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Department
Mathematics & Engineering Courses
Course
MTHE 224
Professor
Prof.
Semester
Fall

Description
Week 2: Probability Goals: • Review counting techniques • Study conditional probability • Study probability distributions for discrete random variables Suggested Textbook Readings: Chapter 2: 2.3-2.5, and Chapter 3: 3.1-3.3 Practice Problems: Section 2.3: 31, 35, 39 Section 2.4: 45, 47, 55, 65 Section 2.5: 71, 73, 81, 85 Section 3.2: 13, 17, 23 Section 3.3: 29, 35, 37, 39 S is set of all possible outcomes in the experiment Week 2: Probability and Discrete Probability Distributions 2 Probability Models The objective of probability is to assign to each event A,an umbr P(A)w hihisa measure of the chance that the event A will occur. To be self-consistent, the assigned probability must satisfy the following axioms. For each rule, draw a Venn diagram representing it. 1. For any sample space S, P(S)=1. 2. For any event A,0 ≤ P(A) ≤ 1. 3. If A and B are mutually exclusive events, then P(A ∪ B)= P(A)+ P(B). 4. If A is the complement of A,t en P(A )=1 − P(A). MTHE 224 Fall 2012 Week 2: Probability and Discrete Probability Distributions 3 diagrams in terms of sets, and then write out the probability of the shaded event. (In the first diagram, the shaded region is the common region of A and B.) A B A B A B For any two eventA s and B, P(A ∪ B)= P(A)+ P(B) − P(A ∩ B). MTHE 224 Fall 2012 Week 2: Probability and Discrete Probability Distributions 4 Counting Techniques When the simple outcomes of an experiment are equally likely, we can employ a simple counting technique to find the probabilities for the compound events. If there are N outcomes in a sample space, and let N(A)e tteubrf outcomes contained in event A, then the probabilities of event A is N(A) P(A)= N Example 2: What is the probability of drawing an ace from a well-shuffled deck of 52 playing cards? However it can be quite difficult to determine the number of outcomes in a sample space by direct enumeration. To handle this kind of problem systematically, it can be useful to construct a tree diagram. Example 3: A consumer testing service rates lawn mowers as being easy, average, or difficult to operate; as being expensive or inexpensive; and as being costly, average, or cheap to repair. In how many different ways can a lawn mower be rated by this testing service? Product Rule If the first element or object of an ordered pair can be selected in n 1 ways, and for each of thesen w1ys, the second element of the pair can be selected in n2ways, then the number of pairs is n 1 2 The product rule can be generalized to k-tuples. MTHE 224 Fall 2012 Week 2: Probability and Discrete Probability Distributions 5 Permutations and Combinations Definition: Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects. The number of permutations of size k that can be constructed from the n objects is P k,n. Pk,n=( n)(n − 1) · ... · (n − k +2)( n − k +1)= n! . (n − k)! Example 4: Suppose Mr Tanner brought seven flavours of ice cream home from the store. His kids DJ, Stephanie, and Michelle all want a different flavour. How many ways are there to give each kid a different flavour? Definition: An unordered subset of size k of n objects is called a combination,t e number of which is denoted ▯n▯ k ▯ n P = n,k= k k! Example 5: Suppose there are seven different flavours of ice cream at the store. If Mr Tanner wants to bring home three different flavoured ice creams for his kids, how many ways are there to do this? Example 6: Five coins are tossed, and number of heads are observed. How many out- comes in each case? MTHE 224 Fall 2012 Week 2: Probability and Discrete Probability Distributions 6 Simulations in MatLab The following is a simulation for tossing three coins and observing the number of heads. H=[0,0,0,0]; ‘‘H’’ store the number of heads observed, for N=1:100 toss 100 times, coins=rand(1,3); Each time we generate a set of three random numbers between 0 and 1, i=0; ‘‘i’’ is used to count the number of heads, for k=1:3 A loop to check the three coins, if coins(k)>0.5 It is a head if the number is greater i=i+1; than 0.5, otherwise it is a tail, end; end; H(i+1)=H(i+1)+1; end display(H); bar(H) As a practice, write a program simulating the five coins in previous example. MTHE 224 Fall 2012 Week 2: Probability and Discrete Probability Distributions 7 Conditional Probability Working with our new ideas of events and sample spaces, we can now define and compute probabilities for more interesting cases. Example 7: (Example 2.24, page 67) Complex components are assembled in a plant that uses two different assembly lines, A an1 A . Li2e A uses o1der equipment than A 2 so it is somewhat slower and less reliable. Suppose on a given day, line A has1as- sembled 8 components, of which 2 have been identified as defective (denoted by D) and 6 as non-defective (UD), whereas the line A ha2 produced 1 defective and 9 non-defective components. Unaware of this information, the sale manager randomly selects one of these 18 components for a demonstration. (a) List all events, and find the corresponding probabilities. (b) Now we add a condition, if a component from line A is s1lected, what is the proba- bility that it is defective? (c) If the selected component turns out to be non-defective, what is the probability that the component is selected from the line A ?2Argue first from intuition, or using a Venn diagram. Law of Conditional Probability For any two eventA s and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by P(A ∩ B) P(A|B)= P(B) MTHE 224 Fall 2012 Week 2: Probability and Discrete Probability Distributions 8 Law of Total Probability Let A 1··· ,A k be mutually exclusive and mutually exhaustive events. Then for any event B, P(B)= P(B|A )P(A1)+ ··1 + P(B|A )P(A ) k k Example 8: (Example 7: Continued) Use Law of Total Probability to find P(D). Example 9: (Example 2.30, page 73) Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. • If an adult has the disease, a positive result will occur 99% of the time, whereas • if an individual has no disease, the test result will show positive only 2% of the time. (a) Now you walk into doctor’s office today, what is the prob
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