ) + P ( B ) - P ( A n B ) )= 1 - P ( A ) - P ( B ) + P ( A ) P ( B )

(as A and B are independent)= Q I - P ( A ) ) ( I - P ( B ) ) = P ( A ) P B ) ,

which implies that nd are independent:El

LetA be the event that thc lifst drawn item is defective and B be the event thatthe second drawn item is defective. Then

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A ~ And Aconstitutes a partition of the sampleSpaceS the theorem of totalprobability becomes applicable and we get

l lLet the events of selecting boxes 1,2,3 and 4 be denoted byB i, B2, Bh and B h respectively.Then

LetA be the event that the chosen ball is red. Note that B l, B2 3 and B4define a partition of the sample space and henceP A ) - P B 1 ) P A / B i ) + P B 2 ) P A / B 2 )

+ P B 3 ) P A / B g ) + P B q )P A / B 4 )

Hence1 2 1 5 1 6 1 9 22P A ) = - - )0 4 1 0 ) - )0 4 10 = 0 = 0.55

The tree diagram for this examplecan be drawn as follows:

Figure 9 he m ia

The problem is to evaluate the probability of B2 givenX i.e. P B i X Bydefinition 1P X ) has already been evaluated in Example 10and s equal to 0.42. Fromthe given data P B ) - 0.2 and X/B2 ) - 0.6.

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Hence TheoremsofProbability

Elr Let Bi i=1,2,3 ) be the event that the vacuum tube h scome from the th

manufacturerand be the event that the tube works for the specified period ofI time. From the given data:

P B 1 ) = 0 .25 ,P B2) = 0 .50 ,P B3) = 0.25P A/B1 ) = 0.1, P A/ B2 ) = 0.2, P A/ B3 0.4.The required probability isP A ) - P B 1 ) P A / B i ) + P B 2 )P A / B 2 ) + P B g )P A / B 3 )

- 0.25 0.1 ) 0.50 0.2 ) 0.25 0.4 )= 0.225.

El4The required probability is P Bl/A )which by Bayes Theorem is given by