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 2012 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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 Spring '08
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