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Probability Theory ShareWordPress Shortcode Are you sure you want to YesNo Your message goes here 3 months ago Reply Are you sure you want to Yes No Your message goes here 3 months ago Reply Are you sure you want to Yes No Your message goes here 2 There was no perceived need for probability theory, so the explanation goes. value of buying a lottery ticket = sum of product of each prize times probability of .


Probability is anattitude of mindtowardsuncertain 8.  Credit of development of probability theory goes to gamblers,and people who bets on horses, who has started discussion withfamous mathematician of that age, after disappointed fromgoddess fortune famous scientists as Gallelio, Pascal, Fermettused their power to solve these problems.

 This undeveloped idea is developed by the scientists as-Laplace, Gauss, Yuler, James Bernoulli. THE THEORY OFPROBABILITY is of interest not only tocard and dice players who were its God fathers but also toall men of action, heads of industries or heads of armieswhose success depends on decision. DEFINITION OFPROBABILITYMathematicalDefinitionStatistical DefinitionAn event happen a times anddoes not happen b timesand all ways are equallylikely then probability ofhappening of an event willbe (a/a+b) andprobability of nothappening of an event willbe (b/a+b).

Probability is calculated onthe basis of available dataor frequencies or pre-experiences.

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Importance ofProbability TheoryBASIS OFSTATISTICALLAWSIMPORTANCEIN GAMES OFCHANCEUSE INSAMPLINGSPECIFICIMPORTANCEIN INSURANCEBUSINESSUSE INECONOMICSAND BUSINESSDECISIONSBASIS OFTESTS OFHYPOTHESISAND TEST OFSIGNIFICANCE 12 19 Apr 2017 - Credit of development of probability theory goes to gamblers, and people who Example:- The probability of my being asked on a date for this .


Theoretical ProbabilityWe assume that all n possible outcomes of aparticular experiment are equally likely, and weassign a probability of 1/n to each possibleoutcomes. Example:- The theoretical probability of rolling a 3on a regular 6 sided die is 1/6.

Relative Probability We conduct an experiment many, many times.

Then we say:- The probability of an event A= How many times A occursHow many trials Relative probability is based on observation or actual measurements. Therelative probability of rolling a 3 is 12/100.

SubjectiveProbability These are values (between 0 and 1 or 0 and 100%) assignedby individuals based on how likely they think events are tooccur.

 Example:- The probability of my being asked on a date forthis weekend is 10% 16. ADDITON PROBABILITYTHEOREM(An Important Theorem of Probability) “The literal meaning of addition theorem is to addthe individual probabilities of two or more events.

The addition theorem in the probabilityconcept is the process of determination of theprobability that either ‘A’ or event ‘B’ occur orboth occur.

The notation between two events‘A’ and ‘B’ the addition is denoted as ‘U’ andpronounced as union.  Then, probability of occurrence of at least oneof these two events is given by the sum of theindividual probabilities.

Proof-Let the event A can occur in p waysand B in q ways then the number of waysin which either event can happen is p+q.

Ifthe total number of possibilities is n, thenby definition of probability, 19. = (p+q)/n = p/n + q/n = P(A)+P(B)P(A or B) = P(A∪B) = P(A)+ P(B)•The probability of either A or Boccursfavourable no.

• Addition theorem probability canbe defined and proved asfollows:-•Where,•P(A)= Probability of occurrence of event ‘A’•P(B)= Probability of occurrence of event ‘B’•P(A∪B)= Probability of occurrence of event ‘A’ or event ‘B’•P(A∩B)= Probability of occurrence of event ‘A’ or event ‘B’ 21 21 Nov 2005 - Statistical desicion theory; Bayesian theory and research in health as describing uncertain knowledge (i.e., Bayesian probability) is central. If you buy any insurance policy, you should expect to lose money in the long run .

Case I –When event are mutually exclusiveLet ‘A’ and ‘B’ are subsets of a finite non empty set ‘S’ thenaccording to the addition rule-P(A∪B)= P(A)+P(B)-P(A).

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Case I I –When event are not mutually exclusiveIf the event ‘A’ and ‘B’ correspond to the two events ‘A’ and ‘B’ ofa random experiment and if the set ‘S’ corresponds to the samplespace ‘S’ of the experiment then the equation (1) become-P(AUB)= P(A)+P(B)-P(A).

•These are event A∪ B refers to themeaning that either event ‘A’ or event ‘B’occurs.

•So we can find the logic behind theaddition probability theorem both mayoccur simultaneously. For example:-The probability of getting spot 2 in a throw of a single dice is 1/6,the probability of getting spot 4 is also 1/6, If it is asked that what isthe probability of getting 2 or 4, it will be 1/6+1/6=2/6 =1/3 on thebasis of addition theorem. Hint:while doing a question it should be kept in mind that addition theoremwill be applicable in all those problems in which either the word ‘or’ has been usedexplicitly or the word ‘or’ is used in analysis of the question 25.

LIMITATION:-•Addition theorem is applicable only when following twoconditions are fulfilled- (a)Events are mutually exclusiveand (b)They all are related to same set. •It should be noted that if two or more events are notmutually exclusive completely, the addition theorem hasto be modified.

Suppose, there are two events A and Band in some occurrence they are not mutually exclusive,then modified formula will be as follows:-• P(AorB)= P(A)+P(B)-P(A&B) 26. Multiplication Law of ProbabilityMultiplication law in probability applies tocombination of events.

When the eventshave to occur together then we make use ofthe multiplication law of probability. Now two cases arise: whetherthe events areindependent or dependent. The events are said tobe independent onlywhen the occurrence ofone event does notchange the probabilityof occurrence of anyother event with it.

The given eventsare said to bedependent when theoccurrence of oneevent changes theprobability ofoccurrence of any 28. RulesThe multiplication rule states that:“The probability of occurrence of given two events or in other words the probability of intersection of two given eventsis equal to the product obtained by finding the product of the probability of occurrence of both events.

”This implies that if A and B are two events given then:P (A and B) = P (A ∩ B) = P (A) * P (B)This rule is applicable in all the cases, that is, when events are independent ordependent. In case when we have dependent events we have to be very careful in determiningthe probability of the second event after the occurrence of first event.

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Itis clear that here the given events are completelyindependent and thus we can multiply theprobabilities of the given two events for findingout the probability of combined events. Now, suppose the candies are taken from thebox without putting the first one back.

It is clearthat the events are dependent and thus we needto find the conditional probability for findingthe probability of occurrence of combinedevent. In such case the multiplication rule is modified as:P (A and B) = P (A ∩ B) = P(A) * P (B|A)Here, P (B|A) is the probabilityof occurrence of the secondevent B when the first event Ahas already occurred.

ProblemsLet us see some examples on the multiplication law of probability.

Example 1:A bag contains 3 pink candies and 7 green candies. Two candies are taken outfrom the bag with replacement.

Solution:Let A = event that first candy is pink and B = event that second candy is pink.

→ P (A) = 3/10 …(i)Since the candies are taken out with replacement, this implies that the given events A and B areindependent. → P (B|A) = P (B) = 3/10 …(ii)Hence by the multiplication law we get,P (A ∩ B) = P (A) * P (B|A)→ P (A ∩ B) = 3/10 * 3/10 using (i) and (ii) = 9/100 = 0.

Example 2:A bag has 4 white cards and 5 blue cards.

We draw two cardsfrom the bag one by one without replacement. Find theprobability of getting both cards white.

Solution:Let A = event that first card is white and B = event that second card is white. Now P (B) = P (B|A) because the events given are dependent on each other. COMBINED USE OF ADDITION ANDMULTIPLICATIONTHEOREMQuestion : A speaks truth in 80% cases, B in 90% cases. In What percentage ofcases are they likely to contradict each other In starting the same fact. Solution :Let (A) and (B) denote the probability that A and B speak the truth.

Then,P(A) = 80/100 = 4/5 P(A) = 1 - P(A) = 1 - 4/5 = 1/5P(B) = 90/100 = 9/10 P(B) = 1 - P(B) = 1 - 9/10 = 1/10They will contradict each other only when one of them speaks the truth and theother speaks a lie. Since, the events are independent, so by using the multiplicationtheorem, we have:1. Probability in the 2nd case = 9/10 * 1/5 = 9/50Since, these cases are mutually exclusive, so by using the additiontheorem. We have the required probability= 4/50 + 9/50 = 13/50 = 26% 34.

In a random experiment, if A and B are two events, then the probability ofoccurrence of event A when event B has already occurred and P(B) ≠ 0, is called theconditional probability and it is denoted by P(AB)P(A/B) = Number of outcomes favourable to A which are also favourable to BNumber of outcomes favourable to BP(A/B) = P(A ∩ B) , P(B) ≠ 0P(B)Similarly, P(B/A) = P(A ∩ B) , P(A) ≠ 0P(A)MULTIPLICATION THEOREMIN CASE OFCONDITIONAL PROBABILITY 35. These are the probabilities calculated onthe basis that something has alreadyhappened.

 The probability that I will pay my electricitybill given that have just been paid.

 The probability that my students will turnupto class given that it is a sunny day. CONDITIONALPROBABILITYUnconditionalProbabilityConditionalProbability 38. Theprobability that it lands with 5 showing up is1/6 this is UNCONDITIONAL PROBABILITY, 39. The probability that it lands with 5 showing up,given that it lands with an odd number showing up, is1/3 this is a CONDITIONAL PROBABILITY.

P(A/B)P(A/B)=P(A∩B) , P(B)≠0P(B)- P(A/B)=Probability of occurrence of event A giventhat event B has already occurred P(B)≠0 41.

- P(B/A)=P(A∩B), P(A)≠0P(A)- P(B/A)=Probability of occurrence event Bgiven that the event A has already occurred. •The conditional probability of an event Agiven that B has occurred lies between 0&1. •P(E∩F)=P(E/F)×P(F)•If E,F,&G are independent given that an event H hasoccurred, then•P(E∩F∩G/H)=P(E/H)×P(F/H)×P(G/H) 44.

Example of ConditionalProbabilityTable shows the result of a class survey:-Find P (wash the dishes/male)Did you wash the dishes last night?YES NOThe condition male limits the sample space to 15 possible outcomes. Therefore, P(washes the dishes/male)= 7/15FEMALE 7 6 13 FEMALESMALE 7 8 15 MALES 45. Bayes’ theorem (also known asBayes’rule or Bayes’law)Is a result in probability theory that relates conditional probabilities.

If A and B denote two events,P(A|B) denotes the conditional probability of A occurring, given thatB occurs. The two conditional probabilities P(A|B) and P(B|A) are in generaldifferent.

Bayes theorem gives a relation between P(A|B) and P(B|A). An important application of Bayes’ theorem is that it gives a rulehow to update or revise the strengths of evidence-based beliefs inlight of new evidencea posteriori.

As a formal theoremBayes’ theorem is valid in all interpretations ofprobability.

However, it plays a central role in the debatearound the foundations of statistics: frequentist andBayesian interpretations disagree about the kinds ofthings to which probabilities should be assigned inapplications. Whereas frequentists assign probabilities torandom events according to their frequencies ofoccurrence or to subsets of populations asproportions of the whole,Bayesians assign probabilities to propositionsthat are uncertain. A consequence is that Bayesians have morefrequent occasion to use Bayes’ theorem.

The articles on Bayesian probability andfrequentist probability discuss these debates atgreaterlength. Suppose we have estimatedprior probabilities for eventswe are concerned with, andthen obtain new information. We would like to a soundmethod to computed revisedor posterior probabilities.

 The probability of two events A and B happening, P(AB), is theprobability of A, P(A), times the P(A B) = P(A)P(B|A) (1) probability of B given that A has occurred, P(B|A). (1) On the other hand, the probability of A and B is also equal to theprobability of B times the probability of A given B.

 P(A B) = P(B)P(A|B) (2) Equating the two yields: P(B)P(A|B) = P(A)P(B|A) (3) and thus P(A|B) = P(A)P(B|A) P(B) (4) This equation, known as Bayes Theorem is the basis of statisticalinference. STATEMENT OF BAYES’THEOREM Bayes’ theorem relates the conditional and marginalprobabilities of stochastic events A and B: P(A|B) = P(B|A) P(A) P(B) Each term in Bayes’ theorem has a conventionalname: 51.  • P(A) is the prior probability or marginal probabilityof A.

It is ”prior” in the sense that it does not takeinto account any information about B Upper Key Stage 1 and Lower Key Stage 2 Probability PowerPoint and BUY NOW to tally red, blue, green and yellow cars as they fly past in the presentation. Essential Question: What examples of Graph Theory exist within the NYC .

 • P(A|B) is the conditional probability of A, given B.

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 • P(B|A) is the conditional probability of B given A.  • P(B) is the prior or marginal probability of B, andacts as a normalizing constant.

, An be a set of mutually exclusiveevents that together form the sample space S.

Let B be any eventfrom the same sample space, such that P(B) >0. Then,P( Ak | B ) = P( Ak ∩ B ) P( A1 ∩ B ) + P( A2 ∩ B ) + .

+ P( An ∩ B )Note: Invoking the fact that P( Ak ∩ B ) = P( Ak )P( B | Ak ), Baye'stheorem can also be expressed asP( Ak | B ) = P( Ak ) P( B | Ak ) P( A1 ) P( B | A1 ) + P( A2 ) P( B | A2 ) + . When to Apply Bayes' Theorem Part of the challenge in applying Bayes' theorem involves recognizingthe types of problems that warrant its use. You should consider Bayes'theorem when the following conditions exist.

: The sample space is partitioned into a set of mutually exclusiveevents A1, A2, .  Within the sample space, there exists an event B, for which P(B)>0.  The analytical goal is to compute a conditional probability of theform: P( Ak | B ).

 You know at least one of the two sets of probabilities describedbelow Probability theory began in seventeenth century France when the two great French Very little computing background is assumed or necessary in order to obtain full the Chance project, which is devoted to providing materials for beginning .

 P( Ak ∩ B ) for each Ak P( Ak ) and P( B | Ak ) for each Ak 54.

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One of the box is selectedat random and a ball is drawn from it. Ifthe ball drawn is red, find the probabilitythat if it is drawn from the first bag.

Solution:-Let A, B, C, & D, be the eventsdefined as follows:-A= 1st box is choosenB= 2nd box is choosenC= 3rd box is choosenD= 4th box is choosen 56.

Since there are three boxes and one of thethree boxes is choosen at random,therefore:-P(A)=P(B)=P(C)= 1/3If A has already occurred, then first box hasbeen chosen which contains 6 red and 4black balls. The probability of drawing a redball from it is 6/10.

So,P(D/A)=6/10Similarly, P(D/B)=5/10 and P(D/C)=4/10We are required to find P(A/D) i.

, given that the ball drawn is red,what is the probability that it is drawn from the first box.

P(D/C)1/3 * 6/10(1/3 * 6/10)+(1/3 * 5/10)+(1/3 * 4/10)=2/5 58. Probability Revision using Bayes’TheoremPriorProbabilitiesNewInformationApplication ofBayes’TheoremPosteriorProbabilities 59.

BIBLIOGRAPHY Reference taken from the book BUSINESS STATISTICSby:- Prof.