Introduction

1. Introduction We will cover 3topics today 1. Set Theory 2. Probability 3. Permutations and Combinations

2. Set Theory An experiment or trial is described as random if the result is not predictable. There are two types of data variables; either discrete or continuous. Our first task with a random experiment is to list (called a set) all the possible outcomes. This is known the sample space. If we can count the number of outcomes in the sample space then the data is discreet (or the sample space is discreet). If we can not, then the data is continuous (i.e. how far can a car travel before the tyres burst). For the single throw of a die we can list the sample space (S) as a set. I.e.

3. Set Theory Set notation is very useful for representing sample spaces and events. If we roll a die and we are looking for the appearance an odd number, then then the outcome is contained within the subset (A) of the sample space (S). Suppose we roll two dice and we are interested in the event that either the sum of the dice equals 8 (A1) or that we roll a 6 (A2), or both. This event is the union of subsets (A3). This subset is also called an event. I.e. the event that we roll a die and we get an odd number. We can write this in notation form as If we are interested in the event that both A1 and A2 occur, then we are looking for the intersection (A4)of A1 and A2.

4. Set Theory We can describe these sets using a Venn diagram. Venn diagrams are diagrams that show all possible relations between a collection of sets. I.e. S A1 A3 A2 S A1 A2 The area in orange is given by The area in yellow is given by

5. Set Theory We can also define some different symbols We can also use a Venn diagram to describe a complimentary set. If Means that A is a member of S. This is not the same as Then the complimentary set A is shown by S Which means that A is a subset of S. We can also define Ac A Which means that A is not a member of S. Empty Set

6. Set Theory Axioms of Probability We can also determine a disjoint set. If two sets A and B (which are both subsets of S) do not have a single element in common then we say that A and B are disjoint sets. For every event ‘A’ in a sample space ‘S’, the probability ‘P(A)’ must satisfy Clearly we can say that If, in a single trial (or experiment) two events can not occur together (i.e. they are disjoint) then the events are said to be mutually exclusive. For n mutually exclusive events

7. Probability From our discussion on sets we can also deduce that :- This brings us to the standard definition of probability. The probability of an event ‘A’ occurring is The probability an event (A) does not occur is The probability either of the events A or B occurs is r is the number of ways in which A can occur. This is called the Addition Theorem for Probability. n is the total number of outcomes.

8. Probability Events A and B are said to be independent if the occurrence of one event does not affect the occurrence of the other event. If A and B are mutually exclusive then In such cases, the probability of the events A and B is occurring defined by Thus, in this case Hence, the probability of the union of the events A and B is defined by

9. Permutations and Combinations How many permutations of the letters a, b, c & d can be made if only two letters are selected? Permutation and Combination are important terms we use when dealing with problems on probability and/or probability distributions. ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc = 12 How many combinations of the letters a, b, c & d can be made if only two letters are selected? Permutation means arrangement Combination means selection ab, ac, ad, bc, bd, cd = 6

10. Permutations and Combinations The notation for a permutation is The next item is chosen from n-1 possibilities leaving n-2 items. If we continue to do this r times we obtain This notation means the number of ways r different items can be selected from n distinct items whilst taking regard of the order of selection. Which is the same as If the items are not replaced, the first item is chosen from n possibilities leaving n-1 items.

11. Permutations and Combinations The notation for a combination is Hence, there are r positions in which we could have selected the first item. Following this logic there are r-1 positions in which we could have picked the second item This notation means the number of ways r different items can be selected from n distinct items without taking regard of the order of selection. If we continue this logic then from nPr permutations there are r! that give the same combination. Hence, we obtain If we are going to select r items without taking regard of the order, then the first item that we selected could have been selected in the first, second, third, …. or last position.

12. Permutations and Combinations Question a) How many different five card hands can be dealt from a standard deck with 52 cards? b) What is the probability that a hand dealt at random consists of 5 spades? This is a combination problem, not a permutation one. Thus, There are 2 598 960 different hands

13. Permutations and Combinations There are 13 spades in the pack hence the number of different hands consisting of 5 spades is To obtain the probability that a random five-card hand contains five spades we deduce that 1287 out of the 2 598 960 possible hands will have 5 spades. Hence, P(five-card spade hand)

14. Permutations and Combinations Question A bag contains 20 balls of which 7 are red (r), 5 are white (w) and 8 are black (b) balls. If three are drawn at random, without replacement, find the probability that a) Two red balls and one black ball are drawn b) One of each colour is drawn c) One or more red balls are drawn d) All are of the same colour

15. Permutations and Combinations The total number of 3 ball selections which can be made is a) The numbers of ways in which two red balls and one black ball can be drawn is 7C2 x 8 Hence, P(2r and 1b) b) One of each colour is drawn The number of ways one of each colour can be chosen is; 7 x 5 x 8 = 280. Hence, P(1r, 1w and 1b)

16. Permutations and Combinations c) The numbers of ways in which no red ball is drawn is 13C3 = 286 from a total of 1140. Hence the probability that a selection contains at least one red ball is Hence, P(≥ 1r) = 1 – P(0.r) d) The probability that either 3 reds or 3 blacks or 3 whites are drawn Hence, P(3r) + P(3w) + P(3b)

17. Conclusion NO seminars this week! We have covered 3topics today 1. Set Theory 2. Probability 3. Permutations and Combinations