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!
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
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 1 1 2 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 1 2 3 i
Assume independene? If so, BC. 9. 8. 7. 504 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 9 8 3 9 2 7 1 8 7.398 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 1 2... n 1) A A A A A A
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 1 2... 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} 1 2 3 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.
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 2 1 2 P A3 A1A 2 P A3 A1A 2 1 2 (note: independent of the seond son) Now, 3 1 2 3 1 2 ( 3 1 2) ( 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 2 1 2 1 2p + ( 1 2p) p p 2 2
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
MARKOV DEPENDENT TRIALS A speial kind of dependene widely used in OR is Markov dependene where: ( k+ 1 k, k 1... 1) ( 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 1 + 1 P f f
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)