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Probability Theory: Cheat Sheet for Exam with Venn Diagram and Examples - Prof. Ryan G. Ma, Exams of Probability and Statistics

A cheat sheet for a probability theory exam, including definitions, formulas, and examples. It covers concepts such as sample space, events, union, intersection, mutually exclusive events, permutations, combinations, bayes' theorem, and venn diagrams. The document also includes a venn diagram and examples to help illustrate the concepts.

Typology: Exams

Pre 2010

Uploaded on 07/30/2009

koofers-user-gi9
koofers-user-gi9 🇺🇸

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Download Probability Theory: Cheat Sheet for Exam with Venn Diagram and Examples - Prof. Ryan G. Ma and more Exams Probability and Statistics in PDF only on Docsity! Stat 225 – Exam 01 Spring 2009 KEY • Show your work on all questions. Unsupported work will not receive full credit. • Report decimal answer to three decimal places. • You are responsible for upholding the Honor Code of Purdue University. This includes protecting your work from other students. • You are allowed the following aids: a one-page 8.5× 11 handwritten (in your hand- writing) cheat sheet, a scientific calculator, and pencils. • Instructors will not interpret questions for you. If you do have questions, wait until you have looked over the whole exam so that you can ask all of your questions at one time. • You must show your student ID (upon request), turn in your cheat sheet and sign the class roster when you turn in your exam to your instructor. • Turn off your cell phone before the exam begins! Problem Points Possible Points Earned 1 20 2 12 3 8 4 12 5 6 6 8 7 12 8 12 9 10 Total 100 1 Problem 1. Matching. Choose the word/phrase from the following list that best fits in the sentences below, and write the corresponding letter in the provided blank. Each word could be used more than once or not at all. (2 points each) A. sample space F. independent K. addition rule B. event G. with replacment L. inclusion-exclusion formula C. union H. without replacement M. partition D. intersection I. permutation N. total probability formula E. mutually exclusive J. combination O. Bayes’ Theorem 1. Events A and B are if knowledge of the occurrence Answer: F of A does not affect the probability that B will occur. 2. A subset chosen at random, without replacement, where the Answer: J order of selection does not matter is a . 3. The set of elements in A or B is the of A and B. Answer: C 4. If A and B are disjoint, then their is empty. Answer: D 5. Rolling a pair of dice is an example of sampling . Answer: G 6. To find P (B) when P (B|A) and P (B|Ac) are known, use Answer: N the . 7. P (A ∪ B) = P (A) + P (B) if A and B are . Answer: E 8. A subset chosen at random, without replacement, where the Answer: I order of selection does matter is a . 9. The is the collection of all possible outcomes Answer: A of an experiment. 10. If P (A|B) is known, use to calculate P (B|A). Answer: O 2 Problem 4. A DVD manufacturer’s entire output is produced by three machines. Ma- chines A, B and C produce 37%, 23%, and 40% of the DVDs, respectively, but 11%, 17% and 12% of the respective DVDs are scratched. A quality control engineer randomly selects a DVD from a recently produced batch. (4 points each) 1. What is the probability that the selected DVD is scratched and was produced by Machine B? Let A, B and C be the events that the DVD is made by machines A, B and C, respectively. Also, let S be the event that the DVD is scratched. Then, according to the multiplication rule, P (S ∩ B) = P (S|B)P (B) = 0.17 · 0.23 = 0.0391. 2. Find the probability that the selected DVD is scratched. Since every DVD is made by exactly one of the three machines, the events A, B and C partition the sample space. Therefore, using total probability we have P (S) = P (S|A)P (A) + P (S|B)P (B) + P (S|C)P (C) = 0.11 · 0.37 + 0.17 · 0.23 + 0.12 · 0.40 = 0.1278 3. Given the selected DVD is scratched, what is the probability it was produced by Machine A? This time, we use Bayes’ theorem. P (A|S) = P (S|A)P (A) P (S) = 0.11 · 0.37 0.1278 = 0.3185. 5 Problem 5. Multiple Choice. The Baumann product of two events A and B, which is denoted by A ⊗ B, is defined as A ⊗ B = [(A ∩ Bc) ∪ (Ac ∩ B)]c. For each question below, choose the most appropriate answer. Hint: A Venn diagram might be helpful. (2 points each) 1. Which of the following best describes A ⊗ B? (a) Everything that is not in A or B. (b) Everything in A or B but not in both. (c) Everything that is not in both A and B. (d) Everything in neither A nor B or in both A and B. 2. Which of the following statements are true? (a) (A ∪ B) ∩ (A ⊗ B) = A ∩ B. (b) A ⊗ ∅ = Ac. (c) (A ⊗ B)c = (Ac ∪ B) ∩ (A ∪ Bc). (d) Both (a) and (b) (e) Both (a) and (c) (f) Both (b) and (c) 3. Which formula could be used to find P (A ⊗ B)? (a) P (A ⊗ B) = 1 − P (A ∪ B) + P (A ∩ B). (b) P (A ⊗ B) = P (Ac ∪ B) + P (A ∩ Bc). (c) P (A ⊗ B) = P (A) + P (B) − 2P (A ∩ B). (d) Both (a) and (b) (e) Both (a) and (c) (f) Both (b) and (c). 6 Problem 6. Consider a game that consists of n rounds and, in each round, the player throws a dart at a target. The player wins a prize if he/she hits a bullseye in at least one of the n rounds. Assume that the throws are independent of one another. Let Bk be the event that a bullseye is hit in round k (k = 1, 2, . . . , n). (4 points each) 1. In a game consisting of n = 3 rounds, with P (Bk) = 0.35 for each k, find the probability that the player wins a prize. Let W be the event that the player wins. Then W = B1∪B2∪B3 so by DeMorgan’s law, the complementation rule, and the assumed independence, we get P (W ) = 1 − P (Bc 1 ∩ Bc 2 ∩ Bc 3 ) = 1 − (1 − 0.35)3 = 1 − 0.653 = 0.7254. Could also use the Inclusion-Exclusion rule but this takes a bit longer. 2. Still assuming P (Bk) = 0.35 for each k, how many rounds n would be necessary to ensure the that player has probability at least 0.95 of winning a prize? Following the same logic as in Part 1 above, it is clear that P (W ) = 1−0.65n. Now set 0.95 ≤ P (W ) = 1 − 0.65n and solve for n: 0.95 ≤ 1 − 0.65n ⇐⇒ 0.65n ≤ 0.05 ⇐⇒ n log 0.65 ≤ log 0.05. Since log 0.65 < 0, dividing both sides by that switches the inequality and we get n ≥ log 0.05/ log 0.65 = 6.95. Therefore, take n = 7. 7
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