Chapter 4: Probability and Probability Distributions
|
|
|
- Shana Floyd
- 6 years ago
- Views:
Transcription
1 Chapter 4: Probability and Probability Distributions 4.1 a. Subjective probability b. Relative frequency c. Classical d. Relative frequency e. Subjective probability f. Subjective probability g. Classical 4.2 Answers will vary depending on person s experience with each situation. c. This is know as the birthday problem. Students will be surprised to learn that in a class of only 23 students the probability of 2 or more students having the same birthday is greater than 50%. 4.3 Using the binomial formula, the probability of guessing correctly or more of the 20 questions is a. Positive Outcomes are 00 to 74; Negative Outcomes are 75 to 99. b. Let M be the number of sets of 20 2-digit numbers out of the 2000 sets in which we have or more positive outcomes (i.e., or more two digit numbers in the range 00 to 74. Then the probability would be approximated by M/ HHH, HHT, HTH, THH, TTH, THT, HTT, TTT 4.6 a. A = {HTT, THT, TTH}; Thus, P(A = 3 8 b. B = {HHH, HHT, HTH, THH, THT, HTT, TTH}; Thus, P(B = 7 8 c. C = {TTT}; Thus, P(C = a. P ( A = = 5 8 P ( B = = 1 8 P ( C = = 7 8 b. Events A and B are not mutually exclusive because B contains A A B = A, which is not the empty set. 4.8 a. P (A B = P (A B/P (B = 3 8 / 7 8 = 3 7 b. P (A C = P (A C/P (C = 0, since A C is empty. c. P (B C = P (B C/P (C = 0, since B C is empty. 4.9 Because P (A B = = P (A A and B are not independent. Because P (A C = = P (A A and C are not independent. Because P (B C = = P (B B and C are not independent. 20
2 4.10 A = {1, 3, 5} and B = {4, 5} a. P (A = P (1 + P (3 + P (5 = = 0.65 P (B = P (4 + P (5 = = 0.40 b. P (A B = P (5 = 0.30 c. P (A B = P (A + P (B P (A B = = 0.75 Alternatively, P (A B = P (1+P (3+P (4+P (5 = = No, since A and B are not mutually exclusive. Also, P (A B = a. A: Generator 1 does not work b. B A: Generator 2 does not work given that Generator 1 does not work c. A B: Generator 1 works or Generator 2 works or Both Generators work 4.13 a. S = {F 1 F 2, F 1 F 3, F 1 F 4, F 1 F 5, F 2 F 1, F 2 F 3, F 2 F 4, F 2 F 5, F 3 F 1, F 3 F 2, F 3 F 4, F 3 F 5, F 4 F 1, F 4 F 2, F 4 F 3, F 4 F 5, F 5 F 1, F 5 F 2, F 5 F 3, F 5 F 4 } b. Let T 1 be the event that the 1st firm chosen is stable and T 2 be the event that the 2nd firm chosen is stable. P (T 1 = 3 5 and P (T 1 = 2 5 P (Both Stable = P (T 1 T 2 = P (T 2 T 1 P (T 1 = ( 2 4 ( 3 5 = 6 20 = 0.30 Alternatively, if we designated F 1 and F 2 as the Shakey firms, then we could go to the list of 20 possible outcomes and identify 6 pairs containing just Stable firms: F 3 F 4, F 3 F 5, F 4 F 3, F 4 F 5, F 5 F 3, F 5 F 4. Thus, the probability that both firms are Stable is 6 20 c. P (One of two firms is Shakey = P (1st chosen is Shakey and 2nd chosen is Stable +P (1st chosen is Stable and 2nd chosen is Shakey = P (T 1 T 2 + P (T 1 T 2 = P (T 2 T 1 P (T 1 + P (T 2 T 1 P (T 1 = ( 3 4 ( ( 2 4 ( 3 5 = = 0.60 Alternatively, in the list of 20 outcomes there are 12 pairs which consist of exactly 1 Shakey firm (F 1 or F 2 and exactly 1 Stable firm (F 3 or F 4 or F 5. Thus, the probability of exactly 1 Shakey firm is d. P (Both Shakey = P (T 1 T 2 = P (T 2 T 1 P (T 1 = ( 1 4 ( 2 5 = 2 20 = 0.10 Alternatively, in the list of 20 outcomes there are 2 pairs in which both firms are Shakey (F 1 or F 2. Thus, the probability that both firms are Shakey is P (A = 192/( = 0.436; P (B = 128/440 = 0.291; P (A B = 48/440 = a. P (A = P (none high + P (little high + P (some high + P (extensive high = = 0.63 P (B = P (low extensive + P (medium extensive + P (high extensive = = 0.36 P (C = P (low none + P (low little + P (medium none + P (medium little = =
3 b. P (A B = P (A B/P (B = 0.22/0.36 = P (A B = P (A B/P (B = 0.41/(1.36 = 0.64 P (B C = P (B C/P (C = 0.14/0.14 = 1 c. P (A B = P (A + P (B P (A B = = 0.77 P (A C = 0; P (B C = a. P(Both customers pay in full = (0.70(0.70 = 0.49 b. P(At least one of 2 customers pay in full = 1 - P(neither customer pays in full = 1 - (1-0.70( = 1 ( = Let A = event customer pays 1st month s bill in full and B = event customer pays 2nd month s bill in full. We are given that P (A = 0.70, P (B A = 0.95, P (B A = 0.90, P (B A = 1 P (B A = = 0.10 a. P (A B = P (B AP (A = (0.95(0.70 = b. P (B A = P (B AP (A = (0.90( = c. P (pay exactly one month in full = 1 P (pays neither month or pays both months = 1 P (A B P (A B = = Let D be the event loan is defaulted, R 1 applicant is poor risk, R 2 fair risk, and R 3 good risk. P (D = 0.01, P (R 1 D = 0.30, P (R 2 D = 0.40, P (R 3 D = 0.30 P (D = 0.01, P (R 1 D = 0.10, P (R 2 D = 0.40, P (R 3 D = 0.50 P (D R 1 = P (R 1 DP (D = (0.30(0.01 P (R 1 DP (D+P (R 1 DP (D (0.30(0.01+(0.10(0.99 = P (R 4.19 P (D R 2 = 2 DP (D = (0.40(0.01 P (R 2 DP (D+P (R 2 DP (D (0.40(0.01+(0.40(0.99 = 0.01 The two probabilities are equal since the proportion of fair risk applicants is the same for both the defaulted and nondefaulted loans Let F be the event fire occurs and T i be the event a type i furnace is in the home for i = 1, 2, 3, 4, where T 4 represent other types. P (T 1 F = = P (F T 1P (T 1 P (F T 1 P (T 1 +P (F T 2 P (T 2 +P (F T 3 P (T 3 +P (F T 4 P (T 4 (0.05(0.30 (0.05(0.30+(0.03(0.25+(0.02(0.+(0.04(0.30 = a. Sensitivity: P (DA C = 50/53 = Specificity: P (DNA RO = 44/47 = b. P (C DA = (0.943( (0.943( (0.021( = c. P (RO DA = 1 P (C DA = =.954 d. P (RO DNA = 4.22 a. P (y = 3 = ( 10 3 (.2 3 (.8 7 = (0.936( (0.936( (0.019( =
4 b. P (y = 2 = ( 4 2 (.4 2 (.6 2 = c. P (y = 12 = ( (.7 12 (.3 4 = a. a. P (y 4 = P (0 + P (1 + P (2 + P (3 + P (4 = = b. P (y > 4 = 1 P (y 4 = = c. P (y 7 = 1 P (8 = 1 ( 8 8 (.4 8 ( = = d. P (y > 6 = P (y = 7 + P (y = 8 = ( 8 7 (.4 7 ( ( 8 8 (.4 8 ( = a. Bar graph of P(y b. P (y 2 = P (y = 0 + P (y = 1 + P (y = 2 = 0.5 c. P (y 7 = P (y = 7 + P (y = 8 + P (y = 9 + P (y = 10 = 0.13 d. P (1 y 5 = P (y = 1 + P (y = 2 + P (y = 3 + P (y = 4 + P (y = 5 = a. P (y 3 = 1 P (y < 3 = 1 ( = 0.64 b. P (2 y 6 = P (y = 2 + P (y = 3 + P (y = 4 + P (y = 5 + P (y = 6 = = 0.60 c. P (y > 8 = P (y = 9 + P (y = 10 = = No, each running of the experiment does not result in one of two possible outcomes No, people may not answer the question Binomial experiment with n=10 and π = a. P (y = 0 = b. P (y = 6 = c. P (y 6 = 1 P (y < 6 = 1 (P (0 + P (1 + P (2 + P (3 + P (4 + P (5 = 1 ( = d. P (y = 10 = Binomial experiment with n=10 and π = a. P (y = 0 = ( 10 0 (.3 0 (.7 10 = P (y = 6 = ( 10 6 (.3 6 (.7 4 = P (y 6 = 1 P (y < 6 = 1 (P (0 + P (1 + P (2 + P (3 + P (4 + P (5 = = P (y = 10 = ( 10 (.3 10 (.7 0 = b. With n = 0 and π =.3 P (y = i=0 ( 0 i (.3 i (.7 0 i. This would be a length calculation. In Section 4.13, we will provide an approximation which will greatly reduce the amount of calculations. However, we can also consider that µ = nπ = (0(.3 = 300, σ = nπ(1 π = 0(.3(.7 = µ ± 3σ = 300 ± (3(14.49 = (256.53, Thus, the chance of observing the event y is very small. 23
5 4.30 a. P (y = 2 = ( 10 2 (.1 2 (.9 8 = (.1 0 (.9 10 ( 10 b. P (y 2 = 1 P (y = 0 P (y = 1 = 1 ( (.1 1 (.9 9 = c. π = P(either outstanding or good = 0.85; P (y = 8 = ( 10 8 ( (. 2 = d. π = P(unsatisfactory = 0.05; P (y = 0 = ( 10 0 ( ( = No. The trials are not identical Binomial with n = 50; π =.17 a. P (y 3 = 3 ( 50 i=0 i (0.07 i ( i = (using a computer program b. The posting of price changes are independent with the same probability 0.07 of being posted incorrectly a = b = a = b = a = z o = 0 b = z o = z o = z o = z o = a. P (500 < y < 696 = P ( < z < = P (0 < z < 1.96 = = b. P (y > 696 = P (z > = P (z > 1.96 = = c. P (304 < y < 696 = P ( < z < = P ( 1.96 < z < 1.96 = = 0.95 d. P (500 k < y < k = P ( 500 k 500 < z < 500+k 500 = P (.01k < z <.01k = From Table 1 we find, P (.845 < z <.845 = Thus,.01k =.845 k = a. y < 130 y < 130 z < 2 b. y > 82.5 y 82.5 > 82.5 z > 1.17 c. P (y < 130 = P (z < 2 = ; P (y > 82.5 = P (z > 1.17 = 1 P (z 1.17 =
6 d. P (y > 106 = P (z > 106 = P (z > 0.4 = 1 P (z 0.4 = P (y < 94 = P (z < 94 = P (z < 0.4 = P (94 < y < 106 = P ( 0.4 < z < 0.4 = = e. P (y > 130 = P (z > 130 = P (z > 2 = 1 P (z 2 = P (y < 70 = P (z < 70 = P (z < 2 = P (70 < y < 130 = P ( 2 < z < 2 = = a. z = 2.33 b. z = a. P (z > 1.96 = = b. P (z > 2.21 = = c. P (z > 2.86 = = d. P (z > 0.73 = 1 = = a. µ = 39; σ = 6; P (y > 50 = P (z > = P (z > 1.83 = = b. Since 55 is = 2.67 std. dev. above µ = 39, thus P (y > 55 = P (z > 2.67 = We would then conclude that the voucher has been lost µ = 500; σ = a. P (y > 600 = P (z > = P (z > 1 = 0.87 b. P (y > 700 = P (z > = P (z > 2 = c. P (y < 450 = P (z < = P (z < 0.5 = d. P (450 < y < 600 = P ( < z < = P ( 0.5 < z < 1 = µ = 0; σ = a. P (y > 200 = P (z > = P (z > 1.43 = b. P (y > 220 = P (z > = P (z > 2 = c. P (y < 120 = P (z < = P (z < 0.86 = d. P ( < y < 200 = P ( 0 < z < = P ( 1.43 < z < 1.43 = µ = 500; σ = a. Find k such that P (y > k = 0.10; P (z > = 0.10 k = (1.285( = or greater. b. P (y < k = 0.40; P (z < = 0.40 k = ( 0.255( = ; This is the 40th percentile No. The sample would be biased toward homes for which the homeowner is at home much of the time. For example, the sample would tend to include more people who work at home and retired persons In order to make the sampling random, the network might choose voters based on draws from a random number table, or more simply choose every nth person exiting. 25
7 4.51 Start at column 2 line 1 we obtain 0, 729, 611, 584, 255, 465, 143, 127, 323, 225, 483, 368, 213, 270, 062, 399, 695, 540, 330, 110, 069, 409, 539, 0, 564. These would be the women selected for the study a. Use a computer program or the random number table to generate 50 numbers. b. Reusing the same numbers may create a bias for a certain segment of the population within each of the precincts The sampling distribution would have a mean of 60 and a standard deviation of 5 16 = If the population distribution is somewhat mound-shaped then the sampling distribution of ȳ should be approximately mound-shaped. In this situation, we would expect approximately 95% of the possible values of ȳ is lie in 60 ± (2(1.25 = (57.5, If the population distribution is somewhat mound-shaped then the sampling distribution of i y i should be approximately mound-shaped with mean of (60(16=960 and a standard deviation of 5 16 = 20. Observing a measurement more than 70 units ( = 3.5 standard deviations away from the mean (960 for a normal distribution is very improbable (P ( z > µ = 930; σ = 130 a. P (800 < y < 1 = P ( < z < = P ( 1 < z < 1.31 = = b. P (y < 800 = P (z < = 0.87 c. P (y > 1200 = P (z > = P (z > 2.08 = = a. 125 ± 32 should contain approximately 68% of the weeks 125 ± 64 should contain approximately 95% of the weeks 125 ± 96 should contain approximately 99.7% of the weeks. b. P (y > 160 = P (z > = P (z > 1.09 = = µ = 2.1; σ = 0.3 a. P (y > 2.7 = P (z > = P (z > 2 = b. P (z > = 0.25 y.75 = (0.6745(0.3 = 2.30 c. Let µ N be the new value of the mean. We need P (y > From Table 1, 0.05 = P (z > and 0.05 = P (y > 2.7 = P ( y µ N 2.7 µ N 0.3 = µ N = 2.7 (0.3(1.645 = Mean for fleet is nµ = (0(2.1 = 3; standard deviation for the fleet is nσ = 0(0.3 = > 2.7 µ N Individual baggage weight has µ = 95; σ = ; Total weight has mean nµ = (200(95 = 19, 000; and standard deviation nσ = 200( = Therefore, P (y > 20, 000 = P (z > 20,000 19, = P (z > 2.02 = No. The last date may not be representative of all days in the month. 26
8 4.61 a. µ = nπ = (00(0.001 = 10 b. σ = 00(0.001(0.999 = 3.16 P (y < 5 = P (y 4 P (z = P (z < 1.74 = c. P (y < 2 = P (y 1 = P (z = P (z < 2.69 = a. P (y > 2265 = P (z > = P (z > 1.47 = b. Approximately normal with mean = 2250 and standard deviation = 10.2 = σȳ = 10.2 = P (ȳ 2268 = P (z = P (z No, there is strong evidence that the new fabric has a greater mean breaking strength a. The normal probability plot and box plots are given here. Normal Probability Plot Sampling Distribution of Sample Mean Sample Mean Sample Mean Quantiles of Standard Normal Note that the plotted points deviate from the straight-line. b. Since the population distribution is skewed to the right for this sample and with the sample size only 2, the sampling distribution for ȳ will be skewed to the right Could use random sampling by first identifying all returns with income greater than $,000. Next, create a computer file with all such returns listed in alphabetical order by last name of client. Randomly select 1% of these returns using a systematic random sampling technique a. The random variable y, representing the number of sales in 5 calls, is binomial with π = The event that the first sale occurs in the first five calls is the complement of the event that no sales occur in the first five calls. Thus, the probability of interest is 1 P (y = 0 = 1 ( 5 0 (.01 0 (.99 5 = b. If the first sale occurs after 10 calls, the first ten calls must have resulted in no sales. Let y be the number of sales in the 10 calls, P (y = 0 = ( 10 0 (.01 0 ( =
9 4.69 n = 400 π = 0.2 a. µ = nπ = 400(.20 = 80; σ = 400(.2(.8 = 8 P (y 25 P (z = P (z b. The ad is not successful. With π =.20, we expect 80 positive responses out of 400 but we observed only 25. The probability of getting so few positive responses is virtually 0 if π =.20. We therefore conclude that π is much less than n = 20000, π = There are 2 possible outcomes and each birth is an independent event. We cannot use the normal approximation because nπ = (20000( = 2 < 5. We can use the binomial formula: P (y 1 = 1 P (y = 0 = 1 ( ( ( = a.,b. The mean and standard deviation of the sampling distribution of ȳ are given when the population distribution has values µ =, σ = : Sample Size Mean Standard Deviation c. As the sample size increases, the sampling distribution of ȳ concentrates about the true value of µ. For n = 5 and 20, the values of ȳ could be a considerable distance from a.-c. The probabilites are given here: Sample Size P (ȳ 105 P (ȳ 95 P (95 ȳ The sample mean s chance of being close to the population mean increases very rapidly as the sample size increases. Of course, the rate of increase also depends on the population standard deviation, σ. 28
Chapter 4: Probability and Probability Distributions
Chapter 4: Probability and Probability Distributions 4.1 How Probability Can Be Used in Making Inferences 4.1 a. Subjective probability b. Relative frequency c. Classical d. Relative frequency e. Subjective
Conditional Probability
Conditional Probability Idea have performed a chance experiment but don t know the outcome (ω), but have some partial information (event A) about ω. Question: given this partial information what s the
Events A and B are said to be independent if the occurrence of A does not affect the probability of B.
Independent Events Events A and B are said to be independent if the occurrence of A does not affect the probability of B. Probability experiment of flipping a coin and rolling a dice. Sample Space: {(H,
Thus, P(F or L) = P(F) + P(L) - P(F & L) = = 0.553
Test 2 Review: Solutions 1) The following outcomes have at least one Head: HHH, HHT, HTH, HTT, THH, THT, TTH Thus, P(at least one head) = 7/8 2) The following outcomes have a sum of 9: (6,3), (5,4), (4,5),
Probability deals with modeling of random phenomena (phenomena or experiments whose outcomes may vary)
Chapter 14 From Randomness to Probability How to measure a likelihood of an event? How likely is it to answer correctly one out of two true-false questions on a quiz? Is it more, less, or equally likely
4/17/2012. NE ( ) # of ways an event can happen NS ( ) # of events in the sample space
I. Vocabulary: A. Outcomes: the things that can happen in a probability experiment B. Sample Space (S): all possible outcomes C. Event (E): one outcome D. Probability of an Event (P(E)): the likelihood
Probability Distributions for Discrete RV
An example: Assume we toss a coin 3 times and record the outcomes. Let X i be a random variable defined by { 1, if the i th outcome is Head; X i = 0, if the i th outcome is Tail; Let X be the random variable
Conditional Probability
Conditional Probability Conditional Probability The Law of Total Probability Let A 1, A 2,..., A k be mutually exclusive and exhaustive events. Then for any other event B, P(B) = P(B A 1 ) P(A 1 ) + P(B
What is Probability? Probability. Sample Spaces and Events. Simple Event
What is Probability? Probability Peter Lo Probability is the numerical measure of likelihood that the event will occur. Simple Event Joint Event Compound Event Lies between 0 & 1 Sum of events is 1 1.5
Mathematics. ( : Focus on free Education) (Chapter 16) (Probability) (Class XI) Exercise 16.2
( : Focus on free Education) Exercise 16.2 Question 1: A die is rolled. Let E be the event die shows 4 and F be the event die shows even number. Are E and F mutually exclusive? Answer 1: When a die is
Quantitative Methods for Decision Making
January 14, 2012 Lecture 3 Probability Theory Definition Mutually exclusive events: Two events A and B are mutually exclusive if A B = φ Definition Special Addition Rule: Let A and B be two mutually exclusive
Why should you care?? Intellectual curiosity. Gambling. Mathematically the same as the ESP decision problem we discussed in Week 4.
I. Probability basics (Sections 4.1 and 4.2) Flip a fair (probability of HEADS is 1/2) coin ten times. What is the probability of getting exactly 5 HEADS? What is the probability of getting exactly 10
Statistics Statistical Process Control & Control Charting
Statistics Statistical Process Control & Control Charting Cayman Systems International 1/22/98 1 Recommended Statistical Course Attendance Basic Business Office, Staff, & Management Advanced Business Selected
Lecture Lecture 5
Lecture 4 --- Lecture 5 A. Basic Concepts (4.1-4.2) 1. Experiment: A process of observing a phenomenon that has variation in its outcome. Examples: (E1). Rolling a die, (E2). Drawing a card form a shuffled
Discrete Probability Distribution
Shapes of binomial distributions Discrete Probability Distribution Week 11 For this activity you will use a web applet. Go to http://socr.stat.ucla.edu/htmls/socr_eperiments.html and choose Binomial coin
9/6/2016. Section 5.1 Probability. Equally Likely Model. The Division Rule: P(A)=#(A)/#(S) Some Popular Randomizers.
Chapter 5: Probability and Discrete Probability Distribution Learn. Probability Binomial Distribution Poisson Distribution Some Popular Randomizers Rolling dice Spinning a wheel Flipping a coin Drawing
Applied Statistics I
Applied Statistics I Liang Zhang Department of Mathematics, University of Utah June 17, 2008 Liang Zhang (UofU) Applied Statistics I June 17, 2008 1 / 22 Random Variables Definition A dicrete random variable
PERMUTATIONS, COMBINATIONS AND DISCRETE PROBABILITY
Friends, we continue the discussion with fundamentals of discrete probability in the second session of third chapter of our course in Discrete Mathematics. The conditional probability and Baye s theorem
The Bernoulli distribution has only two outcomes, Y, success or failure, with values one or zero. The probability of success is p.
The Bernoulli distribution has onl two outcomes, Y, success or failure, with values one or zero. The probabilit of success is p. The probabilit distribution f() is denoted as B (p), the formula is: f ()
CS206 Review Sheet 3 October 24, 2018
CS206 Review Sheet 3 October 24, 2018 After ourintense focusoncounting, wecontinue withthestudyofsomemoreofthebasic notions from Probability (though counting will remain in our thoughts). An important
Deep Learning for Computer Vision
Deep Learning for Computer Vision Lecture 3: Probability, Bayes Theorem, and Bayes Classification Peter Belhumeur Computer Science Columbia University Probability Should you play this game? Game: A fair
2. Conditional Probability
ENGG 2430 / ESTR 2004: Probability and Statistics Spring 2019 2. Conditional Probability Andrej Bogdanov Coins game Toss 3 coins. You win if at least two come out heads. S = { HHH, HHT, HTH, HTT, THH,
Notation: X = random variable; x = particular value; P(X = x) denotes probability that X equals the value x.
Ch. 16 Random Variables Def n: A random variable is a numerical measurement of the outcome of a random phenomenon. A discrete random variable is a random variable that assumes separate values. # of people
Probability. VCE Maths Methods - Unit 2 - Probability
Probability Probability Tree diagrams La ice diagrams Venn diagrams Karnough maps Probability tables Union & intersection rules Conditional probability Markov chains 1 Probability Probability is the mathematics
TA Qinru Shi: Based on poll result, the first Programming Boot Camp will be: this Sunday 5 Feb, 7-8pm Gates 114.
TA Qinru Shi: Based on poll result, the first Programming Boot Camp will be: this Sunday 5 Feb, 7-8pm Gates 114. Prob Set 1: to be posted tomorrow. due in roughly a week Finite probability space S 1) a
Chapter 2 PROBABILITY SAMPLE SPACE
Chapter 2 PROBABILITY Key words: Sample space, sample point, tree diagram, events, complement, union and intersection of an event, mutually exclusive events; Counting techniques: multiplication rule, permutation,
STA 2023 EXAM-2 Practice Problems. Ven Mudunuru. From Chapters 4, 5, & Partly 6. With SOLUTIONS
STA 2023 EXAM-2 Practice Problems From Chapters 4, 5, & Partly 6 With SOLUTIONS Mudunuru, Venkateswara Rao STA 2023 Spring 2016 1 1. A committee of 5 persons is to be formed from 6 men and 4 women. What
To understand and analyze this test, we need to have the right model for the events. We need to identify an event and its probability.
Probabilistic Models Example #1 A production lot of 10,000 parts is tested for defects. It is expected that a defective part occurs once in every 1,000 parts. A sample of 500 is tested, with 2 defective
Chapter 2 Class Notes
Chapter 2 Class Notes Probability can be thought of in many ways, for example as a relative frequency of a long series of trials (e.g. flips of a coin or die) Another approach is to let an expert (such
STAT 430/510 Probability
STAT 430/510 Probability Hui Nie Lecture 3 May 28th, 2009 Review We have discussed counting techniques in Chapter 1. Introduce the concept of the probability of an event. Compute probabilities in certain
CSC Discrete Math I, Spring Discrete Probability
CSC 125 - Discrete Math I, Spring 2017 Discrete Probability Probability of an Event Pierre-Simon Laplace s classical theory of probability: Definition of terms: An experiment is a procedure that yields
CHAPTER - 16 PROBABILITY Random Experiment : If an experiment has more than one possible out come and it is not possible to predict the outcome in advance then experiment is called random experiment. Sample
Lecture 1 : The Mathematical Theory of Probability
Lecture 1 : The Mathematical Theory of Probability 0/ 30 1. Introduction Today we will do 2.1 and 2.2. We will skip Chapter 1. We all have an intuitive notion of probability. Let s see. What is the probability
Basic Statistics for SGPE Students Part II: Probability theory 1
Basic Statistics for SGPE Students Part II: Probability theory 1 Mark Mitchell mark.mitchell@ed.ac.uk Nicolai Vitt n.vitt@ed.ac.uk University of Edinburgh September 2016 1 Thanks to Achim Ahrens, Anna
Key Concepts. Key Concepts. Event Relations. Event Relations
Probability and Probability Distributions Event Relations S B B Event Relations The intersection of two events, and B, is the event that both and B occur when the experient is perfored. We write B. S Event
Discrete random variables and probability distributions
Discrete random variables and probability distributions random variable is a mapping from the sample space to real numbers. notation: X, Y, Z,... Example: Ask a student whether she/he works part time or
Probabilistic models
Probabilistic models Kolmogorov (Andrei Nikolaevich, 1903 1987) put forward an axiomatic system for probability theory. Foundations of the Calculus of Probabilities, published in 1933, immediately became
Lecture 6 Random Variable. Compose of procedure & observation. From observation, we get outcomes
ENM 07 Lecture 6 Random Variable Random Variable Eperiment (hysical Model) Compose of procedure & observation From observation we get outcomes From all outcomes we get a (mathematical) probability model
What is a random variable
OKAN UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE MATH 256 Probability and Random Processes 04 Random Variables Fall 20 Yrd. Doç. Dr. Didem Kivanc Tureli didemk@ieee.org didem.kivanc@okan.edu.tr
Notes slides from before lecture. CSE 21, Winter 2017, Section A00. Lecture 16 Notes. Class URL:
Notes slides from before lecture CSE 21, Winter 2017, Section A00 Lecture 16 Notes Class URL: http://vlsicad.ucsd.edu/courses/cse21-w17/ Notes slides from before lecture Notes March 8 (1) This week: Days
Business Statistics MBA Pokhara University
Business Statistics MBA Pokhara University Chapter 3 Basic Probability Concept and Application Bijay Lal Pradhan, Ph.D. Review I. What s in last lecture? Descriptive Statistics Numerical Measures. Chapter
Probability Pearson Education, Inc. Slide
Probability The study of probability is concerned with random phenomena. Even though we cannot be certain whether a given result will occur, we often can obtain a good measure of its likelihood, or probability.
Mean, Median and Mode. Lecture 3 - Axioms of Probability. Where do they come from? Graphically. We start with a set of 21 numbers, Sta102 / BME102
Mean, Median and Mode Lecture 3 - Axioms of Probability Sta102 / BME102 Colin Rundel September 1, 2014 We start with a set of 21 numbers, ## [1] -2.2-1.6-1.0-0.5-0.4-0.3-0.2 0.1 0.1 0.2 0.4 ## [12] 0.4
STA 2023 EXAM-2 Practice Problems From Chapters 4, 5, & Partly 6. With SOLUTIONS
STA 2023 EXAM-2 Practice Problems From Chapters 4, 5, & Partly 6 With SOLUTIONS Mudunuru Venkateswara Rao, Ph.D. STA 2023 Fall 2016 Venkat Mu ALL THE CONTENT IN THESE SOLUTIONS PRESENTED IN BLUE AND BLACK
CISC 1100/1400 Structures of Comp. Sci./Discrete Structures Chapter 7 Probability. Outline. Terminology and background. Arthur G.
CISC 1100/1400 Structures of Comp. Sci./Discrete Structures Chapter 7 Probability Arthur G. Werschulz Fordham University Department of Computer and Information Sciences Copyright Arthur G. Werschulz, 2017.
MATH1231 Algebra, 2017 Chapter 9: Probability and Statistics
MATH1231 Algebra, 2017 Chapter 9: Probability and Statistics A/Prof. Daniel Chan School of Mathematics and Statistics University of New South Wales danielc@unsw.edu.au Daniel Chan (UNSW) MATH1231 Algebra
Lecture Notes 1 Basic Probability. Elements of Probability. Conditional probability. Sequential Calculation of Probability
Lecture Notes 1 Basic Probability Set Theory Elements of Probability Conditional probability Sequential Calculation of Probability Total Probability and Bayes Rule Independence Counting EE 178/278A: Basic
ECE 450 Lecture 2. Recall: Pr(A B) = Pr(A) + Pr(B) Pr(A B) in general = Pr(A) + Pr(B) if A and B are m.e. Lecture Overview
ECE 450 Lecture 2 Recall: Pr(A B) = Pr(A) + Pr(B) Pr(A B) in general = Pr(A) + Pr(B) if A and B are m.e. Lecture Overview Conditional Probability, Pr(A B) Total Probability Bayes Theorem Independent Events
Lecture 3 - Axioms of Probability
Lecture 3 - Axioms of Probability Sta102 / BME102 January 25, 2016 Colin Rundel Axioms of Probability What does it mean to say that: The probability of flipping a coin and getting heads is 1/2? 3 What
Discrete Random Variable
Discrete Random Variable Outcome of a random experiment need not to be a number. We are generally interested in some measurement or numerical attribute of the outcome, rather than the outcome itself. n
SDS 321: Introduction to Probability and Statistics
SDS 321: Introduction to Probability and Statistics Lecture 2: Conditional probability Purnamrita Sarkar Department of Statistics and Data Science The University of Texas at Austin www.cs.cmu.edu/ psarkar/teaching
Introduction Probability. Math 141. Introduction to Probability and Statistics. Albyn Jones
Math 141 to and Statistics Albyn Jones Mathematics Department Library 304 jones@reed.edu www.people.reed.edu/ jones/courses/141 September 3, 2014 Motivation How likely is an eruption at Mount Rainier in
Probability Pr(A) 0, for any event A. 2. Pr(S) = 1, for the sample space S. 3. If A and B are mutually exclusive, Pr(A or B) = Pr(A) + Pr(B).
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License. Your use of this material constitutes acceptance of that license and the conditions of use of materials on this
27 Binary Arithmetic: An Application to Programming
27 Binary Arithmetic: An Application to Programming In the previous section we looked at the binomial distribution. The binomial distribution is essentially the mathematics of repeatedly flipping a coin
Notes slides from before lecture. CSE 21, Winter 2017, Section A00. Lecture 15 Notes. Class URL:
Notes slides from before lecture CSE 21, Winter 2017, Section A00 Lecture 15 Notes Class URL: http://vlsicad.ucsd.edu/courses/cse21-w17/ Notes slides from before lecture Notes March 6 (1) This week: Days
Sec$on Summary. Assigning Probabilities Probabilities of Complements and Unions of Events Conditional Probability
Section 7.2 Sec$on Summary Assigning Probabilities Probabilities of Complements and Unions of Events Conditional Probability Independence Bernoulli Trials and the Binomial Distribution Random Variables
Conditional Probability and Bayes Theorem (2.4) Independence (2.5)
Conditional Probability and Bayes Theorem (2.4) Independence (2.5) Prof. Tesler Math 186 Winter 2019 Prof. Tesler Conditional Probability and Bayes Theorem Math 186 / Winter 2019 1 / 38 Scenario: Flip
6.2 Introduction to Probability. The Deal. Possible outcomes: STAT1010 Intro to probability. Definitions. Terms: What are the chances of?
6.2 Introduction to Probability Terms: What are the chances of?! Personal probability (subjective) " Based on feeling or opinion. " Gut reaction.! Empirical probability (evidence based) " Based on experience
(i) Given that a student is female, what is the probability of having a GPA of at least 3.0?
MATH 382 Conditional Probability Dr. Neal, WKU We now shall consider probabilities of events that are restricted within a subset that is smaller than the entire sample space Ω. For example, let Ω be the
Probability (10A) Young Won Lim 6/12/17
Probability (10A) Copyright (c) 2017 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later
Chapter 13, Probability from Applied Finite Mathematics by Rupinder Sekhon was developed by OpenStax College, licensed by Rice University, and is
Chapter 13, Probability from Applied Finite Mathematics by Rupinder Sekhon was developed by OpenStax College, licensed by Rice University, and is available on the Connexions website. It is used under a
Chapter 2 Solutions Page 12 of 28
Chapter 2 Solutions Page 12 of 28 2.34 Yes, a stem-and-leaf plot provides sufficient information to determine whether a dataset contains an outlier. Because all individual values are shown, it is possible
Bayes Rule for probability
Bayes Rule for probability P A B P A P B A PAP B A P AP B A An generalization of Bayes Rule Let A, A 2,, A k denote a set of events such that S A A2 Ak and Ai Aj for all i and j. Then P A i B P Ai P B
Outline. 1. Define likelihood 2. Interpretations of likelihoods 3. Likelihood plots 4. Maximum likelihood 5. Likelihood ratio benchmarks
Outline 1. Define likelihood 2. Interpretations of likelihoods 3. Likelihood plots 4. Maximum likelihood 5. Likelihood ratio benchmarks Likelihood A common and fruitful approach to statistics is to assume
Math 243 Section 3.1 Introduction to Probability Lab
Math 243 Section 3.1 Introduction to Probability Lab Overview Why Study Probability? Outcomes, Events, Sample Space, Trials Probabilities and Complements (not) Theoretical vs. Empirical Probability The
MODULE 2 RANDOM VARIABLE AND ITS DISTRIBUTION LECTURES DISTRIBUTION FUNCTION AND ITS PROPERTIES
MODULE 2 RANDOM VARIABLE AND ITS DISTRIBUTION LECTURES 7-11 Topics 2.1 RANDOM VARIABLE 2.2 INDUCED PROBABILITY MEASURE 2.3 DISTRIBUTION FUNCTION AND ITS PROPERTIES 2.4 TYPES OF RANDOM VARIABLES: DISCRETE,
CLASS 6 July 16, 2015 STT
CLASS 6 July 6, 05 STT-35-04 Plan for today: Preparation for Quiz : Probability of the union. Conditional Probability, Formula of total probability, ayes Rule. Independence: Simple problems (solvable by
Instructor Solution Manual. Probability and Statistics for Engineers and Scientists (4th Edition) Anthony Hayter
Instructor Solution Manual Probability and Statistics for Engineers and Scientists (4th Edition) Anthony Hayter 1 Instructor Solution Manual This instructor solution manual to accompany the fourth edition
Introduction to Probability and Sample Spaces
2.2 2.3 Introduction to Probability and Sample Spaces Prof. Tesler Math 186 Winter 2019 Prof. Tesler Ch. 2.3-2.4 Intro to Probability Math 186 / Winter 2019 1 / 26 Course overview Probability: Determine
LECTURE NOTES by DR. J.S.V.R. KRISHNA PRASAD
.0 Introduction: The theory of probability has its origin in the games of chance related to gambling such as tossing of a coin, throwing of a die, drawing cards from a pack of cards etc. Jerame Cardon,
STAT/MA 416 Answers Homework 4 September 27, 2007 Solutions by Mark Daniel Ward PROBLEMS
STAT/MA 416 Answers Homework 4 September 27, 2007 Solutions by Mark Daniel Ward PROBLEMS 2. We ust examine the 36 possible products of two dice. We see that 1/36 for i = 1, 9, 16, 25, 36 2/36 for i = 2,
Conditional Probability 2 Solutions COR1-GB.1305 Statistics and Data Analysis
Conditional Probability 2 Solutions COR-GB.305 Statistics and Data Analysis The Birthday Problem. A class has 50 students. What is the probability that at least two students have the same birthday? Assume
4th IIA-Penn State Astrostatistics School July, 2013 Vainu Bappu Observatory, Kavalur
4th IIA-Penn State Astrostatistics School July, 2013 Vainu Bappu Observatory, Kavalur Laws of Probability, Bayes theorem, and the Central Limit Theorem Rahul Roy Indian Statistical Institute, Delhi. Adapted
Probability Distributions. Conditional Probability.
Probability Distributions. Conditional Probability. CSE21 Winter 2017, Day 21 (B00), Day 14 (A00) March 6, 2017 http://vlsicad.ucsd.edu/courses/cse21-w17 Probability Rosen p. 446, 453 Sample space, S:
Statistical Experiment A statistical experiment is any process by which measurements are obtained.
(التوزيعات الا حتمالية ( Distributions Probability Statistical Experiment A statistical experiment is any process by which measurements are obtained. Examples of Statistical Experiments Counting the number
Term Definition Example Random Phenomena
UNIT VI STUDY GUIDE Probabilities Course Learning Outcomes for Unit VI Upon completion of this unit, students should be able to: 1. Apply mathematical principles used in real-world situations. 1.1 Demonstrate
(a) Fill in the missing probabilities in the table. (b) Calculate P(F G). (c) Calculate P(E c ). (d) Is this a uniform sample space?
Math 166 Exam 1 Review Sections L.1-L.2, 1.1-1.7 Note: This review is more heavily weighted on the new material this week: Sections 1.5-1.7. For more practice problems on previous material, take a look
Expected Value. Lecture A Tiefenbruck MWF 9-9:50am Center 212 Lecture B Jones MWF 2-2:50pm Center 214 Lecture C Tiefenbruck MWF 11-11:50am Center 212
Expected Value Lecture A Tiefenbruck MWF 9-9:50am Center 212 Lecture B Jones MWF 2-2:50pm Center 214 Lecture C Tiefenbruck MWF 11-11:50am Center 212 http://cseweb.ucsd.edu/classes/wi16/cse21-abc/ March
Stat 2300 International, Fall 2006 Sample Midterm. Friday, October 20, Your Name: A Number:
Stat 2300 International, Fall 2006 Sample Midterm Friday, October 20, 2006 Your Name: A Number: The Midterm consists of 35 questions: 20 multiple-choice questions (with exactly 1 correct answer) and 15
Chapter. Probability
Chapter 3 Probability Section 3.1 Basic Concepts of Probability Section 3.1 Objectives Identify the sample space of a probability experiment Identify simple events Use the Fundamental Counting Principle
Semester I Review. The authors also kept track of the color of the first born in each litter. (B = brown, G = gray, W = white, and T = tan)
Answer Key A.P. Statistics Semester I Review In the paper Reproduction in Laboratory colonies of Bank Vole, the authors presented the results of a study of litter size. (A vole is a small rodent with a
Lower bound for sorting/probability Distributions
Lower bound for sorting/probability Distributions CSE21 Winter 2017, Day 20 (B00), Day 14 (A00) March 3, 2017 http://vlsicad.ucsd.edu/courses/cse21-w17 Another application of counting lower bounds Sorting
NLP: Probability. 1 Basics. Dan Garrette December 27, E : event space (sample space)
NLP: Probability Dan Garrette dhg@cs.utexas.edu December 27, 2013 1 Basics E : event space (sample space) We will be dealing with sets of discrete events. Example 1: Coin Trial: flipping a coin Two possible
3rd IIA-Penn State Astrostatistics School July, 2010 Vainu Bappu Observatory, Kavalur
3rd IIA-Penn State Astrostatistics School 19 27 July, 2010 Vainu Bappu Observatory, Kavalur Laws of Probability, Bayes theorem, and the Central Limit Theorem Bhamidi V Rao Indian Statistical Institute,
324 Stat Lecture Notes (1) Probability
324 Stat Lecture Notes 1 robability Chapter 2 of the book pg 35-71 1 Definitions: Sample Space: Is the set of all possible outcomes of a statistical experiment, which is denoted by the symbol S Notes:
Part (A): Review of Probability [Statistics I revision]
![Part (A): Review of Probability [Statistics I revision]](/thumbs/93/111285649.jpg "Part (A): Review of Probability [Statistics I revision]")
Part (A): Review of Probability [Statistics I revision] 1 Definition of Probability 1.1 Experiment An experiment is any procedure whose outcome is uncertain ffl toss a coin ffl throw a die ffl buy a lottery
Answer. Find the gradient of the curve y x at x 4
(ISO/IEC - 7-5 Certified) SUMMER 8 EXAMINATION ject Name: Applied Mathematics Model wer ject Code: Important Instructions to eaminers: ) The answers should be eamined by key words and not as word-to-word
CS4705. Probability Review and Naïve Bayes. Slides from Dragomir Radev
CS4705 Probability Review and Naïve Bayes Slides from Dragomir Radev Classification using a Generative Approach Previously on NLP discriminative models P C D here is a line with all the social media posts
Formal Modeling in Cognitive Science Lecture 19: Application of Bayes Theorem; Discrete Random Variables; Distributions. Background.
Formal Modeling in Cognitive Science Lecture 9: ; Discrete Random Variables; Steve Renals (notes by Frank Keller) School of Informatics University of Edinburgh s.renals@ed.ac.uk February 7 Probability
Formal Modeling in Cognitive Science
Formal Modeling in Cognitive Science Lecture 9: Application of Bayes Theorem; Discrete Random Variables; Steve Renals (notes by Frank Keller) School of Informatics University of Edinburgh s.renals@ed.ac.uk
Probability Distributions. Conditional Probability
Probability Distributions. Conditional Probability Russell Impagliazzo and Miles Jones Thanks to Janine Tiefenbruck http://cseweb.ucsd.edu/classes/sp16/cse21-bd/ May 16, 2016 In probability, we want to
Exam III Review Math-132 (Sections 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 8.1, 8.2, 8.3)
1 Exam III Review Math-132 (Sections 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 8.1, 8.2, 8.3) On this exam, questions may come from any of the following topic areas: - Union and intersection of sets - Complement of
Introduction to bivariate analysis
Introduction to bivariate analysis When one measurement is made on each observation, univariate analysis is applied. If more than one measurement is made on each observation, multivariate analysis is applied.
Rules of Probability
( ) ( ) = for all Corollary: Rules of robablty The probablty of the unon of any two events and B s roof: ( Φ) = 0. F. ( B) = ( ) + ( B) ( B) If B then, ( ) ( B). roof: week 2 week 2 2 Incluson / Excluson
Homework 4 Solution, due July 23
Homework 4 Solution, due July 23 Random Variables Problem 1. Let X be the random number on a die: from 1 to. (i) What is the distribution of X? (ii) Calculate EX. (iii) Calculate EX 2. (iv) Calculate Var
Discussion 03 Solutions
STAT Discussion Solutions Spring 8. A new flavor of toothpaste has been developed. It was tested by a group of people. Nine of the group said they liked the new flavor, and the remaining indicated they
II. Introduction to probability, 2
GEOS 33000/EVOL 33000 5 January 2006 updated January 10, 2006 Page 1 II. Introduction to probability, 2 1 Random Variables 1.1 Definition: A random variable is a function defined on a sample space. In
Introduction to bivariate analysis
Introduction to bivariate analysis When one measurement is made on each observation, univariate analysis is applied. If more than one measurement is made on each observation, multivariate analysis is applied.
Expected Value 7/7/2006
Expected Value 7/7/2006 Definition Let X be a numerically-valued discrete random variable with sample space Ω and distribution function m(x). The expected value E(X) is defined by E(X) = x Ω x m(x), provided
TOPIC 12 PROBABILITY SCHEMATIC DIAGRAM
TOPIC 12 PROBABILITY SCHEMATIC DIAGRAM Topic Concepts Degree of Importance References NCERT Book Vol. II Probability (i) Conditional Probability *** Article 1.2 and 1.2.1 Solved Examples 1 to 6 Q. Nos