GENG200 Problems.docx-Microsoft Word Home Insert Page LayoutReferences MailingsR

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GENG200 Problems.docx-Microsoft Word Home Insert Page LayoutReferences MailingsReviewViewFoxit PDF Find ac Replace Calibri (Body) Paste Normal No Spa. Heading 1 Heading 2TitleSubtitle SubtieEm."-Change . F Format Painter Blu-abax, X'Aa, styles . 1 Select- Font Paragraph Styles Editing 8 1-9 110 12, ,13. 14-1 ,15-1 ,16' ,17 ,18-1-19 Given an unfair coin with two sides marked one and two with 40% and 60% chance respectively. ie. when toasting the coin there is 40% chance to get the face marked one on top and 60% to get the second one 0.4 0.6 Assume you will betoasting the coin as many times as possibleto gettace twoon top ooou keep toasting the coin until you get face marked two, then you stop) We are interested in the number of times you should toast the coin to get face marked twofor the first time, let X be the random variable describing this number. 01. Write down the sample space associated with this random experiment Q2, write down the possible volues of × Q3. What is the probability to getface two on top from the first trial? 04. Whot is the probability togetface two on top (for the first timej atthe third trial? 05. Determine the proba bility mass function, PMF, of X a4. Determine the expected value of X and Variance of X 05. Assume that the probability of getting face marked twoon top is denoted by p, LetX be the number of trials you toasted the coin togetface twoon top for the first time. Tryto Euess the general form of the PMF of X (as function of pl by coin toetface two on top for the fimtip Page: 3 ot 3 words: 463 English (United States) 7:57 PM 10/17/2017

Explanation / Answer

Question 1 .

Sample space = {1,2,3,4,5.....}

Question 2

Possible value of X are X >0 and X E N

Question 3

Pr(Getting 2 on first trial ) = Pr(2) = 0.6

Question 4

Pr (X = 3) = Pr(First fail) * Pr(Second Fail) * Pr( Third time 2 on top) = 0.4 * 0.4 * 0.6 = 0.096

Question 5

PMF (X) = p(X) = (1-p)X-1 p = 0.4X-1 0.6

Question 6

E(X) = 1/p = 1/0.6 = 5/3

Var(X) = (1-p)/p2 = (1- 0.6)/ 0.62 = 1.111

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