CONDITIONAL PROBABILITY (continued) ODDS RATIO ( ) P( H ) Provides nformation on how much more likely that an event A occurs than that it does not.

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Thomas Morris Morton
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1 CONDITIONAL PROBABILITY (ontinued) ODDS RATIO Def: The odds ratio of an event A is defined by 1 Provides nformation on how muh more likely that an event A ours than that it does not. Its value an hange with updated information. So use Bayes formula and reompute the ratio: P H E P E H P H P( H E) P E P( H ) P E H P E P H E P H P E H P H E P H P E H Note the old value is modified by the ratio of the onditional probability!

2 INDEPENDENCE If P( E) P( F) P EF, E and F are independent. If events are not independent, they are dependent. Independene is different from mutually exlusive. Proposition: If E and F are independent, then so are E and show this. + + P E P EF P EF P E P F P EF rearranging P EF P E 1 P F P E P F! F. Use basi priniples to Three events are independent iff: P( E) P( F) P( G) P( E) P( F) P( F) P( G) P( E) P( G) P EFG P EF P FG P EG (, ) B C B C Similarly, (, ) (, ) P B A C P B C P B A P B P C A B P C B P C A P C

3 This result extends to (... ) A 1 n i i 1 You an show this easily by n (... ) (... ) 1A2 1A2 An P( A1... An) P( A1)... A A n 1 Ex: Let a ball be drawn from an urn ontaining 4 balls, numbered 1-4. Assume that all outomes in S { 1, 2, 3, 4} are equally likely. Look at the events: {, } the ball is a 1 or a 2, {, }, {, } A 1 2 B 1 3 C 1 4 Are they independent?,, P B P C 1 2 B 1 4 P B C P C P BC P B P C BC 1 4 P B P C Now, look at the onditional probabilities: Is P( C AB) P( C) P( C A) P( C B)??? P C 1/ 2 P C A P C B No ---- P C A, B 1!!! Ex: Suppose an automobile aident on a street in Austin where Car 1 stops suddenly and is hit from behind by Car 2 on a rainy day. Suppose that three persons a,b, witness the aident as they are on their way to lass. Suppose that the probability that eah witness has orretly observed that Car 1 stopped suddenly is estimated by having the witnesses observe a number of ontrived inidents about whih eah is questioned. Assume that it P a. 9, P b. 8, P. 7. is learned that Let A,B,C denote the events that a,b, will state that Car 1 stopped suddenly. {, :, } S z z z z y n i i

4 Assume independene? If so, BC Now, remember that if A,B,C are independent, so are A,B, and C. Thus, the probability that exatly two of the witnesses will say that Car 1 stopped suddenly an be alulated easily as: (. )(. )(. ) + (. )(. )(. ) + (. )(. )(. ) BC AB C A BC What about the probability that at least two will say that Car 1 stopped suddenly? (.902) INDEPENDENT TRIALS Def: Independent trials are experiments onsisting of independent events suh that all subexperiments are idential (same sample spae and same probability funtion) Ex: Given and infinite no. of trials, P(suess) p, P(failure)1-p. What is the probability of k i) exatly k suesses in n trials? p ( 1 p) n k ii) all trials are a suess in n? n p iii) all trials are a suess if n is not finite? If E i is defined to be a failure, n n P E1 E2... En p. Now use ontinuity: P Ei P lim Ei i 1 n i 1 n n 0, p < 1 limp Ei lim p n i 1 n 1, p 1 DEPENDENT TRIALS Events whih are not independent are dependent. By the multipliative rule: ( 1) ( 2 1) ( 3 1 2)... ( n n 1) A A A A A A

5 Ex: Given an urn with M balls, M W are white. Let a sample of size n< M W be drawn w/out replaement. What is the probability that all the balls are white? Let A i be the event that the ball drawn on the ith draw is white. Need the probability of the intersetion of the A i ( i Ai 1) A A w ( ) ( ) M i 1 M i 1 (... A ) A 1 2 n w w 1... ( w ( )) ( )... ( ) M M M n 1 M M 1 M n 1 Ex: The first hild born to a woman was a boy with hemophilia. The woman s family had no history of hemophilia. Being onerned about having a seond hild, she reasoned that her son was a mutant beause he did not inherit the trait from her. The probability of a seond hild having the disease, if he is a boy, is the same as the probability of him being a mutant (a very small no.). What do you think???? Consider the history: Given the 3-tuple: { z, z, z } { Mother, son 1, son 2} z 1 true (t) or false (f) depending on whether the mother is a arrier. z 2 or z 3 true (t) or false (f) depending on whether the ith son has the disease Now look at the events A i. In the sample spae S of these events, the A i orrespond to the value of z. We want: ( 3 2) A ( 2 3) A 2 The probability funtion on the subsets of S must be defined.

6 Assume: Mother has no history of disease. A boy has an X and a Y hromosome. If he has hemophilia, he has an X hromosome whih has a gene ausing the disease. Let p be the probability of a mutation of the X hromosome into X. A woman arries two X hromosomes. At least one of them must be mutated if she transmits the disease. (Assume that the probability of her having two mutated hromosomes is negligible.) Let A 1 denote the mother is a arrier, et. 2 ( 1) ( 1 ) 1 1 p 2p 1 2p If she has a mutant gene, the first son will (will not) inherit it with: ( 2 1) ( 2 1) A 1 2 A 1 2 If she does not have a mutant gene, the first son s probabilities of having (not having) the hromosome are: P A2 A 1 p P A2 A 1 1 p Now look at the seond son s probabilities of inheriting or having the hromosome (and similarly not inheriting/not having the hromosome: P A3 A1A2 P A3 A1A P A3 A1A 2 P A3 A1A (note: independent of the seond son) Now, ( 3 1 2) ( ) A A A A p ( 2 3) ( 1 2 3) + ( 1 2 3) A A A A A A A A A 1 p ( 1) ( 2 1) ( 3 1 2) ( 1 ) ( 2 1 ) ( 3 1 2) A A A + A A A p + ( 1 2p) p p 2 2

7 So, ( 2) ( 2 1) ( 1) ( 2 1 ) ( 1 ) + A A ( 3 2) A 1 2p + p 1 2p 2p 2 p2 1 (Is this what you would have expeted?) 2p 4

8 MARKOV DEPENDENT TRIALS A speial kind of dependene widely used in OR is Markov dependene where: ( k+ 1 k, k ) ( k+ 1 k) A A A A Suppose that there are two outomes (events) suh that the probability of one is p and the other is 1-p. If trials are independent, we all this a Bernoulli proess. So we only need to ompute probabilities of adjaent events in order to represent the proess. Let (, ) ( k+ 1 k ) P s s s A s denote the probability of a suess on the k+1 st trial, given a suess at k th trial. Likewise, the probability of a failure at the k+1 st trial, given a suess at k th trial: ( k+ 1 k ) (, ) (, ) f A s P s f 1 P s s P f, s 1 P f, f, et Now onsider a sequene of trials. Suppose you desire the probability of suess at the kth trial, p k (s), k1,2,...n and the probability of failure be 1- p k (s). In general, for Markov proesses, we know: so ( k) ( k 1) ( k k 1) ( k 1) ( k k 1) A + A, () () () ( )( ) p s p s P s, s + 1 p s 1 p f, f k k 1 k 1 whih an be simplified. This is a differene equation: () () (, ) (, ) (, ) pk s pk 1 s P s s + P f f P f f

9 We an solve for any p k (s) in terms of p 1 (s) and the transition probabilities. If we know the value of p 1 (s), we an alulate the rest onditional on the outome of the first trial. These probabilities must be represented as P (, ) Bernoulli trials. For example, k i j where i and j are either s or f for P k (s,s) onditional probability of suess at the (k+1)st trial, given suess at the first trial, et. (Note that this is after k transitions)

1.3 Complex Numbers; Quadratic Equations in the Complex Number System*

1.3 Complex Numbers; Quadratic Equations in the Complex Number System* 04 CHAPTER Equations and Inequalities Explaining Conepts: Disussion and Writing 7. Whih of the following pairs of equations are equivalent? Explain. x 2 9; x 3 (b) x 29; x 3 () x - 2x - 22 x - 2 2 ; x