Lecture 14. Text: A Course in Probability by Weiss 5.6. STAT 225 Introduction to Probability Models February 23, Whitney Huang Purdue University
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1 Lecture 14 Text: A Course in Probability by Weiss 5.6 STAT 225 Introduction to Probability Models February 23, 2014 Whitney Huang Purdue University 14.1
2 Agenda 14.2
3 Review So far, we have covered Bernoulli and Binomial distribution: number of successes in an independent trials (sampling with replacement) with fixed sample size 14.3
4 Review So far, we have covered Bernoulli and Binomial distribution: number of successes in an independent trials (sampling with replacement) with fixed sample size Hypergeometric distribution: number of successes in a dependent trials (sampling without replacement) with fixed sample size 14.3
5 Review So far, we have covered Bernoulli and Binomial distribution: number of successes in an independent trials (sampling with replacement) with fixed sample size Hypergeometric distribution: number of successes in a dependent trials (sampling without replacement) with fixed sample size Poisson distribution: number of successes (events) occurring in a fixed interval of time and/or space without fixed sample size 14.3
6 Review So far, we have covered Bernoulli and Binomial distribution: number of successes in an independent trials (sampling with replacement) with fixed sample size Hypergeometric distribution: number of successes in a dependent trials (sampling without replacement) with fixed sample size Poisson distribution: number of successes (events) occurring in a fixed interval of time and/or space without fixed sample size 14.3
7 Review So far, we have covered Bernoulli and Binomial distribution: number of successes in an independent trials (sampling with replacement) with fixed sample size Hypergeometric distribution: number of successes in a dependent trials (sampling without replacement) with fixed sample size Poisson distribution: number of successes (events) occurring in a fixed interval of time and/or space without fixed sample size In some cases, we want to know the sample size necessary to get a certain number of successes distribution: number of trials until the 1 st success (including the success) 14.3
8 Review So far, we have covered Bernoulli and Binomial distribution: number of successes in an independent trials (sampling with replacement) with fixed sample size Hypergeometric distribution: number of successes in a dependent trials (sampling without replacement) with fixed sample size Poisson distribution: number of successes (events) occurring in a fixed interval of time and/or space without fixed sample size In some cases, we want to know the sample size necessary to get a certain number of successes distribution: number of trials until the 1 st success (including the success) Negative Binomial distribution: number of trials until the r th success (including the r th success) 14.3
9 Review So far, we have covered Bernoulli and Binomial distribution: number of successes in an independent trials (sampling with replacement) with fixed sample size Hypergeometric distribution: number of successes in a dependent trials (sampling without replacement) with fixed sample size Poisson distribution: number of successes (events) occurring in a fixed interval of time and/or space without fixed sample size In some cases, we want to know the sample size necessary to get a certain number of successes distribution: number of trials until the 1 st success (including the success) Negative Binomial distribution: number of trials until the r th success (including the r th success) In both and Binomial distribution, the trials are independent 14.3
10 : Characteristics of the : Let X be a r.v. The definition of X: The number of trials it takes to get the 1 st success 14.4
11 : Characteristics of the : Let X be a r.v. The definition of X: The number of trials it takes to get the 1 st success The support: x = 1, 2, 14.4
12 : Characteristics of the : Let X be a r.v. The definition of X: The number of trials it takes to get the 1 st success The support: x = 1, 2, Its parameter(s) and definition(s): p: the probability of success in a single trial 14.4
13 : Characteristics of the : Let X be a r.v. The definition of X: The number of trials it takes to get the 1 st success The support: x = 1, 2, Its parameter(s) and definition(s): p: the probability of success in a single trial The probability mass function (pmf): p X (x) = p(1 p) x 1 for x = 1, 2, 14.4
14 : Characteristics of the : Let X be a r.v. The definition of X: The number of trials it takes to get the 1 st success The support: x = 1, 2, Its parameter(s) and definition(s): p: the probability of success in a single trial The probability mass function (pmf): p X (x) = p(1 p) x 1 for x = 1, 2, The expected value: E[X] = 1 p 14.4
15 : Characteristics of the : Let X be a r.v. The definition of X: The number of trials it takes to get the 1 st success The support: x = 1, 2, Its parameter(s) and definition(s): p: the probability of success in a single trial The probability mass function (pmf): p X (x) = p(1 p) x 1 for x = 1, 2, The expected value: E[X] = 1 p The variance: Var(X) = 1 p p
16 Properties of distribution Tail Probability: P(X > x) = (1 p) x Example 14.5
17 Properties of distribution Tail Probability: P(X > x) = (1 p) x Example P(X 4) = P(X > 3) = (1 p)
18 Properties of distribution Tail Probability: P(X > x) = (1 p) x Example P(X 4) = P(X > 3) = (1 p) 3 P(3 X 9) = P(X 3) P(X 10) = P(X > 2) P(X > 9) = (1 p) 2 (1 p)
19 Properties of distribution Tail Probability: P(X > x) = (1 p) x Example P(X 4) = P(X > 3) = (1 p) 3 P(3 X 9) = P(X 3) P(X 10) = P(X > 2) P(X > 9) = (1 p) 2 (1 p) 9 Memoryless Property: P(X > s + t X > s) = P(X > t) Example 14.5
20 Properties of distribution Tail Probability: P(X > x) = (1 p) x Example P(X 4) = P(X > 3) = (1 p) 3 P(3 X 9) = P(X 3) P(X 10) = P(X > 2) P(X > 9) = (1 p) 2 (1 p) 9 Memoryless Property: P(X > s + t X > s) = P(X > t) Example P(X > 15 X > 9) = P(X > 6) = (1 p)
21 Properties of distribution Tail Probability: P(X > x) = (1 p) x Example P(X 4) = P(X > 3) = (1 p) 3 P(3 X 9) = P(X 3) P(X 10) = P(X > 2) P(X > 9) = (1 p) 2 (1 p) 9 Memoryless Property: P(X > s + t X > s) = P(X > t) Example P(X > 15 X > 9) = P(X > 6) = (1 p) 6 P(X 8 X 5) = 1 P(X > 8 X 5) = 1 P(X > 8 X > 6) = 1 P(X > 2) = 1 (1 p)
22 Example 36 Suppose Dunphy is really bad at tossing a Frisbee. Suppose Dunphy hits pedestrians at a rate of 1 out of 5 people that walk past the campus mall. Let X be of number of tosses it takes to hit the 1 st pedestrian. 1 Name distribution (with parameter(s)) 14.6
23 Example 36 Suppose Dunphy is really bad at tossing a Frisbee. Suppose Dunphy hits pedestrians at a rate of 1 out of 5 people that walk past the campus mall. Let X be of number of tosses it takes to hit the 1 st pedestrian. 1 Name distribution (with parameter(s)) 2 What is the probability that his first accidental hitting is the 5 th person that walks by? 14.6
24 Example 36 Suppose Dunphy is really bad at tossing a Frisbee. Suppose Dunphy hits pedestrians at a rate of 1 out of 5 people that walk past the campus mall. Let X be of number of tosses it takes to hit the 1 st pedestrian. 1 Name distribution (with parameter(s)) 2 What is the probability that his first accidental hitting is the 5 th person that walks by? 3 What is the probability that he does not hit anyone in the first 10 tosses? 14.6
25 Example 36 Suppose Dunphy is really bad at tossing a Frisbee. Suppose Dunphy hits pedestrians at a rate of 1 out of 5 people that walk past the campus mall. Let X be of number of tosses it takes to hit the 1 st pedestrian. 1 Name distribution (with parameter(s)) 2 What is the probability that his first accidental hitting is the 5 th person that walks by? 3 What is the probability that he does not hit anyone in the first 10 tosses? 4 What is the probability that it takes no more than 20 tosses to hit the first pedestrian given that he does not hit anyone in the first 10 tosses? 14.6
26 Example 36 cont d Solution. 1 X Geo(p =.2) 14.7
27 Example 36 cont d Solution. 1 X Geo(p =.2) 2 P(X = 5) = (0.2)(1 0.2) 4 =
28 Example 36 cont d Solution. 1 X Geo(p =.2) 2 P(X = 5) = (0.2)(1 0.2) 4 = P(X > 10) = (1.2) 10 =
29 Example 36 cont d Solution. 1 X Geo(p =.2) 2 P(X = 5) = (0.2)(1 0.2) 4 = P(X > 10) = (1.2) 10 = P(X 20 X > 10) = 1 P(X > 20 X > memoryless propetry 10) ===== 1 P(X > 10) = =
30 Practice HW 2, Problem 12 You randomly call friends who could be potential partners for a dance. You think that they all respond to your requests independently of each other, and you estimate that each one is 7% likely to accept your request. Let X denote the number of calls to successfully get a date. 1 What are the distribution and parameters of X? 14.8
31 Practice HW 2, Problem 12 You randomly call friends who could be potential partners for a dance. You think that they all respond to your requests independently of each other, and you estimate that each one is 7% likely to accept your request. Let X denote the number of calls to successfully get a date. 1 What are the distribution and parameters of X? 2 How many calls do you expect to make to get a date? 14.8
32 Practice HW 2, Problem 12 You randomly call friends who could be potential partners for a dance. You think that they all respond to your requests independently of each other, and you estimate that each one is 7% likely to accept your request. Let X denote the number of calls to successfully get a date. 1 What are the distribution and parameters of X? 2 How many calls do you expect to make to get a date? 3 What is the probability you get a date on the 12 th call? 14.8
33 Practice HW 2, Problem 12 You randomly call friends who could be potential partners for a dance. You think that they all respond to your requests independently of each other, and you estimate that each one is 7% likely to accept your request. Let X denote the number of calls to successfully get a date. 1 What are the distribution and parameters of X? 2 How many calls do you expect to make to get a date? 3 What is the probability you get a date on the 12 th call? 4 What is the probability it takes you at least 10 calls to get a date? 14.8
34 Practice HW 2, Problem 12 You randomly call friends who could be potential partners for a dance. You think that they all respond to your requests independently of each other, and you estimate that each one is 7% likely to accept your request. Let X denote the number of calls to successfully get a date. 1 What are the distribution and parameters of X? 2 How many calls do you expect to make to get a date? 3 What is the probability you get a date on the 12 th call? 4 What is the probability it takes you at least 10 calls to get a date? 5 What is the probability it takes you no more than 15 calls to get a date? 14.8
35 Practice HW 2, Problem 12 You randomly call friends who could be potential partners for a dance. You think that they all respond to your requests independently of each other, and you estimate that each one is 7% likely to accept your request. Let X denote the number of calls to successfully get a date. 1 What are the distribution and parameters of X? 2 How many calls do you expect to make to get a date? 3 What is the probability you get a date on the 12 th call? 4 What is the probability it takes you at least 10 calls to get a date? 5 What is the probability it takes you no more than 15 calls to get a date? 6 Given you ve already called 8 people and still do not have a date, what is the probability it will take less than 14 calls? 14.8
36 Practice HW 2, Problem 12 Solution. 1 X Geo(p = 0.07) 14.9
37 Practice HW 2, Problem 12 Solution. 1 X Geo(p = 0.07) 2 E[X] = 1 p = =
38 Practice HW 2, Problem 12 Solution. 1 X Geo(p = 0.07) 2 E[X] = 1 p = = P(X = 12) = (0.07)(1 0.07) 11 =
39 Practice HW 2, Problem 12 Solution. 1 X Geo(p = 0.07) 2 E[X] = 1 p = = P(X = 12) = (0.07)(1 0.07) 11 = P(X 10) = P(X > 9) = (1 0.07) 9 =
40 Practice HW 2, Problem 12 Solution. 1 X Geo(p = 0.07) 2 E[X] = 1 p = = P(X = 12) = (0.07)(1 0.07) 11 = P(X 10) = P(X > 9) = (1 0.07) 9 = P(X 15) = 1 P(X > 15) = 1 (1 0.07) 15 =
41 Practice HW 2, Problem 12 Solution. 1 X Geo(p = 0.07) 2 E[X] = 1 p = = P(X = 12) = (0.07)(1 0.07) 11 = P(X 10) = P(X > 9) = (1 0.07) 9 = P(X 15) = 1 P(X > 15) = 1 (1 0.07) 15 = P(X < 14 X > 8) = 1 P(X > 13 X > 8) = 1 P(X > 5) = 1 (1 0.07) 5 =
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