Math 215 Phase Iv Individual Proj.
Math 215-1202A-02
Discrete Mathematics
Phase V Individual Project
May 14, 2012
Phase V Part I. * A = {1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36} * B = { 2,4,6,8,10,11,13,15,17,20,22,24,26,28,29,31,33,35} * C = {0,00} * D = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36} * E = {1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35} * F = {1,2,3,4,5,6,7,8,9,10,11,12}
Using the above sets we will determine the following; * A⋃B = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36} * A⋂D = {12,14,16,18,30,32,34,36} * B⋂C = {⌀} this would be a disjointed equation * C⋃E = {0,00,1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35} * B⋂F = {2,4,6,8,10,11}
* E⋂F = { 1,3,5,7,9,11}
Spin – PK | Number | Odd/even | Color | 1 | 5 | Red | Odd | 2 | 22 | Black | Even | 3 | 2 | Black | Even | 4 | 30 | Red | Even | 5 | 16 | Red | Even | 6 | 4 | Black | Even | 7 | 16 | Red | Even | 8 | 24 | Black | Even | 9 | 7 | Red | Odd | 10 | 15 | Black | Odd |
Bet1
Bet1
lose lose win win Bet4
Bet4
Game over
Game over cont cont
Bet3
Bet3
gameover gameover Game over
Game over cont cont cont cont
Bet 2
Bet 2
Bet 2
Bet 2 cont cont
Cont.
Cont.
With the probability of the roulette wheel the following outcomes can be concluded; * With 38 places on the wheel 18 of them are odd this would give you an 18/38 or 9/19 chance of landing on an odd number. * There are only 2 places on the wheel that are green so this would give a 3/38 or a 1/19 chance of landing on a green spot. * There are 18 red numbers on the wheel and 2 green numbers giving a 20/38 or a 10/19 chance of landing on red or green. * With the 18 red numbers 8 of them are even giving a 26/38 chance or a 13/19 chance of landing on red or even. With the proposed license plate schemes,
Discrete Mathematics
Phase V Individual Project
May 14, 2012
Phase V Part I. * A = {1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36} * B = { 2,4,6,8,10,11,13,15,17,20,22,24,26,28,29,31,33,35} * C = {0,00} * D = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36} * E = {1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35} * F = {1,2,3,4,5,6,7,8,9,10,11,12}
Using the above sets we will determine the following; * A⋃B = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36} * A⋂D = {12,14,16,18,30,32,34,36} * B⋂C = {⌀} this would be a disjointed equation * C⋃E = {0,00,1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35} * B⋂F = {2,4,6,8,10,11}
* E⋂F = { 1,3,5,7,9,11}
Spin – PK | Number | Odd/even | Color | 1 | 5 | Red | Odd | 2 | 22 | Black | Even | 3 | 2 | Black | Even | 4 | 30 | Red | Even | 5 | 16 | Red | Even | 6 | 4 | Black | Even | 7 | 16 | Red | Even | 8 | 24 | Black | Even | 9 | 7 | Red | Odd | 10 | 15 | Black | Odd |
Bet1
Bet1
lose lose win win Bet4
Bet4
Game over
Game over cont cont
Bet3
Bet3
gameover gameover Game over
Game over cont cont cont cont
Bet 2
Bet 2
Bet 2
Bet 2 cont cont
Cont.
Cont.
With the probability of the roulette wheel the following outcomes can be concluded; * With 38 places on the wheel 18 of them are odd this would give you an 18/38 or 9/19 chance of landing on an odd number. * There are only 2 places on the wheel that are green so this would give a 3/38 or a 1/19 chance of landing on a green spot. * There are 18 red numbers on the wheel and 2 green numbers giving a 20/38 or a 10/19 chance of landing on red or green. * With the 18 red numbers 8 of them are even giving a 26/38 chance or a 13/19 chance of landing on red or even. With the proposed license plate schemes,