PROBABILITY: CONDITIONING, BAYES THEOREM [DEVORE 2.4]
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![PROBABILITY: CONDITIONING, BAYES THEOREM [DEVORE 2.4] CONDITIONAL PROBABILITY: Let events E, F be events in the sample space Ω of an experiment.](/docs-images/95/122760177/images/1-1.jpg "Then: The conditional probability of F given E, P(F E), is the probability of event F assuming E has already occurred.")
1 PROBABILITY: CONDITIONING, BAYES THEOREM [DEVORE 2.4] CONDITIONAL PROBABILITY: Let events E, F be events in the sample space Ω of an experiment. Then: The conditional probability of F given E, P(F E), is the probability of event F assuming E has already occurred. The conditional probability of E given F, P(E F ), is the probability of event E assuming F has already occurred. WARNING: Order matters: in general, P(F E) P(E F ) CONDITIONAL PROBABILITY: Let events E, F be events in the sample space Ω such that P(E) > 0. Then: P(If E then F ) = P(F given E) = P(F E) := E F E = P(E F ) P(E) INTERSECTION OF EVENTS: Let events E, F be events in the sample space Ω of an experiment. Then: P(E F ) = P(E) P(F E) = P(F ) P(E F ) PARTITIONING A SAMPLE SPACE: Events E 1, E 2,..., E k Ω partition sample space Ω if: k E 1, E 2,..., E k are pairwise disjoint AND E i = Ω Think of the sample space Ω as a puzzle & the partitioning events E 1, E 2,..., E k as the puzzle pieces. Events E 1, E 2, E 3, E 4 partition sample space Ω. LAW OF TOTAL PROBABILITY (LTP): Let events E 1,..., E k partition sample space Ω. Then k P(F ) = P(F E i) = P(F E i) P(E i) In practice, it s less tedious to work with the middle form involving intersections rather than right-most form. BAYES THEOREM: Let events E 1,..., E k Ω partition sample space Ω. Then P(E j F ) = P(F Ej) k P(F E i) = P(F Ej) P(Ej) k P(F E i) P(E i) for j = 1, 2,..., k In practice, it s less tedious to work with the middle form involving intersections rather than right-most form. Moreover, it s often far easier to directly work with the Law of Total Probability than directly with Bayes Theorem.
2 PROBABILITY: PROBABILITY TREES [DEVORE 2.4] Events F 1, F 2, F 3 must partition the sample space Ω. Events E 1, E 2, E 3 must partition the sample space Ω.
3 PROBABILITY: JOINT PROBABILITY TABLES [DEVORE 2.4] 2-STAGE EXPERIMENTS & 2-WAY JOINT PROBABILITY TABLES: F F c TOTAL E P(E F ) P(E F c ) P(E) E c P(E c F ) P(E c F c ) P(E c ) TOTAL P(F ) P(F c ) (DON T CARE) F 1 F 2 F 2 TOTAL E 1 P(E 1 F 1) P(E 1 F 2) P(E 1 F 3) P(E 1) E 2 P(E 2 F 1) P(E 2 F 2) P(E 2 F 3) P(E 2) E 3 P(E 3 F 1) P(E 3 F 2) P(E 3 F 3) P(E 3) TOTAL P(F 1) P(F 2) P(F 3) (DON T CARE) Events E 1, E 2, E 3 must partition sample space Ω. Events F 1, F 2, F 3 must partition sample space Ω. 3-STAGE EXPERIMENTS & 3-WAY JOINT PROBABILITY TABLES: E G G c TOTAL F P(E F G) P(E F G c ) P(E F ) F c P(E F c G) P(E F c G c ) P(E F c ) TOTAL P(E G) P(E G c ) (DON T CARE) E c G G c TOTAL F P(E c F G) P(E c F G c ) P(E c F ) F c P(E c F c G) P(E c F c G c ) P(E c F c ) TOTAL P(E c G) P(E c G c ) (DON T CARE) E 1 G 1 G 2 G 3 TOTAL F 1 P(E 1 F 1 G 1) P(E 1 F 1 G 2) P(E 1 F 1 G 3) P(E 1 F 1) F 2 P(E 1 F 2 G 1) P(E 1 F 2 G 2) P(E 1 F 2 G 3) P(E 1 F 2) F 3 P(E 1 F 3 G 1) P(E 1 F 3 G 2) P(E 1 F 3 G 3) P(E 1 F 3) TOTAL P(E 1 G 1) P(E 1 G 2) P(E 1 G 3) (DON T CARE) E 2 G 1 G 2 G 3 TOTAL F 1 P(E 2 F 1 G 1) P(E 2 F 1 G 2) P(E 2 F 1 G 3) P(E 2 F 1) F 2 P(E 2 F 2 G 1) P(E 2 F 2 G 2) P(E 2 F 2 G 3) P(E 2 F 2) F 3 P(E 2 F 3 G 1) P(E 2 F 3 G 2) P(E 2 F 3 G 3) P(E 2 F 3) TOTAL P(E 2 G 1) P(E 2 G 2) P(E 2 G 3) (DON T CARE) E 3 G 1 G 2 G 3 TOTAL F 1 P(E 3 F 1 G 1) P(E 3 F 1 G 2) P(E 3 F 1 G 3) P(E 3 F 1) F 2 P(E 3 F 2 G 1) P(E 3 F 2 G 2) P(E 3 F 2 G 3) P(E 3 F 2) F 3 P(E 3 F 3 G 1) P(E 3 F 3 G 2) P(E 3 F 3 G 3) P(E 3 F 3) TOTAL P(E 3 G 1) P(E 3 G 2) P(E 3 G 3) (DON T CARE) Events E 1, E 2, E 3 must partition sample space Ω. Events F 1, F 2, F 3 must partition sample space Ω. Events G 1, G 2, G 3 must partition sample space Ω.
4 EX 2.4.1: Consider the following system of connected components shown below. Components 1 & 2 are connected in series with component 3 connected to them in parallel. For the entire system to function, either components 1 & 2 must function or component 3 must function. The experiment consists of observing each component s condition and labeling it as either (S)uccess or (F )ailure. So, for example, the outcome for components 1 & 2 functioning but 3 not functioning would be denoted as SSF. (a) Determine the sample space Ω for the experiment. (b) Find the probability that Component 1 functions. (c) Find the probability that Component 2 functions. (d) Find the probability that the entire system functions. (e) Find the probability that Component 1 functions given the entire system functions. Interpret result in context. (f) Find the probability that if Component 2 functions then Component 1 functions. Interpret result in context. (g) Find the probability that Components 1 & 2 both function given entire system functions. Interpret result. (h) Find the probability that Component 1 functions or Component 2 functions given entire system functions.
5 EX 2.4.2: The incidence of a virus in a village is 10% and that a test correctly identifies the virus 95% of the time. Assume that false positives occur 8% of the time. (a) Identify the events & probabilities that are given. (b) Construct a probability tree representing the experiment. Label all relevant events & probabilities. (c) What is the probability of a person in the village having the virus? (d) What is the probability of a person testing positive for the virus given the person has the virus? (e) What is the probability that if a person has the virus then the person tests negative for the virus? (f) What is the probability that a person has the virus and tests negative for the virus? (g) What is the probability that a person has the virus given the person tests positive for the virus? (h) What is the probability that a person does not have the virus given the person tests negative for the virus?
6 EX 2.4.3: Two hundred engineering students were randomly selected and their height & eye color were recorded and tallied: 8% of the short students have blue eyes 13.5% of the tall students have blue eyes 11.5% of the short students have brown eyes 10.5% of the tall students have brown eyes 9% of the short students have green eyes 22% of the tall students have green eyes 15.5% of the short students have hazel eyes 10% of the tall students have hazel eyes (a) Construct the joint probability table, label all relevant events & probabilities, including the column totals & row totals. (b) Find the probability that a student is tall and have brown eyes. (c) Find the probability that a student is short. (d) Find the probability that a student has hazel eyes. (e) Find the probability that a student has green eyes given the student is tall. (f) Find the probability that if a student has blue eyes then the student is short. (g) Find the probability that a student has blue eyes or hazel eyes. (h) Find the probability that a student has green eyes or hazel eyes given the student is tall. (i) Find the probability that a student is tall given the student has green eyes or hazel eyes. (j) Find the probability that a student has green eyes or blue eyes or brown eyes given the student is short.
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