Probability Exercises

1. Probability Exercises

2. Experiment • Something capable of replication under stable conditions. • Example: Tossing a coin

3. Sample Space • The set of all possible outcomes of an experiment. • A sample space can be finite or infinite, & discrete or continuous.

4. A set is discrete ifyou can put you finger on one element after another & not miss any in between. • That’s not possible if a set is continuous.

5. Example: discrete, finite sample space • Experiment: Tossing a coin once. • Sample space: {H, T}. • This sample space is discrete; you can put your finger on one element after the other & not miss any. • This sample space is also finite. There are just two elements; that’s a finite number.

6. Example: discrete, infinite sample space • Experiment: Eating potato chips: • Sample space: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...} • This sample space is discrete; you can put your finger on one element after another & not miss any in between. • This sample space is infinite, however, since there are an infinite number of possibilities.

7. Example: continuous sample space. • Experiment: Burning a light bulb until it burns out. Suppose there is a theoretical maximum number of hours that a bulb can burn & that is 10,000 hours. • Sample space: The set of all real numbers between 0 & 10,000. • Between any two numbers you can pick in the sample space, there is another number. • For example, the bulb could burn for 99.777 hours or 99.778. But it could also burn for 99.7775, which is in between. You cannot put your finger on one element after another & not miss any in between. • This sample space is continuous. • It is also infinite, since there are an infinite number of possibilities. • All continuous sample spaces are infinite.

8. Eventa subset of the outcomes of an experiment • Example: • Experiment: tossing a coin twice • Sample space (all possible outcomes) = {HH, HT, TH, TT} • An event could be that you got at least one head on the two tosses. • So the event would be {HH, HT, TH}

9. Subjective versus objective probability

10. Subjective vs. Objective Probability • Subjective Probability is probability in lay terms. Something is probable if it is likely. • Example: I will probably get an A in this course. • Objective Probability is what we’ll use in this course. • Objective Probability is the relative frequency with which something occurs over the long run. • What does that mean?

11. Objective Probability: Developing the Idea • Suppose we flip a coin & get tails. Then the relative frequency of heads is 0/1 = 0. • Suppose we flip it again & get tails again.Our relative frequency of heads is 0/2 = 0. • We flip it 8 more times & get a total of 6 tails & 4 heads.The relative freq of heads is 4/10 = 0.4 • We flip it 100 times & get 48 heads.The relative freq of heads is 48/100 = 0.48 • We flip it 1000 times & get 503 heads.The relative freq of heads is 503/1000 = 0.503 • If the coin is fair, & we could flip it an infinite number of times, what would the relative frequency of heads be? • 0.5 or 1/2 • That’s the relative freq over the long run or probability of heads.

12. Two Basic Properties of Probability • 1. 0 < Pr(E) < 1 for every subset of the sample space S • 2. Pr(S) = 1

13. Counting Rules • We’ll look at three counting rules. • 1. Basic multiplication rule • 2. Permutations • 3. Combinations

14. Multiplication Rule Example

15. Multiplication RuleExample • Suppose we toss a coin 3 times & examine the outcomes. • (One possible outcome would be HTH.) • How many outcomes are possible?

16. We have 2 possibilities for the 1st toss, H & T. • H T

17. We can pair each of these with 2 possibilities. • H T • H T H T

18. That gives 4 possibilities on the 2 tosses: HH, • H T • H T H T

19. HT, • H T • H T H T

20. TH • H T • H T H T

21. And TT • H T • H T H T

22. If we toss the coin a 3rd time, we can pair each of 4 possibilities with a H or T. • H T • H T H T • H T H T H T H T

23. So for 3 tosses, we have 8 possibilities: HHH, • H T • H T H T • H T H T H T H T

24. HHT, • H T • H T H T • H T H T H T H T

25. HTH, • H T • H T H T • H T H T H T H T

26. HTT, • H T • H T H T • H T H T H T H T

27. THH, • H T • H T H T • H T H T H T H T

28. THT, • H T • H T H T • H T H T H T H T

29. TTH, • H T • H T H T • H T H T H T H T

30. and TTT. • H T • H T H T • H T H T H T H T

31. Again, we had 8 possibilities for 3 tosses. • 2 x 2 x 2 = 23 = 8

32. In general, • If we have an experiment with kparts (such as 3 tosses) • and each part has n possible outcomes (such as heads & tails), • then the total number of possible outcomes for the experiment is • nk • This is the simplest multiplication rule.

33. As a variation, suppose that we have an experiment with 2 parts, the 1st part has m possibilities, & the 2nd part has n possibilities. • How many possibilities are there for the experiment?

34. In particular, we might have a coin & a die. • So one possible outcome could be H6 (a head on the coin & a 1 on the die). • How many possible outcomes does the experiment have?

35. We have 2 x 6 = 12 possibilities. • Return to our more general question about the 2-part experiment with m & n possibilities for each part. • We now see that the total number of outcomes for the experiment is • mn

36. If we had a 3-part experiment, the 1st part has m possibilities, the 2nd part has n possibilities, & the 3rd part has p possibilities, how many possible outcomes would the experiment have? • m n p

37. Permutations Counting Rules

38. Permutations • Suppose we have a horse race with 8 horses: A,B,C,D,E,F,G, and H. • We would like to know how many possible arrangements we can have for the 1st, 2nd, & 3rd place horses. • One possibility would be G D F.(F D G would be a different possibility because order matters here.)

39. We have 8 possible horses we can pick for 1st place. • Once we have the 1st place horse, we have 7 possibilities for the 2nd place horse. • Then we have 6 possibilities for the 3rd place horse. • So we have 8 • 7 • 6 = 336 possibilities. • This is the number of permutations of 8 objects (horses in this case), taken 3 at a time.

40. How many possible arrangements of all the horses are there?In other words, how many permutations are there of 8 objects taken 8 at a time? • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 40,320 • This is 8 factorial or 8!

41. We’d like to develop a general formula for the number of permutations of n objects taken k at a time. • Let’s work with our horse example of 8 horses taken 3 at a time. • The number of permutations was 8 • 7 • 6.

42. We can multiply our answer (8 • 7 • 6) by 1 & still have the same answer. Note that this is Multiplying we have

43. So the number of permutations of n objects taken k at a time is

44. Combinations Counting Rules

45. Combinations • Suppose we want to know how many different poker hands there are. • In other words, how many ways can you deal 52 objects taken 5 at a time. • Keep in mind that if we have 5 cards dealt to us, it doesn’t matter what order we get them. It’s the same hand. • So while order matters in permutations, order does not matter in combinations.

46. Let’s start by asking a different question. What is the number of permutations of 52 cards taken 5 at a time? nPk = 52P5 = 52 • 51 • 50 • 49 • 48 = 311,875,200 When we count the number of permutations, we are counting each reordering separately .

47. However, each reordering should not be counted separately for combinations. • We need to figure out how many times we counted each group of 5 cards. • For example, we counted the cards ABCDE separately as ABCDE, BEACD, CEBAD, DEBAC, etc. • How many ways can we rearrange 5 cards? • We could rearrange 8 horses 8! ways. • So we can rearrange 5 cards 5! = 120 ways.

48. That means that we counted each group of cards 120 times. • So the number of really different poker hands is the number of permutations of 52 cards taken 5 at a time divided by the 120 times that we counted each hand.

49. So the number of combinations of 52 cards taken 5 at a time is

50. The number of combinations of n objects taken k at a time is