Sample Space and Probability: Understanding Probability Models and Axioms

This tutorial explores the concepts of sample space, probability models, probabilistic models, and probability axioms. It covers the basics of assigning probabilities, defining events, and understanding the fundamental axioms of probability. Examples and questions are provided to aid in comprehension.

• Probability
• Sample Space
• Models
• Axioms
• Tutorial

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1. Tutorial 1: Sample Space and Probability 1 Baoxiang WANG bxwang@cse Spring 2017

2. Probability models A probability model is an assignment of probabilities to every element of the sample space. Probabilities are nonnegative and add up to one. Examples S = { HH, HT, TH, TT } models a pair of coins with equally likely outcomes

3. Elements of a Probabilistic Model Event: a subset of sample space. ? is a set of possible outcomes Example. ? = ??,?? , the event that the two coins give the same side. The probability law assigns our knowledge or belief to an event ? a number ? ? 0. It specifies the likelihood of any outcome.

4. Probability Axioms 1. (Non-negativity) ?(?) 0, for every event ?. 2. (Additivity) For any two disjoint events ? and ?, ? ? ? = ? ? + ? ? In general, if ?1, ?2, are disjoint events, then ? ?1 ?2 = ? ?1 + ? ?2 + 3. (Normalization) ?( ) = 1.

5. Question If 4 boys and 4 girls randomly sit in a row, what is the probability of the following seating arrangement (a) the boys and the girls are each to sit together The total number of all arrangements is 8!=40320 The number of target arrangements is 4! 4! 2=1152 The probability #{?????? ????????????} #{??? ????????????} 1152 40320= 1 35 =

6. Question If 4 boys and 4 girls randomly sit in a row, what is the probability of the following seating arrangement (b) only the boys must sit together The total number of all arrangements is 8!=40320 The number of target arrangements is 4! 5! =2880 The probability #{?????? ????????????} #{??? ????????????} 2880 40320= 1 14 =

7. Question If 4 boys and 4 girls randomly sit in a row, what is the probability of the following seating arrangement (c) no two people of the same sex are allowed to sit together The total number of all arrangements is 8!=40320 The number of target arrangements is 4! 4! 2=1152 The probability #{?????? ????????????} #{??? ????????????} 1152 40320= 1 35 =

8. Question Eight balls are randomly withdrawn from an urn that contains 8 red, 8 blue, and 8 green balls. Find the probability that (a) 4 red, 2 blue, and 2 green balls are withdrawn The total number of all combinations is 24 8=735471 The number of target combinations is 8 4 8 2 8 2=54880 The probability #{?????? ????????????} #{??? ????????????} 54880 735471 =

9. Question Eight balls are randomly withdrawn from an urn that contains 8 red, 8 blue, and 8 green balls. Find the probability that (b) at least 3 red balls are withdrawn The total number of all combinations is 24 8=735471 The number of target combinations is 24 8 8 2 16 8 1 16 7 16 8=406857 6 The probability #{?????? ????????????} #{??? ????????????} 12329 22287 =

10. Question Eight balls are randomly withdrawn from an urn that contains 8 red, 8 blue, and 8 green balls. Find the probability that (c) all withdrawn balls are of the same color The total number of all combinations is 24 8=735471 The number of target combinations is 3 The probability #{?????? ????????????} #{??? ????????????} 1 = 245247

11. Question Suppose P(A) = 2 3and P(B) = 2 5. Show that 1 15 P(A B) 2 5 Best case: P(A B) 2 B A 5 Worst case: P(AUB)=P(A)+P(B)-P(A B) 1 B A P(A B) 1 15

12. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (a) a royal flush? (A hand of 10, J, Q, K, A of the same suit.) There are 4 suits in total. So the probability is 4 52 5 ? =

13. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (b) a straight flush? (A hand of five adjacent values of the same suit, but not a royal flush. Note that A, 2, 3, 4, 5 counts as a straight flush but K, A, 2, 3, 4 doesn t.) For each suit, the number of straight flush is 9 There are 4 suits, so the total number of straight flush 9 4 = 36

14. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (b) a straight flush? (A hand of five adjacent values of the same suit, but not a royal flush. Note that A, 2, 3, 4, 5 counts as a straight flush but K, A, 2, 3, 4 doesn t.) The probability is 36 52 5 ? =

15. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (c) a flush? (A hand of the same suit, but it is not a straight flush or royal flush.) In one suit, the number of special flush is 10. The number of flush in one suit 13 5 10 = 1227

16. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (c) a flush? (A hand of the same suit, but it is not a straight flush or royal flush.) The probability is ? =4 1227 52 5

17. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (d) a full house? (A hand with three cards of the same value, plus two cards with the same value as each other.) ? = the value of three cards ? = the value of two cards All possible pairs of {X,Y} 13 2 2 = 156

18. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (d) a full house? (A hand with three cards of the same value, plus two cards with the same value as each other.) Choose three cards of the same value 4 3 Choose two cards of the same value 4 2

19. Question Assume that all 52 what is the probability of being dealt poker hands are equally likely, 5 (d) a full house? (A hand with three cards of the same value, plus two cards with the same value as each other.) The probability is 156 4 3 4 52 5 2 ? =

20. Conditional Probability Definition. Conditional probability of ? given ? is ? ? ? =? ? ? , ? ? where we assume that ?(?) > 0. If ?(?) = 0: then ? ? ? is undefined. Fact. ? ? ? form a legitimate probability law satisfying the three axioms.

21. Question Let ? and ? be events with ?(?) > 0 and ?(?) > 0. We say that an event ? suggests an event ? if ?(?|?) > ?(?), and does not suggest event ? if ?(?|?) < ?(?). (a) Show that ? suggests ? if and only if ? suggests ?. ? ? ? = ?(? ?)/?(?) ? suggests ? if and only if ? ? ? > ? ? ? ? Equivalent to ? suggesting ?

22. Question Let ? and ? be events with ?(?) > 0 and ?(?) > 0. We say that an event ? suggests an event ? if ?(?|?) > ?(?), and does not suggest event ? if ?(?|?) < ?(?). (b) Show that ? suggests ? if and only if ??does not suggest ?. Assume that ? ??> 0. ??does not suggest ? if and only if ? ? ??< ? ? ? ?? We have ? ? ??= ? ? ? ? ? And ? ??= 1 ?(?)

23. Question Let ? and ? be events with ?(?) > 0 and ?(?) > 0. We say that an event ? suggests an event ? if ?(?|?) > ?(?), and does not suggest event ? if ?(?|?) < ?(?). (b) Show that ? suggests ? if and only if ??does not suggest ?. Assume that ? ??> 0. Substituting in the previous inequality ? ? ? ? ? < ? ? (1 ? ? ) Or ? ? ? > ? ? ? ? Equivalent to ? suggesting ?

24. Question We know that a treasure is located in one of two places, with probabilities ? and 1 ?, respectively, where 0 < ? < 1. We search the first place and if the treasure is there, we find it with probability ? > 0. Show that the event of not finding the treasure in the first place suggests that the treasure is in the second place Let ? and ? be the events ? = {the treasure is in the second place} ? = {not find the treasure in the first place}

25. Question We know that a treasure is located in one of two places, with probabilities ? and 1 ?, respectively, where 0 < ? < 1. We search the first place and if the treasure is there, we find it with probability ? > 0. Show that the event of not finding the treasure in the first place suggests that the treasure is in the second place By the total probability theorem ? ? = ? ??? ? ??+ ? ? ? ? ? = ? 1 ? + (1 ?)

26. Question We know that a treasure is located in one of two places, with probabilities ? and 1 ?, respectively, where 0 < ? < 1. We search the first place and if the treasure is there, we find it with probability ? > 0. Show that the event of not finding the treasure in the first place suggests that the treasure is in the second place By the total probability theorem ? ? ? =?(? ?) ?(?) 1 ? 1 ??> 1 ? = ?(?) 1 ? = ? 1 ? + (1 ?) =

27. Thanks Question are welcome.