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f2011lectexamp14

# f2011lectexamp14 - *Whenever the rst word of a problem is...

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*Whenever the first word of a problem is pre- ceeded by an asterisk, the problem is for enrich- ment purposes only. Chapter 1 Lecture Examples: FALL 2011 1. A CM has a sample space that consists of four elements, denoted: a, b, c and d. As- suming the ELC, find the probabilities of each of the following events. (a) A = { a } (b) B = { a, b } (c) C = { b, c, d } 2. Refer to the previous problem. Now, in- stead of the ELC, assume that the probabil- ities of a, b, c and d follow the ratio 9:3:3:1. (Note: If interested, see Mendelian inheri- tance in Wikipedia for a discussion of the 9:3:3:1 ratio, as well as the 1:2:1 and the 3:1 ratios.) (a) Determine the probabilities of the in- dividual outcomes a, b, c and d. (b) Calculate the probabilities of the events A , B and C given in the pre- vious problem. 3. You are given the following information: the events A and B are disjoint; P ( A ) = 0 . 40 ; and P ( B ) = 0 . 25 . Calculate the fol- lowing probabilities. (a) P ( A or B ) . (b) P ( A c ) . (c) P ( B c ) . 4. You are given the following information: P ( A ) = 0 . 25 ; P ( B ) = 0 . 45 ; P ( AB ) = 0 . 20 . Calculate P ( A or B ) . 5. What is wrong with each of the following? (a) P ( A ) = 0 . 20 ; P ( B ) = 0 . 55 ; and P ( AB ) = 0 . 25 . (b) P ( A ) = 0 . 60 ; P ( B ) = 0 . 55 ; and A and B are disjoint. 6. Consider a sample space with three mem- bers: 1, 2 and 3. Assume the ELC and i.i.d. trials. The following table helps to visual- ize the results of the first two trials: X 2 X 1 1 2 3 1 (1,1) (1,2) (1,3) 2 (2,1) (2,2) (2,3) 3 (3,1) (3,2) (3,3) The nine entries in this table are equally likely. Define X = X 1 + X 2 , the total of the num- bers obtained in the first two trials. Find the sampling distribution of X . 7. Consider a sample space with five mem- bers: 0, 1, 2, 3 and 4. Assume the ELC and i.i.d. trials. The following table helps to visualize the results of the first two trials: X 2 X 1 0 1 2 3 4 0 (0,0) (0,1) (0,2) (0,3) (0,4) 1 (0,1) (1,1) (1,2) (1,3) (1,4) 2 (0,2) (2,1) (2,2) (2,3) (2,4) 3 (0,3) (3,1) (3,2) (3,3) (3,4) 4 (0,4) (4,1) (4,2) (4,3) (4,4) The 25 entries in this table are equally likely. Define X = X 1 X 2 , the product of the num- bers obtained in the first two trials. Find the sampling distribution of X . 1

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Chapter 1 Lecture Examples: Continued 8. Consider a sample space with three mem- bers: 1, 2 and 3. Do not assume the ELC. Instead assume the following: P (1) = 0 . 2 , P (2) = 0 . 1 and P (3) = 0 . 7 . Assume i.i.d. trials. The following table helps to visualize the results of the first two trials: X 2 X 1 1 2 3 1 (1,1) (1,2) (1,3) 2 (2,1) (2,2) (2,3) 3 (3,1) (3,2) (3,3) Note that these nine entries are not equally likely. Define X = X 1 + X 2 , the total of the num- bers obtained in the first two trials. Find the sampling distribution of X . 9. Refer to the previous question. Let X = X 1 + X 2 + X 3 , the total of the numbers obtained in the first three trials. Find the sampling distribution of X . 10. Consider the CM: Cast a balanced die five times and compute the sum, X , of the five numbers obtained. Assume independence of casts. The table below presents a huge amount of information: the exact probabil- ities for X ; the computer simulation ap- proximations based on m = 100,000 runs; the nearly certain interval for each P ( X = x ) . Note that every nearly certain inter- val contains the exact probability; i.e. every one is correct.
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f2011lectexamp14 - *Whenever the rst word of a problem is...

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