0

0

2099 days ago,
602 views

PowerPoint PPT Presentation
CMSC 203. Discrete Structures. 2. Discrete Probability. All that you have found out about tallying constitutes the premise for registering the likelihood of occasions to happen.In the accompanying, we will utilize the idea test for a technique that yields one of a given arrangement of conceivable outcomes.This set of conceivable results is known as the specimen space of the experiment.An occasion is a subset of the example

Presently it's a great opportunity to take a gander at… Discrete Probability Discrete Structures

Discrete Probability Everything you have found out about checking constitutes the reason for figuring the likelihood of occasions to happen. In the accompanying, we will utilize the thought explore for a methodology that yields one of a given arrangement of conceivable results. This arrangement of conceivable results is known as the example space of the investigation. An occasion is a subset of the example space. Discrete Structures

Discrete Probability If all results in the example space are similarly likely, the accompanying meaning of likelihood applies: The likelihood of an occasion E, which is a subset of a limited specimen space S of similarly likely results, is given by p(E) = |E|/|S|. Likelihood values run from 0 (for an occasion that will never happen) to 1 (for an occasion that will dependably happen at whatever point the trial is completed). Discrete Structures

Discrete Probability Example I: A urn contains four blue balls and five red balls. What is the likelihood that a ball browsed the urn is blue? Arrangement: There are nine conceivable results, and the occasion "blue ball is picked" involves four of these results. Along these lines, the likelihood of this occasion is 4/9 or around 44.44%. Discrete Structures

Discrete Probability Example II: What is the likelihood of winning the lottery 6/49, that is, picking the right arrangement of six numbers out of 49? Arrangement: There are C(49, 6) conceivable results. Just a single of these results will really make us win the lottery. p(E) = 1/C(49, 6) = 1/13,983,816 Discrete Structures

Complimentary Events Let E be an occasion in an example space S. The likelihood of an occasion –E, the complimentary occasion of E, is given by p(- E) = 1 – p(E). This can without much of a stretch be appeared: p(- E) = (|S| - |E|)/|S| = 1 - |E|/|S| = 1 – p(E). This lead is valuable on the off chance that it is simpler to decide the likelihood of the complimentary occasion than the likelihood of the occasion itself. Discrete Structures

Complimentary Events Example I: An arrangement of 10 bits is arbitrarily created. What is the likelihood that no less than one of these bits is zero? Arrangement: There are 2 10 = 1024 conceivable results of creating such a grouping. The occasion –E, "none of the bits is zero" , incorporates just a single of these results, to be specific the succession 1111111111. Subsequently, p(- E) = 1/1024. Presently p(E) can without much of a stretch be registered as p(E) = 1 – p(- E) = 1 – 1/1024 = 1023/1024. Discrete Structures

Complimentary Events Example II: What is the likelihood that no less than two out of 36 individuals have a similar birthday? Arrangement: The example space S includes all potential outcomes for the birthdays of the 36 individuals, so |S| = 365 36 . Give us a chance to consider the occasion –E ("no two individuals out of 36 have a similar birthday"). –E incorporates P(365, 36) results (365 potential outcomes for the main individual's birthday, 364 for the second, et cetera). At that point p(- E) = P(365, 36)/365 36 = 0.168, so p(E) = 0.832 or 83.2% Discrete Structures

Discrete Probability Let E 1 and E 2 be occasions in the example space S. At that point we have: p(E 1 E 2 ) = p(E 1 ) + p(E 2 ) - p(E 1 E 2 ) Does this help you to remember something? Obviously, the standard of incorporation rejection . Discrete Structures

Discrete Probability Example: What is the likelihood of a positive whole number chosen indiscriminately from the arrangement of positive whole numbers not surpassing 100 to be detachable by 2 or 5? Arrangement: E 2 : "number is distinguishable by 2" E 5 : "whole number is distinct by 5" E 2 = {2, 4, 6, … , 100} |E 2 | = 50 p(E 2 ) = 0.5 Discrete Structures

Discrete Probability E 5 = {5, 10, 15, … , 100} |E 5 | = 20 p(E 5 ) = 0.2 E 2 E 5 = {10, 20, 30, … , 100} |E 2 E 5 | = 10 p( E 2 E 5 ) = 0.1 p( E 2 E 5 ) = p( E 2 ) + p( E 5 ) – p( E 2 E 5 ) p( E 2 E 5 ) = 0.5 + 0.2 – 0.1 = 0.6 Discrete Structures

Discrete Probability What happens if the results of an examination are not similarly likely? All things considered, we relegate a likelihood p(s) to every result s S, where S is the specimen space. Two conditions must be met: (1): 0 p(s) 1 for each sS, and (2): sS p(s) = 1 This implies, as we definitely know, that (1) every likelihood must be an incentive in the vicinity of 0 and 1, and (2) the probabilities must indicate 1, since one of the results is ensured to happen. Discrete Structures

Discrete Probability How would we be able to get these probabilities p(s) ? The likelihood p( s ) appointed to a result s rises to the furthest reaches of the quantity of times s happens separated by the quantity of times the trial is performed. When we know the probabilities p(s), we can register the likelihood of an occasion E as takes after: p(E) = sE p(s) Discrete Structures

Discrete Probability Example I: A pass on is one-sided so that the number 3 seems twice as regularly as each other number. What are the probabilities of every single conceivable result? Arrangement: There are 6 conceivable results s 1 , … , s 6 . p(s 1 ) = p(s 2 ) = p(s 4 ) = p(s 5 ) = p(s 6 ) p(s 3 ) = 2p(s 1 ) Since the probabilities must indicate 1, we have: 5p(s 1 ) + 2p(s 1 ) = 1 7p(s 1 ) = 1 p(s 1 ) = p(s 2 ) = p(s 4 ) = p(s 5 ) = p(s 6 ) = 1/7, p(s 3 ) = 2/7 Discrete Structures

Discrete Probability Example II: For the one-sided kick the bucket from Example I, what is the likelihood that an odd number shows up when we roll the bite the dust? Arrangement: E odd = {s 1 , s 3 , s 5 } Remember the equation p(E) = sE p(s). p(E odd ) = sE odd p(s) = p(s 1 ) + p(s 3 ) + p(s 5 ) p(E odd ) = 1/7 + 2/7 + 1/7 = 4/7 = 57.14% Discrete Structures

Conditional Probability If we flip a coin three circumstances, what is the likelihood that an odd number of tails shows up (occasion E) , if the principal hurl is a tail (occasion F) ? In the event that the primary hurl is a tail, the conceivable arrangements are TTT, TTH, THT, and THH. In two out of these four cases, there is an odd number of tails. Hence, the likelihood of E, under the condition that F happens, is 0.5. We call this contingent likelihood . Discrete Structures

Conditional Probability If we need to figure the contingent likelihood of E given F, we utilize F as the specimen space. For any result of E to happen under the condition that F likewise happens, this result should likewise be in E F. Definition: Let E and F be occasions with p(F) > 0. The contingent likelihood of E given F, meant by p(E | F), is characterized as p(E | F) = p(E F)/p(F) Discrete Structures

Conditional Probability Example: What is the likelihood of an irregular piece string of length four contains no less than two continuous 0s, given that its first piece is a 0 ? Arrangement: E: "bit string contains no less than two successive 0s" F: "first piece of the string is a 0" We know the recipe p(E | F) = p(E F)/p(F) . E F = {0000, 0001, 0010, 0011, 0100} p(E F) = 5/16 p(F) = 8/16 = 1/2 p(E | F) = (5/16)/(1/2) = 10/16 = 5/8 = 0.625 Discrete Structures

Independence Let us come back to the case of flipping a coin three circumstances. Does the likelihood of occasion E (odd number of tails) rely on upon the event of occasion F (first hurl is a tail) ? At the end of the day, is it the case that p(E | F) p(E) ? We really find that p(E | F) = 0.5 and p(E) = 0.5, so we say that E and F are free occasions . Discrete Structures

Independence Because we have p(E | F) = p(E F)/p(F), p(E | F) = p(E) if and just if p(E F) = p(E)p(F). Definition: The occasions E and F are said to be free if and just if p(E F) = p(E)p(F). Clearly, this definition is symmetrical for E and F. On the off chance that we have p(E F) = p(E)p(F), then it is likewise genuine that p(F | E) = p(F). Discrete Structures

Independence Example: Suppose E is the occasion that a haphazardly produced bit string of length four starts with a 1, and F is the occasion that an arbitrarily created bit string contains a significantly number of 0s. Are E and F autonomous? Arrangement: Obviously, p(E) = p(F) = 0.5. E F = {1111, 1001, 1010, 1100} p(E F) = 0.25 p(E F) = p(E)p(F) Conclusion: E And F are autonomous . Discrete Structures

Bernoulli Trials Suppose an explore different avenues regarding two conceivable results ,, for example, flipping a coin. Every execution of such a trial is known as a Bernoulli trial . We will call the two conceivable results a win or a disappointment , individually. On the off chance that p is the likelihood of a win and q is the likelihood of a disappointment, clearly p + q = 1. Discrete Structures

Bernoulli Trials Often we are keen on the likelihood of precisely k victories when an investigation comprises of n free Bernoulli trials . Case: A coin is one-sided so that the likelihood of head is 2/3. What is the likelihood of precisely four heads to come up when the coin is hurled seven circumstances? Discrete Structures

Bernoulli Trials Solution: There are 2 7 = 128 conceivable results. The quantity of conceivable outcomes for four heads among the seven trials is C(7, 4). The seven trials are free, so the likelihood of each of these results is (2/3) 4 (1/3) 3 . Therefore, the likelihood of precisely four heads to show up is C(7, 4)(2/3) 4 (1/3) 3 = 560/2187 = 25.61% Discrete Structures

Bernoulli Trials Theorem: The likelihood of k triumphs in n free Bernoulli trials, with likelihood of accomplishment p and likelihood of disappointment q = 1 – p, is C(n, k)p k q n-k . See the reading material for the confirmation. We mean by b(k; n, p) the likelihood of k achievements in n free Bernoulli trials with likelihood of progress p and likelihood of disappointment q = 1 – p. Considered as capacity of k, we call b the binomial circulation . Discrete Structures

SPONSORS

No comments found.

SPONSORS

SPONSORS