# Probability Dissertation Example

- Discipline:
- Economics
- Chapter:
- Abstract
- Level:
- Bachelor's
- Pages:
- 2
- Words:
- 550

Name Instructor Course Date of submission Probability Question 1: Using the AND rule A dies has got six faces and every face has got number of dots ranging from one dot to six dots Once you toss two dies and count the number of dots on the faces of the dies then you get the following combination of integers with tabulated sum below it. All the outcomes are independent. (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Summation dies 2 dies 1 sum(+) 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Possible number of outcomes we have that is our sample space = 36 Possible number of times we can get a total of 11 dots on the faces of the two dies =2 Therefore Probability of getting a sum of 11 dots =favourable outcomestotal possible outcomes = 236 = 118 or 0.055556 Use of OR rule and all the outcomes are mutually exclusiveProbability of getting 11 dots = probability that dies 1 has 6 dots and dies 2 has 5 dots or probability that dies 1 has 6 dots and dies 2 has 5 dots Probability of dies having 5 dots = 1/6 Probability of dies having 6 dots = 1/6 Probability of 11 dots = (16 *16 )+ (16 *16 ) = 118 or 0.055556 Question 2 219075104775Pr (P|E) = 0.9 pr(P|E)= 0.9 pr(P|E)= 0.9 0Pr (P|E) = 0.9 pr(P|E)= 0.9 pr(P|E)= 0.9 2066925-257175P P 1314450-4762500 1609724153670Pr (-P|E)= 0.1 0Pr (-P|E)= 0.1 1104900153670E E 13144502095500-65722559055Pr (E)= 0.6 0Pr (E)= 0.6 15144751021080025431752087880-P 0-P 2343150830580P 00P 2438400230505-P 0-P 11620491554480-E 0-E -47625706755O O 15144751744980133350830580013335020955 1438275175895Pr (P|(¬E)...

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