Stat447F08HW2Sol

Stat447F08HW2Sol - 1 Out of text problem If A is the event...

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Unformatted text preview: 1. Out of text problem: If A is the event that the first roll is red, and B is the event that the second roll is red, and C is the event that die A was used, and 1 is the event that the coin shows head, and 2 is the event that the coin shows tail, we have 1 = 2 = , 1 = , 2 = 2 1 2 3 1 3 1 = , 2 = . 3 3 2 1 a) Proof: From Theorem 2.12 in the text book we know that, 1 2 1 1 1 P = P D1 1 + P 2 2 = + = 2 3 2 3 2 b) 1 = = 3 3 2 2 4 9 2 = = 3 3 1 1 1 9 = 1 1 + 2 2 1 4 1 1 5 = + = 2 9 2 9 18 1 From part a) we know that P = . Similarly we have, 2 1 2 1 1 1 P = P D1 1 + P 2 2 = + = 2 3 2 3 2 5 5 = = 18 = 1 9 2 c) We realize that C = D1 , so we have = 1 = 1 1 2 2 1 2 = 1 1 1 = = 3 3 2 9 2 4 = = 9 = 5 5 18 2. CB 2.110 From the question we know that, = 0.25 = 0.20 = 0.40, = 0.75 = 0.20, = 0.40 = 0.25, = 0.15 So we can see that, = + + + = 0.25 0.20 + 0.20 0.40 + 0.40 0.25 + 0.75 0.15 = 0.3425 | = , = 0.25 0.20 = = 0.1460 0.3425 , = 0.20 0.40 = = 0.2336 0.3425 | = | = , = 0.40 0.250 = = 0.2920 0.3425 | = , 0.75 0.15 = = 0.3425 = 0.3285 (a) From the above we know that | is the biggest, so the most likely cause of explosion is sabotage. (b) From the above we know that | is the smallest, so the least likely cause of the explosion is static electricity. 3. CB 2.112 (a) If 1 is the event that the student gets the one of the cards writing "I smoke marijuana at least once a week", and 2 is the event that the student gets one of the cards writing "I do not smoke marijuana at least once a week", then we have, 1 = 12 20 = , 2 = 5 3 20-12 20 = 2 5 = |1 1 = 3 = 5 (b) From part (a) we know that, Substituting = 106 250 + |2 2 1 + 1 - 2 2 2 1 + 1 - = + 5 5 5 = 5 - 2 to the formula above we gain, 106 = 5 - 2 = 0.12 250 4. CB 2.120 component 1 component 2 parallel subsystem 0.75 0.60 A B C D E 0.98 0.99 0.65 0.70 0.60 If the system works, then component 1, component 2 and the parallel subsystem must all works. P parallel subsystem works = at least one of A, B, C, D and E works = 1 - none of A, B, C, D or E works = 1 - 1 - 0.75 1 - 0.60 1 - 0.65 1 - 0.70 1 - 0.60 = 0.9958 Thus we know that, = = 1 2 = 0.98 0.99 0.9958 = 0.9661 5. CB 3.4 (a) According to Definition 3.2 in the text book, 5 5 1= Thus = 1 15 =1 = =1 = 15 . (b) Not required (c) According to Definition 3.2 in the text book, 1= =1 = =1 2 = 6 + 1 2 + 1 6 Thus, + 1 2 + 1 (d) According to Definition 3.2 in the text book, + + = 1= =1 = =1 1 4 1 = 3 Thus c=3. 6. CB 3.6 It's obvious that when c=0, f(x) cannot serve as the values of the probability distribution of a random variable. When 0, + + = Because + 1 =1 =1 diverges, + =1 =1 = + = =1 + 1 =1 = 1 regardless of the value of c. So whatever c is, + does not equal to 1, which indicate that f(x) =1 cannot serve as the values of the probability distribution of a random variable. 7. CB 3.12 (a) 5 1 1 2 < 6 = 6 - 2 = - = 6 3 2 (b) 1 1 1 = 4 = 4 - lim () = - = 4 2 3 6 (c) From the distribution function we can see that the random variable X can only have four values: 1,4, 6 and 10. Similarly as what we did in part (b) we know that, 1 1 -0= 1 3 3 5 1 1 = 6 = 6 - lim () = - = 6 6 2 3 5 1 = 10 = 10 - lim () = 1 - = 10 6 6 So we know that the probability distribution is, = 1 = 1 - lim () = 1 3 , when x=1 or 6; () = 1 6 , when x=4 or 10. 8. CB 3.22 (a) According to Theorem 3.6 in the text book we know that, 4 1= = 4 0 Thus we know gain = . 4 1 (b) 1 < = 4 1 4 0 4 4 1 1 > 1 = = 2 1 4 = 0 - 1 4 1 9. CB 3.28 0 = 0, = , 3 0; 0 < < 1; 1 2; 1 0 3 = + () - = 1 3 11 0 3 = , 3 1 + 1 2 3 = -1 3 , 2 < < 4; 4. 1, 10. CB 3.34 5 = 5 = 1 - 9 16 = 52 25 > 8 = 1 - 8 = 1 - 8 = 1 - 1 - 11. CB 3.35 1- = = 9 2 9 9 = 82 64 = 18 -3 , > 3; 0, 5 elsewhere. 5 16 25 3 3 + + 9 > 8 = () = 18 -3 = 64 8 8 5 = () = 18 -3 = ...
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