Toán học - Chapter 7: Discrete probability

Discrete ProbabilityChapter 7With Question/Answer AnimationsCopyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.Chapter SummaryIntroduction to Discrete ProbabilityProbability TheoryBayes’ TheoremExpected Value and VarianceAn Introduction to Discrete ProbabilitySection 7.1Section SummaryFinite ProbabilityProbabilities of Complements and Unions of EventsProbabilistic ReasoningProbability of an Event We first study Pierre-Simon Laplace’s classical theory of probability, which he introduced in the 18th century, when he analyzed games of chance.We first define these key terms:An experiment is a procedure that yields one of a given set of possible outcomes.The sample space of the experiment is the set of possible outcomes.An event is a subset of the sample space.Here is how Laplace defined the probability of an event: Definition: If S is a finite sample space of equally likely outcomes, and E is an event, that is, a subset of S, then the probability of E is p(E) = |E|/|S|.For every event E, we have 0 ≤ p(E) ≤ 1. This follows directly from the definition because 0 ≤ p(E) = |E|/|S| ≤ |S|/|S| ≤ 1, since 0 ≤ |E| ≤ |S|.Pierre-Simon Laplace (1749-1827)Applying Laplace’s Definition Example: An urn contains four blue balls and five red balls. What is the probability that a ball chosen from the urn is blue? Solution: The probability that the ball is chosen is 4/9 since there are nine possible outcomes, and four of these produce a blue ball. Example: What is the probability that when two dice are rolled, the sum of the numbers on the two dice is 7? Solution: By the product rule there are 62 = 36 possible outcomes. Six of these sum to 7. Hence, the probability of obtaining a 7 is 6/36 = 1/6. Applying Laplace’s Definition Example: In a lottery, a player wins a large prize when they pick four digits that match, in correct order, four digits selected by a random mechanical process. What is the probability that a player wins the prize? Solution: By the product rule there are 104 = 10,000 ways to pick four digits. Since there is only 1 way to pick the correct digits, the probability of winning the large prize is 1/10,000 = 0.0001. A smaller prize is won if only three digits are matched. What is the probability that a player wins the small prize? Solution: If exactly three digits are matched, one of the four digits must be incorrect and the other three digits must be correct. For the digit that is incorrect, there are 9 possible choices. Hence, by the sum rule, there a total of 36 possible ways to choose four digits that match exactly three of the winning four digits. The probability of winning the small price is 36/10,000 = 9/2500 = 0.0036.Applying Laplace’s Definition Example: There are many lotteries that award prizes to people who correctly choose a set of six numbers out of the first n positive integers, where n is usually between 30 and 60. What is the probability that a person picks the correct six numbers out of 40? Solution: The number of ways to choose six numbers out of 40 is C(40,6) = 40!/(34!6!) = 3,838,380. Hence, the probability of picking a winning combination is 1/ 3,838,380 ≈ 0.00000026. Can you work out the probability of winning the lottery with the biggest prize where you live?Applying Laplace’s Definition Example: What is the probability that the numbers 11, 4, 17, 39, and 23 are drawn in that order from a bin with 50 balls labeled with the numbers 1,2, , 50 if The ball selected is not returned to the bin.The ball selected is returned to the bin before the next ball is selected. Solution: Use the product rule in each case.Sampling without replacement: The probability is 1/254,251,200 since there are 50 ∙49 ∙47 ∙46 ∙45 = 254,251,200 ways to choose the five balls.b) Sampling with replacement: The probability is 1/505 = 1/312,500,000 since 505 = 312,500,000.The Probability of Complements and Unions of Events Theorem 1: Let E be an event in sample space S. The probability of the event = S − E, the complementary event of E, is given by Proof: Using the fact that | | = |S| − |E|, The Probability of Complements and Unions of Events Example: A sequence of 10 bits is chosen randomly. What is the probability that at least one of these bits is 0? Solution: Let E be the event that at least one of the 10 bits is 0. Then is the event that all of the bits are 1s. The size of the sample space S is 210. Hence,The Probability of Complements and Unions of Events Theorem 2: Let E1 and E2 be events in the sample space S. Then Proof: Given the inclusion-exclusion formula from Section 2.2, |A ∪ B| = |A| + | B| − |A ∩ B|, it follows that Example: What is the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5? Solution: Let E1 be the event that the integer is divisible by 2 and E2 be the event that it is divisible 5? Then the event that the integer is divisible by 2 or 5 is E1 ∪ E2 and E1 ∩ E2 is the event that it is divisible by 2 and 5. It follows that: p(E1 ∪ E2) = p(E1) + p(E2) – p(E1 ∩ E2) = 50/100 + 20/100 − 10/100 = 3/5.The Probability of Complements and Unions of EventsMonty Hall Puzzle Example: You are asked to select one of the three doors to open. There is a large prize behind one of the doors and if you select that door, you win the prize. After you select a door, the game show host opens one of the other doors (which he knows is not the winning door). The prize is not behind the door and he gives you the opportunity to switch your selection. Should you switch? Solution: You should switch. The probability that your initial pick is correct is 1/3. This is the same whether or not you switch doors. But since the game show host always opens a door that does not have the prize, if you switch the probability of winning will be 2/3, because you win if your initial pick was not the correct door and the probability your initial pick was wrong is 2/3.132(This is a notoriously confusing problem that has been the subject of much discussion . Do a web search to see why!) Probability TheorySection 7.2Section SummaryAssigning ProbabilitiesProbabilities of Complements and Unions of EventsConditional Probability IndependenceBernoulli Trials and the Binomial DistributionRandom VariablesThe Birthday ProblemMonte Carlo AlgorithmsThe Probabilistic Method (not currently included in the overheads)Assigning Probabilities Laplace’s definition from the previous section, assumes that all outcomes are equally likely. Now we introduce a more general definition of probabilities that avoids this restriction.Let S be a sample space of an experiment with a finite number of outcomes. We assign a probability p(s) to each outcome s, so that:i. 0 ≤ p(s) ≤ 1 for each s Î Sii. The function p from the set of all outcomes of the sample space S is called a probability distribution.Assigning Probabilities Example: What probabilities should we assign to the outcomes H(heads) and T (tails) when a fair coin is flipped? What probabilities should be assigned to these outcomes when the coin is biased so that heads comes up twice as often as tails? Solution: For a fair coin, we have p(H) = p(T) = ½. For a biased coin, we have p(H) = 2p(T). Because p(H) + p(T) = 1, it follows that 2p(T) + p(T) = 3p(T) = 1. Hence, p(T) = 1/3 and p(H) = 2/3. Uniform Distribution Definition: Suppose that S is a set with n elements. The uniform distribution assigns the probability 1/n to each element of S. (Note that we could have used Laplace’s definition here.) Example: Consider again the coin flipping example, but with a fair coin. Now p(H) = p(T) = 1/2. Probability of an Event Definition: The probability of the event E is the sum of the probabilities of the outcomes in E.Note that now no assumption is being made about the distribution. Example Example: Suppose that a die is biased so that 3 appears twice as often as each other number, but that the other five outcomes are equally likely. What is the probability that an odd number appears when we roll this die? Solution: We want the probability of the event E = {1,3,5}. We have p(3) = 2/7 and p(1) = p(2) = p(4) = p(5) = p(6) = 1/7. Hence, p(E) = p(1) + p(3) + p(5) = 1/7 + 2/7 + 1/7 = 4/7.Probabilities of Complements and Unions of EventsComplements: still holds. Since each outcome is in either E or , but not both, Unions: also still holds under the new definition. Combinations of Events Theorem: If E1, E2, is a sequence of pairwise disjoint events in a sample space S, thensee Exercises 36 and 37 for the proofConditional Probability Definition: Let E and F be events with p(F) > 0. The conditional probability of E given F, denoted by P(E|F), is defined as: Example: A bit string of length four is generated at random so that each of the 16 bit strings of length 4 is equally likely. What is the probability that it contains at least two consecutive 0s, given that its first bit is a 0? Solution: Let E be the event that the bit string contains at least two consecutive 0s, and F be the event that the first bit is a 0. Since E ⋂ F = {0000, 0001, 0010, 0011, 0100}, p(E⋂F)=5/16.Because 8 bit strings of length 4 start with a 0, p(F) = 8/16= ½. Hence,Conditional Probability Example: What is the conditional probability that a family with two children has two boys, given that they have at least one boy. Assume that each of the possibilities BB, BG, GB, and GG is equally likely where B represents a boy and G represents a girl. Solution: Let E be the event that the family has two boys and let F be the event that the family has at least one boy. Then E = {BB}, F = {BB, BG, GB}, and E ⋂ F = {BB}.It follows that p(F) = 3/4 and p(E⋂F)=1/4. Hence, Independence Definition: The events E and F are independent if and only if Example: Suppose E is the event that a randomly generated bit string of length four begins with a 1 and F is the event that this bit string contains an even number of 1s. Are E and F independent if the 16 bit strings of length four are equally likely? Solution: There are eight bit strings of length four that begin with a 1, and eight bit strings of length four that contain an even number of 1s.Since the number of bit strings of length 4 is 16, Since E⋂F = {1111, 1100, 1010, 1001}, p(E⋂F) = 4/16=1/4. We conclude that E and F are independent, because p(E⋂F) =1/4 = (½) (½)= p(E) p(F) p(E⋂F) = p(E)p(F).p(E) = p(F) = 8/16 = ½. Independence Example: Assume (as in the previous example) that each of the four ways a family can have two children (BB, GG, BG,GB) is equally likely. Are the events E, that a family with two children has two boys, and F, that a family with two children has at least one boy, independent? Solution: Because E = {BB}, p(E) = 1/4. We saw previously that that p(F) = 3/4 and p(E⋂F)=1/4. The events E and F are not independent since p(E) p(F) = 3/16 ≠ 1/4= p(E⋂F) .Pairwise and Mutual Independence Definition: The events E1, E2, , En are pairwise independent if and only if p(Ei⋂Ej) = p(Ei) p(Ej) for all pairs i and j with i ≤ j ≤ n. The events are mutually independent if whenever ij, j = 1,2,., m, are integers with 1 ≤ i1 ai i := i + 1 m := aj for k := 0 to j − i − 1 aj-k := aj-k-1 ai := m{Now a1,,an is in increasing order} At step i for i = 2, .,n, insertion sort inserts the ith element in the original list into the correct position in the sorted list of the first i -1 elements.continued → Average-Case Complexity of Insertion Sort Solution: Let X be the random variable equal to the number of comparisons used by insertion sort to sort a list of a1, a2, ., an distinct elements. E(X) is the average number of comparisons.Let Xi be the random variable equal to the number of comparisons used to insert ai into the proper position after the first i −1 elements a1, a2, ., ai-1 have been sorted. Since X = X2 + X3 + ∙∙∙ + Xn, E(X) = E(X2 + X3 + ∙∙∙ + Xn) = E(X2) + E(X3) + ∙∙∙ + E(Xn).To find E(Xi) for i = 2,3,,n, let pj(k) be the probability that the largest of the first j elements in the list occurs at the kth position, that is, max(a1, a2, ., aj ) = ak, where 1 ≤ k ≤ j.Assume uniform distribution; pj(k) = 1/j .Then Xi(k) = k. continued → Average-Case Complexity of Insertion Sort Since ai could be inserted into any of the first i positionsIt follows thatHence, the average-case complexity is .The Geometric Distribution Definition 2: A random variable X has geometric distribution with parameter p if p(X = k) = (1 − p)k-1p for k = 1,2,3,, where p is a real number with 0 ≤ p ≤ 1. Theorem 4: If the random variable X has the geometric distribution with parameter p, then E(X) = 1/p. Example: Suppose the probability that a coin comes up tails is p. What is the expected number of flips until this coin comes up tails?The sample space is {T, HT, HHT, HHHT, HHHHT, }.Let X be the random variable equal to the number of flips in an element of the sample space; X(T) = 1, X(HT) = 2, X(HHT) = 3, etc. By Theorem 4, E(X) = 1/p.see text for full details Independent Random Variables Definition 3: The random variables X and Y on a sample space S are independent if p(X = r1 and Y = r2) = p(X = r1)∙ p(Y = r2). Theorem 5: If X and Y are independent variables on a sample space S, then E(XY) = E(X)E(Y).see text for the proof Variance Deviation: The deviation of X at s ∊ S is X(s) − E(X), the difference between the value of X and the mean of X. Definition 4: Let X be a random variable on the sample space S. The variance of X, denoted by V(X) is That is V(X) is the weighted average of the square of the deviation of X. The standard deviation of X, denoted by σ(X) is defined to be Theorem 6: If X is a random variable on a sample space S, then V(X) = E(X2) − E(X)2. Corollary 1: If X is a random variable on a sample space S and E(X) = μ , then V(X) = E((X −μ)2). see text for the proof see text for the proof. Variance Example: What is the variance of the random variable X, where X(t) = 1 if a Bernoulli trial is a success and X(t) = 0 if it is a failure, where p is the probability of success and q is the probability of failure? Solution: Because X takes only the values 0 and 1, it follows that X2(t) = X(t). Hence, Variance of the Value of a Die: What is the variance of a random variable X, where X is the number that comes up when a fair die is rolled? Solution: We have V(X) = E(X2) − E(X)2 . In an earlier example, we saw that E(X) = 7/2. Note that E(X2) = 1/6(12 + 22 + 32 +42 + 52 + 62) = 91/6. We conclude that V(X) = E(X2) − E(X)2 = p − p2 = p(1 − p) = pq.Variance Bienaymé‘s Formula: If X and Y are two independent random variables on a sample space S, then V(X + Y) = V(X) + V(Y). Furthermore, if Xi, i = 1,2, ,n, with n a positive integer, are pairwise independent random variables on S, then V(X1 + X2 + ∙∙∙ + Xn) = V(X1) + V(X2) + ∙∙∙ + V(Xn). Example: Find the variance of the number of successes when n independent Bernoulli trials are performed, where on each trial, p is the probability of success and q is the probability of failure. Solution: Let Xi be the random variable with Xi ((t1, t2, ., tn)) = 1 if trial ti is a success and Xi ((t1, t2, ., tn)) = 0 if it is a failure. Let X = X2 + X3 + . Xn. Then X counts the number of successes in the n trials. By Bienaymé ‘s Formula, it follows that V(X)= V(X1) + V(X2) + ∙∙∙ + V(Xn). By the previous example ,V(Xi) = pq for i = 1,2, ,n. Hence, V(X) = npq. Irenée-Jules Bienaymé (1796-1878)see text for the proof Chebyshev’s Inequality Chebyschev’s Inequality: Let X be a random variable on a sample space S with probability function p. If r is a positive real number, then p(|X(s) − E(X)| ≥ r ) ≤ V(X)/r2. Example: Suppose that X is a random variable that counts the number of tails when a fair coin is tossed n times. Note that X is the number of successes when n independent Bernoulli trials, each with probability of success ½ are done. Hence, (by Theorem 2) E(X) = n/2 and (by Example 18) V(X) = n/4. By Chebyschev’s inequality with r = √n, p(|X(s) − n/2 | ≥ √n ) ≤ (n/4 )(√n )2 = ¼. This means that the probability that the number of tails that come up on n tosses deviates from the mean , n/2, by more than √n is no larger than ¼. Pafnuty Lvovich Chebyshev (1821-1894)see text for the proof