Chapter 2. Conditional Probability

Transcription

1 Chapter. Conditional Probability The probabilities assigned to various events depend on what is known about the experimental situation when the assignment is made. For a partiular event A, we have used A) to represent the probability assigned to A; we now think of A) as the original or unonditional probability of the event A.. The definition of onditional probability In this setion, we examine how to information an event B has ourred affets the probability assigned to A. We will use the notation A B) to represent the onditional probability of A given that the event B has ourred. Conditioning is one of the fundamental tools of probability: probably the most fundamental tool. It is espeially helpful for alulating the probabilities of intersetions, suh as A B), whih themselves are ritial for the useful Partition Theorem. Additionally, the whole field of stohasti proesses is based on the idea of onditional probability. What happens next in a proess depends, or is onditional, on what has happened beforehand.

2 ependent events. Suppose A and B are two events on the same sample spae. There will often be dependene between A and B. This means that if we know that B has ourred, it hanges our knowledge of the hane that A will our. Example: Toss a die one. However, if we know that B has ourred, then there is an inreased hane that A has ourred: Conditioning as reduing the sample spae Example. The ar survey in Examples of basi probability alulations also asked respondents whih they valued more highly in a ar: ease of parking, or style/prestige. Here are the responses: Suppose we pik a respondent at random from all those in the table.

3 Let event A = respondent thinks that prestige is more important. Suppose we redue our sample spae from

4 This is our definition of onditional probability: efinition: Let A and B be two events with B)>0. The onditional probability that event A ours, given that event B has ourred, is written A B), and is given by Note: Follow the reasoning above arefully. It is important to understand why the onditional probability is the probability of the intersetion within the new sample spae Conditioning on event B means hanging the sample spae to B. Think of A B) as the hane of getting an A, from the set of B's only. The Multipliation Rule For any events A and B,

5 New statement of the Partition Theorem (The Law of Total Probability) The Multipliation Rule gives us a new statement of the Partition Theorem (Total Probability Theorem) Both formulations of the Partition Theorem are very widely used, but espeially the onditional formulation Examples of onditional probability and partitions Example. A news magazine publishes three olumns entitled Art (A), Books (B), and Cinema (C). Reading habits of a randomly seleted reader with respet to these olumns are Read regularly A B C A B A C B C A B C Probability We thus have A B) 0.08 A B) 0.8 B) 0. A ( B C)) A B C) 0.55 B C) 0.7 A ( A B C)) A reads at least one) A A B C) A B C) A) A B C) 0.9 ( A B) C)) A B C) 0.59 C) 0.7

6 Example. Four individuals have responded to a request by a blood bank for blood donations. None of them has donated before, so their blood types are unknown. Suppose only type A+ is desired and only one of the four atually has this type. If the potential donors are seleted in random order for typing, what is the probability that at least three individuals must by typed to obtain the desired type? Solution. Making the identifiation B={first type not A+} and A={seond type not A+}, B)=/. Given that the first type is not A+, two of the three individuals left are not A+, so A B)=/. The multipliation rule now gives at least three individuals are typed)=a B)=A B)B) =/*/=0.5 The multipliation rule is most useful when the experiment onsists of several stages in suession. The onditioning event B then desribes the outome of the first stage and A the outome of the seond, so that A B)-onditioning on what ours first-will often be known. stages. The rule is easily extended to experiments involving more than two More than two events To find A A A ), we an apply the multipliation rule suessively: Where A ours first, followed by A, and finally A.

7 Example 5. For the blood typing experiment of the above example, third type is A) third is first isn' t seond isn' t) seond isn' t first isn' t) first isn' t) 0.5 Example 6: A box ontains w white balls and r red balls. raw balls without replaement. What is the probability of getting the sequene white, red, white? Solution: Example 7. Tom gets the bus to ampus every day. The bus is on time with probability 0.6, and late with probability 0.. The sample spae an be written as We an formulate events as follows: T = on time ; L = late. From the information given, the events have probabilities: T) = 0.6 ; L) = 0. Question(a) o the events T and L form a partition of the sample spae? Explain why or why not. Solution. Yes. They over all possible journeys (probabilities sum to ), and there is no overlap in the events by definition.

8 The buses are sometimes rowded and sometimes noisy, both of whih are problems for Tom as he likes to use the bus journeys to do his Stats assignments. When the bus is on time, it is rowded with probability 0.5. When it is late, it is rowded with probability 0.7. The bus is noisy with probability 0.8 when it is rowded, and with probability 0. when it is not rowded. Question(b) Formulate events C and N orresponding to the bus being rowded and noisy. o the events C and N form a partition of the sample spae? Explain why or why not. Solution. Let C = rowded, N = noisy. C and N do NOT form a partition of. It is possible for the bus to be noisy when it is rowded, so there must be some overlap between C and N. Question() Write down probability statements orresponding to the information given above. Your answer should involve two statements linking C with T and L, and two statements linking N with C. Solution. Questin(d) Find the probability that the bus is rowded.

9 Question(e) Find the probability that the bus is noisy. Example 8. A hain of video stores sells three different brands of VCRs. Of its VCR sales, 50% are brand (the least expensive), 0% are brand, and 0% are brand. Eah manufaturer offers a -year warranty on parts and labor. It is known that 5% of brand s VCRs require warranty repair work, whereas the orresponding perentages for brands and are 0% and 0%, respetively. Question(a) What is the probability that a randomly seleted purhaser has bought a brand VCR that will need repair while under warranty? Question(b) What is the probability that a randomly seleted purhaser has a VCR that will need repair while under warranty? Question() If a ustomer returns to the store with a VCR that needs warranty repair work, what is the probability that it is a brand VCR? A brand VCR? A brand VCR? Solution. Let (a) A i ={brand i is purhased}, for i=,,and. B={needs repair}, B ={doesn t need repair}. Then A )=0.5, A )=0., A )=0. B A )=0.5, B A )=0., B A )=0.. A B )=B A ) A )=0.5*0.5=0.5 (b) B)=(brand and repair) or (brand and repair) or (brand and repair)) = A B )+ A B )+ A B )= =0.05 A 0.5 () A B )= B) A 0. 6, A B )= B) B) 0.05 B) 0.05 A B )=- A B )- A B )=0.

10 . Statistial Independene Two events A and B are statistially independent if the ourrene of one does not affet the ourrene of the other. We use this as our definition of statistial independene. For more than two events, we say: Statistial independene for alulating the probability of an intersetion We usually have two hoies.

11 . IF A and B are statistially independent, then. If A and B are not known to be statistially independent, we usually have to use onditional probability and the multipliation rule: This still requires us to be able to alulate A B). Note: If events are physially independent, then they will also be statistially independent. Pairwise independene does not imply mutual independene Example 9. A jar ontains balls: one red, one white, one blue, and one red, white& blue. raw one ball at random. So A, B and C are NOT mutually independent, despite being pairwise independent.

12 Example 0. It is known that 0% of a ertain ompany s washing mahines require servie while under warranty, whereas only 0% of its dryers need suh servie. If someone purhases both a washer and a dryer made by this ompany, what is the probability that both mahines need warranty servie? Let A denote the event that the washer needs servie while under warranty, Let B defined analogously for the dryer. Then A)=0., B)=0.. Assuming that the two mahines funtion independently of one another, the desired probability is P ( A B) A) B) The probability that neither mahine needs servie is P ( A' B') A') B') (0.7)(0.9) 0.6. Example. A system onsists of four omponents, as illustrated in Fig.The entire system will work if either the - subsystem works or if the - subsystem works (sine the two subsystems are onneted in parallel). Sine the two omponents in eah subsystem are onneted in series, a subsystem will work only if both its omponents work. If omponents work or fail independently of one another and if eah works with probability 0.9, what is the probability that the entire system will work (the system reliability oeffiient)? Letting A i (i=,,,) be the event that the ith omponent works, the mutually independent. A i s are The event that the - subsystem works in A A, and similarly, A A denotes the event that the - subsystem works. The event that the entire system works is A A ) ( A ), so P[( A A ) ( A A )] A A ) A A ) P[( A ( A A ) ( A A ) A ) A ) A ) A ) A ) A ) A ) (0.9)(0.9) (0.9)(0.9) (0.9)(0.9)(0.9)(0.9) A )]

13 Example. Suppose that a mahine produes a defetive item with probability p (0<p<) and produes a nondefetive item with probability -p. Suppose further that six items produed by the mahine are seleted at random and inspeted, and that the results (defetive or nondefetive) for these six items are independent. We shall determine the probability that exatly two of the six items are defetive. Solution. It an be assumed that the sample spae S ontains all possible arrangements of six items, eah one of whih might be either defetive or nondefetive. Let j denote the event that the jth item in the sample is defetive, then j is the event that this item is nondefetive. Sine the outomes for the six different items are independent, the probability of obtaining any partiular sequene of defetive and nondefetive items will simply be the produt of the individual probabilities for the items. For example, 5 ) 6 ) ) ) ) ( p) p( p)( p) p( p) p ( p). 5 ) It an be seen that the probability of any other partiular sequene in S ontaining two defetive items and four nondefetive items will also be p ( p). 6 ) Sine there are 6 distint arrangements of two defetive items and four nondefetive items. 6 The probability of obtaining exatly two defetives is p ( p).. Bayes' Theorem: inverting onditional probabilities Then This is the simplest form of Bayes' Theorem, named after Thomas Bayes (70-6), English lergyman and founder of Bayesian Statistis. Bayes' Theorem allows us to invert the onditioning, i.e. to

14 express B A) in terms of A B). This is very useful. For example, it might be easy to alulate, later event earlier event); but we might only observe the later event and wish to dedue the probability that the earlier event ourred, earlier event later event) Full statement of Bayes' Theorem: Example. The ase of the Perfidious Gardener. Mr Smith owns a hysterial rosebush. It will die with probability / if watered, and with probability / if not watered. Worse still, Smith employs a perfidious gardener who will fail to water the rosebush with probability /. Smith returns from holiday to find the rosebush... EA!! What is the probability that the gardener did not water it?

15 So the gardener failed to water the rosebush with probability /. Example. The ase of the efetive Kethup Bottle. Kethup bottles are produed in different fatories, aounting for 50%, 0%, and 0% of the total output respetively. The perentage of defetive bottles from the fatories is respetively 0.%, 0.6%, and.%. A statistis leturer who eats only kethup finds a defetive bottle in her wig. What is the probability that it ame from Fatory? Information given: