Difference between revisions of "Homework3 ECE302Fall2008sanghavi" - Rhea

Revision as of 14:16, 16 September 2008

Instructions

Homework 3 can be downloaded here on the ECE 302 course website.

Problem 1: Monte Hall, twisted

http://nostalgia.wikipedia.org/wiki/Monty_Hall_problem Explains the original Monty Hall problem and then the problem considering two contestants are involved.

HW3.1.a Zhongtian Wang_ECE302Fall2008sanghavi

HW3.1.a Shao-Fu Shih_ECE302Fall2008sanghavi

HW3.1.a Beau "ballah-fo-life" Morrison_ECE302Fall2008sanghavi

HW3.1.a Suan-Aik Yeo_ECE302Fall2008sanghavi

HW3.1.b Zhongtian Wang_ECE302Fall2008sanghavi

HW3.1.b Spencer Mitchell_ECE302Fall2008sanghavi

HW 3.1b Sahil Khosla_ECE302Fall2008sanghavi

Problem 1: Monte Hall, twisted

HW3.1.b Anand Gautam_ECE302Fall2008sanghavi This scenario is almost the same like the original problem. Just that your friend is kicked out by the host. The probability of you picking the car if you swap is 2/3 and not swapping is 1/3.

HW3.1.b Steve Streeter_ECE302Fall2008sanghavi The best way to look at this part of the problem is to divide it into 3 events. A: [You-Car,Friend-Goat], B: [You-Goat,Friend-Car], C: [Both picked goats]. From there you can use the condition that your friend was kicked off. So, for event A, if you switch, P(winning) = 0. Not switching: P(winning) = 1. You can determine the probabilities of other events in the same fashion.

HW3.1.b Kushagra Kapoor_ECE302Fall2008sanghavi This problem can be divided into 4 parts

A: P[pick car initially] = 1/3

  P[win car without swapping/you picked car initially] = 1/3

B: P[pick goat initially] = 2/3

  P[win car without swapping/you picked goat initially] = 0

C: P[pick goat initially] = 2/3

  P[win car after swapping/you picked goat initially] = 2/3

D: P[pick car initially] = 1/3

  P[win car after swapping/you picked car initially] = 0

So P[win car without swapping] = 1/3 and P[win car after swapping] = 2/3.

Problem 2: A Bayesian Proof

HW3.2 - Steve Anderson_ECE302Fall2008sanghavi

HW3.2 Tiffany Sukwanto_ECE302Fall2008sanghavi

HW3.2 Sang Mo Je_ECE302Fall2008sanghavi

Problem 3: Internet Outage

HW3.3 Gregory Pajot_ECE302Fall2008sanghavi

HW3.3 Monsu Mathew_ECE302Fall2008sanghavi

HW3.3 Joe Romine_ECE302Fall2008sanghavi

HW3.3 Katie Pekkarinen_ECE302Fall2008sanghavi

Problem 4: Colored Die

HW3.4.a Seraj Dosenbach_ECE302Fall2008sanghavi

HW3.1.b Anand Gautam For part a, the chance of drawing each die is independent of the other one. So the probability of red on the first roll is just the product of the chance of drawing a red face on each die. hope that makes sense.

HW3.4.b Joon Young Kim_ECE302Fall2008sanghavi

[HW 3.4.c Junzhe Geng] In this problem, we are asked to find the possibility of selecting a 3-red-face die when the first three rolls all give red faces. which is to find P(3-R-F|3R) according to formula: P(3-R-F|3R)=P(3-R-F n 3R)/P(3R), P(3-R-F n 3R) is easy to find. to find P(3R), we need to calculate each cases: selected 1-R-F(one red face), 2-R-F, or 3-R-F thus, P(3R)=P(3R n 1-R-F)+P(3R n 2-R-F)+P(3R n 3-R-F) each of those are not hard to find. my final result was 3/4. Hope it is right.

HW3.4 Aishwar Sabesan _ECE302Fall2008sanghavi

HW3.4c AJ Hartnett_ECE302Fall2008sanghavi

HW3.4 Jaewoo Choi_ECE302Fall2008sanghavi

Problem 5: Fuzzy Logic

3.5 - Divyanshu Kamboj_ECE302Fall2008sanghavi

       A		                B		

P(A=1) = p P(B=1) = p P(A=0) = 1-p P(B=0) = 1-p

P(A=1,C=1) = P(A=1) . P(C=1) = p.P(A=1,B=0) = p^2.(1-p) (1) P(A=1,B=0) = P(A=1) . P(B=0) = p.(1-p) (2)

Since, (1) & (2) are not equal to each other, A & C are not independent of each other when bits are biased.

3.5 - Katie Pekkarinen_ECE302Fall2008sanghavi

3.5 - Caleb Ayew-ew_ECE302Fall2008sanghavi

3.5 - Brian Thomas_ECE302Fall2008sanghavi