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Chapter 6 Section 3 Assignment of Probabilities. Sample Space and Probabilities Sample Space : S = { s 1, s 2, s 3, …, s N-1, s N } where s 1, s 2, s.

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Presentation on theme: "Chapter 6 Section 3 Assignment of Probabilities. Sample Space and Probabilities Sample Space : S = { s 1, s 2, s 3, …, s N-1, s N } where s 1, s 2, s."— Presentation transcript:

1 Chapter 6 Section 3 Assignment of Probabilities

2 Sample Space and Probabilities Sample Space : S = { s 1, s 2, s 3, …, s N-1, s N } where s 1, s 2, s 3, …, s N-1, s N are all the possible outcomes of the experiment Each outcome can be assigned a number that represents the probability of an outcome.

3 Probability Distribution OutcomeProbability s1s1 p1p1 s2s2 p2p2 s3s3 p3p3 sNsN pNpN

4 Fundamental Properties of Probability Distributions 1.0 < p 1 < 1 0 < p 2 < 1 0 < p 3 < 1 0 < p N < 1 2. p 1 + p 2 + p 3 + … + p N-1 + p N = 1

5 Assigning Probabilities 1.Theoretically A.Using Logic and/or Counting Techniques B.Best method for assigning probabilities C.See Example 1 & 2 on page 273 2.Empirically (i.e. Experimentally) A.Works only when the observed trial are representative of the sample space. B.See Example 3 on pages 273 and 274

6 Exercise 3 (page 281) Red die and green die are tossed and the number of the side facing up are observed. Sample Space: S = { (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) }

7 Exercise 3 continued Note that each pair is equally likely to be picked if we randomly select a pair, thus the probability distribution is: OutcomeProbability s 1 = (1,1)p 1 = 1/36 s 2 = (1,2)p 2 = 1/36 s 3 = (1,3)p 3 = 1/36 s 36 = (6,6)p 36 = 1/36

8 Addition Principle Let the event E be defined as E = { s, t, u, …, z } Then Pr( E ) = Pr(s) + Pr(t) + Pr(u) + … + Pr(z)

9 Exercise 3 Part (a) Let E = {‘the numbers add up to 8’} E = { (2,6), (3,5), (4,4), (5,3), (6,2) } Find the Pr( E ) Pr( E ) = Pr( (2,6) ) + Pr( (3,5) ) + Pr( (4,4) ) + Pr( (5,3) ) + Pr( (6,2) ) Pr( E ) = 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36  0.1389

10 Inclusion – Exclusion Principle as Applied to Probabilities Let E and F be events, then Pr( E  F ) = Pr ( E ) + Pr ( F ) – Pr( E  F ) Venn Diagrams and Probabilities We can use Venn Diagrams and place numbers that represent the probability that the elements in the basic region will occur.

11 Exercise 17 (page 282) Let E and F be events. Pr( E ) = 0.6, Pr( F ) = 0.5, and Pr( E  F ) = 0.4 (a)Find Pr( E  F ) (b)Find Pr( E  F)

12 Odds When the odds in favor of an event a to b, then the probability of the event can be found by p = a / ( a + b ) a represents the number of times that the event occurs b represents the number of times that the event does not occur

13 Converting an Odds to a Probability Example: If the odds of a horse to win a race is 1 to 60, then then probability of the horse winning is: a = 1 and b = 60 p = a / (a + b) = 1 / (1 + 60) = 1/61  0.0164

14 Converting a Probability to an Odds List the probability as a decimal number Convert the decimal number to a fraction. Reduce the fraction (when possible). Split the value in the denominator to the sum of two values, where the first value in the sum is the value in the numerator. Get the values of a and b from the resulting fraction.


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