f2011lectexamp14

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
*Whenever the frst 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, fnd 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 .Ca lcu la tetheFo l - lowing probabilities. (a) P ( A or B ) . (b) P ( A c ) . (c) P ( B c ) . 4. You are given the Following inFormation: P ( A . 25 ; P ( B . 45 ; P ( AB 0 . 20 te P ( A or B ) . 5. What is wrong with each oF the Following? (a) P ( A . 20 ; P ( B . 55 ;a n d P ( AB . 25 . (b) P ( A . 60 ; P ( B . 55 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 frst 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. Defne X = X 1 + X 2 ,thetotaloFthenum- bers obtained in the frst two trials. ±ind the sampling distribution oF X . 7. Consider a sample space with fve 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 frst 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. Defne X = X 1 X 2 ,theproductoFthenum- bers obtained in the frst two trials. ±ind the sampling distribution oF X . 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 Frst 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. DeFne X = X 1 + X 2 ,thetotalofthenum- bers obtained in the Frst two trials. ±ind the sampling distribution of X . 9. Refer to the previous question. Let X = X 1 + X 2 + X 3 ,theto ta lo fthenumbe r s obtained in the Frst three trials. ±ind the sampling distribution of X . 10. Consider the CM: Cast a balanced die Fve times and compute the sum, X ,oftheFve numbers obtained. Assume independence of casts. The table below presents a huge amount of information: the exact probabil- ities for X ;t h ec om p u t e rs im u l a t i o na p - proximations based on m = 100,000 runs; the nearly certain interval for each P ( X = x ) .N o t e t h a te v e r yn e a r l yc e r t a i n i n t e r - val contains the exact probability; i.e. every one is correct.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 26

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online