ISM_chapter3 - Chapter 3 Discrete Random Variables and...

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31 Chapter 3: Discrete Random Variables and Their Probability Distributions 3.1 P ( Y = 0) = P (no impurities) = .2, P ( Y = 1) = P (exactly one impurity) = .7, P ( Y = 2) = .1. 3.2 We know that P ( HH ) = P ( TT ) = P ( HT ) = P ( TH ) = 0.25. So, P ( Y = -1) = .5, P ( Y = 1) = .25 = P ( Y = 2). 3.3 p (2) = P ( DD ) = 1/6, p (3) = P ( DGD ) + P ( GDD ) = 2(2/4)(2/3)(1/2) = 2/6, p (4) = P ( GGDD ) + P ( DGGD ) + P ( GDGD ) = 3(2/4)(1/3)(2/2) = 1/2. 3.4 Define the events: A : value 1 fails B : valve 2 fails C : valve 3 fails ) ( ) 2 ( C B A P Y P = = = .8 3 = 0.512 ) ( ) ( )) ( ( ) 0 ( C B P A P C B A P Y P = = = = .2(.2 + .2 - .2 2 ) = 0.072. Thus, P ( Y = 1) = 1 - .512 - .072 = 0.416. 3.5 There are 3! = 6 possible ways to assign the words to the pictures. Of these, one is a perfect match, three have one match, and two have zero matches. Thus, p (0) = 2/6, p (1) = 3/6, p (3) = 1/6. 3.6 There are 2 5 = 10 sample points, and all are equally likely: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5). a. p (2) = .1, p (3) = .2, p (4) = .3, p (5) = .4. b. p (3) = .1, p (4) = .1, p (5) = .2, p (6) = .2, p (7) = .2, p (8) = .1, p (9) = .1. 3.7 There are 3 3 = 27 ways to place the three balls into the three bowls. Let Y = # of empty bowls. Then: p (0) = P (no bowls are empty) = 27 6 27 ! 3 = p (2) = P(2 bowls are empty) = 27 3 p (1) = P(1 bowl is empty) = 1 27 18 27 3 27 6 = . 3.8 Note that the number of cells cannot be odd. p (0) = P (no cells in the next generation) = P (the first cell dies or the first cell splits and both die) = .1 + .9(.1)(.1) = 0.109 p (4) = P (four cells in the next generation) = P(the first cell splits and both created cells split) = .9(.9)(.9) = 0.729. p (2) = 1 – .109 – .729 = 0.162. 3.9 The random variable Y takes on vales 0, 1, 2, and 3. a. Let E denote an error on a single entry and let N denote no error. There are 8 sample points: EEE , EEN , ENE , NEE , ENN , NEN , NNE , NNN . With P ( E ) = .05 and P ( N ) = .95 and assuming independence: P ( Y = 3) = (.05) 3 = 0.000125 P ( Y = 2) = 3(.05)2(.95) = 0.007125 P ( Y = 1) = 3(.05) 2 (.95) = 0.135375 P ( Y = 0) = (.95) 3 = 0.857375.
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32 Chapter 3: Discrete Random Variables and Their Probability Distributions Instructor’s Solutions Manual b. The graph is omitted. c. P ( Y > 1) = P ( Y = 2) + P ( Y = 3) = 0.00725. 3.10 Denote R as the event a rental occurs on a given day and N denotes no rental. Thus, the sequence of interest is RR , RNR , RNNR , RNNNR , … . Consider the position immediately following the first R : it is filled by an R with probability .2 and by an N with probability .8. Thus, P ( Y = 0) = .2, P ( Y = 1) = .8(.2) = .16, P ( Y = 2) = .128, … . In general, P ( Y = y ) = .2(.8) y , y = 0, 1, 2, … . 3.11 There is a 1/3 chance a person has O + blood and 2/3 they do not. Similarly, there is a 1/15 chance a person has O blood and 14/15 chance they do not. Assuming the donors are randomly selected, if X = # of O + blood donors and Y = # of O blood donors, the probability distributions are 0 1 2 3 p ( x ) (2/3) 3 = 8/27 3(2/3) 2 (1/3) = 12/27 3(2/3)(1/3) 2 =6/27 (1/3) 3 = 1/27 p ( y ) 2744/3375 196/3375 14/3375 1/3375 Note that Z = X + Y = # will type O blood. The probability a donor will have type O blood is 1/3 + 1/15 = 6/15 = 2/5. The probability distribution for Z is 0 1 2 3 p ( z ) (2/5) 3 = 27/125 3(2/5) 2 (3/5) = 54/27 3(2/5)(3/5) 2 =36/125 (3/5) 3 = 27/125 3.12 E ( Y ) = 1(.4) + 2(.3) + 3(.2) + 4(.1) = 2.0 E (1/ Y ) = 1(.4) + 1/2(.3) + 1/3(.2) + 1/4(.1) = 0.6417 E ( Y 2 – 1) = E ( Y 2 ) – 1 = [1(.4) + 2 2 (.3) + 3 2 (.2) + 4 2 (.1)] – 1 = 5 – 1 = 4.
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This note was uploaded on 03/18/2009 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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ISM_chapter3 - Chapter 3 Discrete Random Variables and...

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