Probability Homework Solutions

Homework Assignment 1 – Probability for IBA This assignment is voluntary: if you want to participate, hand in your (own) solutions, handwritten on one A4 (with your name, student number and group on top) at the start of the tutorial on Thursday 28th of November 2013. We consider the population all youngsters (age 18-22) in The Netherlands and the events F = “the youngster has a Facebook account” and T = “the youngster has a Twitter account”. The following information is the result of an extensive survey: 70% of all youngsters have a Facebook account, 40% has a Twitter account and 20% has both. Draw a Venn diagram for the population and the events and answer the following questions. Motivate each of the answers briefly (preferably using probability rules): a. b. c. d. e. What is the probability that a Dutch youngster has a Facebook or a Twitter account? What is the probability that a Dutch youngster has only a Facebook account? What is the probability that a Dutch youngster has no Facebook nor a Twitter account? What is the probability that a Facebook user has a Twitter account as well? Are the events F and T independent? In the above mentioned survey the researchers also informed about their social status: 60% were studying (event A) and 40% were not. The researchers reported furthermore that 80% of the students used a smartphone (event B), where only 55% of the non-students used such a device. Express the given probabilities in A and B ( ( ) ) and solve f and g (clearly state your motivation!) f. compute the probability that the youngster uses a smartphone g. We meet a youngster having a smartphone: what is the probability he is studying? Compare the answer to the 60% proportion of students in the population and explain the difference. Grades: a b c d e f g total 1 1 1 1 1 2 3 10 Solutions homework assignment 1 ( ) ( Given is that ( ) (“Read” the rules of probability in the Venn diagram) ( ) ( ) ( ) ( ) (general addition rule) ) ( ) ( ) b. ( (Use the Venn-diagram: you can split up F in ( ( ) ( ) c. (Using the complement rule and the result of a.) ) T F a. d. e. f. ( | ) ( F and TC FC and TC ) and ( ).) ) ( ) ) ( ) are independent if ( ( ), ) ( ) but ( ( ) (or, using the result of d., because 28.6% = ( | ) ( ) ) ( ) ( | ) ( | ) Then ( ) is the probability of a nonstudent Then ( ) See the tree diagram below Or use the Venn diagram ( ) ( ) ( ( ) ( | ) ( A ) ) ( | ) AC B A and B g. FC and T F and T ( | ) ( AC and B ) ( ) Due to the fact that students use smartphones more often, the proportion of students amongst the smartphone users is larger than the proportion of students in the population. 0.80 0.6 Student (A) 0.20 Arbitrary youngster 0.4 Nonstudent (AC) 0.55 0.45 Has smartphone (B) Has no smartphone (BC) Has smartphone (B) Has no smartphone (BC) Homework Assignment 3 – Probability for IBA Hand in your (own) solutions, handwritten on one A4 (with your name, student number and group on top), at the start of the tutorial on Thursday 12th of December 2013. When solving the exercises below only use a simple calculator, the formula sheet and the standard normal table. Applying the Normal distribution A project-leader has the supervision of two (completely different) building projects, one in The Netherlands and one in Belgium. The experienced project-leader expects a net profit of k€ 10 for the project and a standard deviation of k€ 5 in The Netherlands and an expected profit of k€ 12 with standard deviation k€ 12 in Belgium. Assume that the net profit X in the Netherlands and the net profit Y in Belgium are both normally distributed. a. Compute for both projects (separately) the probability of loss (= negative profit). b. Compute the 99th percentile c for X: 𝑃(𝑋 ≤ 𝑐) = 0.99 c. Compute 𝑃(𝑋 + 𝑌 > 30), the probability that the total profit of the two projects is larger than k€ 30 d. Compute 𝑃(𝑋 ≤ 𝑌) , the probability that the project in NL has a lower profit than the one in Belgium. The project-leader estimates that 80% of (all of) his projects has a positive profit. Next year he has to supervise 100 projects and we wonder how many of them will have a positive profit. e. Give a probability model (random variable and its distribution) for this situation and approximate the probability that at most 75 of the projects have a positive profit. Grades: a b c d e total 2 2 2 2 2 10 Solutions homework assignment 3 (In these solutions Z is a standard normal variable) a. X is Normal(10, 5) and the z-score of 0 is = = −2 So 𝑃(𝑋 < 0) = 𝑃(𝑍 ≤ −2) = 0.0228 Similarly 𝑃(𝑌 < 0) = 𝑃 (𝑍 < ) = 𝑃(𝑍 < 1) = 0.1587 b. The z-score of c is and from the standard normal table we know: 𝑃(𝑍 ≤ 2.33) = 0.01 𝑐−10 If 𝑃(𝑋 ≤ 𝑐) = 𝑃 (𝑍 ≤ 5 ) = 0.99, then = 2.33, so 𝑐 = 5 2.33 + 10 = 21.65 ( c. X + Y is normal as well with μ = E(X + Y) = E(X) + E(Y) = 10 + 12 = 22 and 𝜎 = √𝜎 + 𝜎 = √5 + 12 = 13 ) 𝑃(𝑋 + 𝑌 > 30) = 𝑃 (𝑍 > ) ≈ 1 − 𝑃(𝑍 ≤ 0.62) = 1 − 0.7324 ≈ 26.8% d. 𝑃(𝑋 > 𝑌) = 𝑃(𝑋 − 𝑌 > 0), where X - Y is normal as well with μ = E(X) +-E(Y) = 10 - 12 = -2 and 𝜎 = √𝜎 + 𝜎 = √5 + 12 = 13 ( ) 𝑃(𝑋 ≤ 𝑌) = 𝑃 (𝑍 ≤ ) ≈ 𝑃(𝑍 ≤ 0.15) = 0.5596 ≈ 56.0% e. Model X = “the number of profitable projects among 100 projects” X is binomial (n = 100, p = 0.8) (assuming that the profits are independent, all with p = 0.8) X is approximately normal with 𝜇 = 𝐸(𝑋) = 𝑛𝑝 = 80 and 𝜎 = √𝑛𝑝(1 − 𝑝) = √16 = 4 𝑃(𝑋 ≤ 75) = 𝑃 (𝑍 ≤ ) = 𝑃(𝑍 ≤ −1.25) = 0.1056 (With continuity correction: 𝑃(𝑋 ≤ 75) = 𝑃(𝑋 ≤ 75.5) ≈ 𝑃(𝑍 ≤ −1.13) = 0.1292)

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Стратмор и его высокопоставленные посетители попадали в шифровалку и уходили незаметно для остальных сотрудников. Лифт спускался на пятьдесят ярдов вниз и затем двигался вбок по укрепленному туннелю еще сто девять ярдов в подземное помещение основного комплекса агентства.

Лифт, соединяющий шифровалку с основным зданием, получал питание из главного комплекса, и оно действовало, несмотря на отключение питания шифровалки. Стратмору, разумеется, это было хорошо известно, но даже когда Сьюзан порывалась уйти через главный выход, он не обмолвился об этом ни единым словом.

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