Finding Probability Using Tree Diagrams and Outcome Tables

Finding Probability Using Tree Diagrams and Outcome Tables Chapter 4.5 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U Author: G. Greer (ideas from K. Myers)

H H T H T Tree Diagrams • if you flip a coin twice, you can model the possible outcomes using a tree diagram or an outcome table resulting in 4 possible outcomes T

H H H H H H T T T T T T Tree Diagrams Continued • if you rolled 1 die and then flipped a coin you have 12 possible outcomes (1,H) 1 (1,T) (2,H) 2 (2,T) (3,H) 3 (3,T) (4,H) (4,T) 4 (5,H) 5 (5,T) (6,H) 6 (6,T)

Sample Space • the sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin • clearly there are 12 possible outcomes (6 x 2) • there are 3 possible outcomes for an odd die and a head • so the probability is 3 in 12 or ¼ • P(odd roll, head) = ¼

Multiplicative Principle for Counting • The total number of outcomes is the product of the possible outcomes at each step in the sequence • if a is selected from A, and b selected from B… • n (a,b) = n(A) x n(B) • (this assumes that each outcome has no influence on the next outcome) • How many possible three letter words are there? • you can choose 26 letters for each of the three positions, so there are 26 x 26 x 26 = 17576 • how many possible license plates are there in Ontario? • 26 x 26 x 26 x 26 x 10 x 10 x 10

Independent and Dependent Events • two events are independent of each other if an occurence in one event does not change the probability of an occurrence in the other • what is the probability of getting heads when you know in advance that you will throw an even die? • these are independent events, so knowing the outcome of the second does not change the probability of the first

Multiplicative Principle for Probability of Independent Events • we know that if A and B are independent events, then… • P(B | A) = P(B) • if this is not true, then the events are dependent • we can also prove that if two events are independent the probability of both occurring is… • P(A and B) = P(A) · P(B)

R R B G R B B G R G B G Example • a sock drawer has a red, a green and a blue sock • you pull out one sock, replace it and pull another out • draw a tree diagram representing the possible outcomes • what is the probability of drawing 2 red socks? • these are independent events

Example • if you draw a card, replace it and draw another, what is the probability of two aces? • 4/52 x 4/52 • independent events • if you draw a card and then draw a second card (no replacement), what is the probability of two aces? • 4/52 x 3/51 • second event depends on first event • the sample space is reduced by the first event

Predicting Outcomes • Mr. Greer is going fishing • he finds that he catches fish 70% of the time when the wind is out of the east • he also finds that he catches fish 50% of the time when the wind is out of the west • if there is a 60% chance of a west wind today, what are his chances of having fish for dinner? • we will start by creating a tree diagram

Tree Diagram P=0.3 fish dinner 0.5 west 0.6 0.5 P=0.3 bean dinner fish dinner 0.7 P=0.28 0.4 east 0.3 bean dinner P=0.12

Continued… • P(east, catch) = P(east) x P(catch | east) • = 0.4 x 0.7 = 0.28 • P(west, catch) = P(west) x P(catch | west) • = 0.6 x 0.5 = 0.30 • Probability of a fish dinner: 0.28 + 0.3 = 0.58 • So Mr. Greer has a 58% chance of catching a fish for dinner

Exercises • read the examples on pages 239-244 • try page 245# 1,2,4,7,8,10 • check each answer with the back of the book to make sure that you are understanding these concepts

Counting Techniques and Probability Strategies - Permutations Chapter 4.6 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U Author: G. Greer (ideas from K. Myers)

Arrangements of objects • suppose you have three people in a line • how many different arrangements are there? • it turns out that there are 6 • how many arrangements are there for blocks of different colors • how many for 4 blocks • how many for 5 blocks • how many for 6 blocks • what is the pattern?

Selecting When Order Matters • when order matters, we have fewer choices for later places in the arrangements • for the problem of 3 people: • for person 1 we have 3 choices • for person 2 we have 2 choices left • for person 3 we have one choice left • the number of possible arrangements for 3 people is 3 x 2 x 1 = 6 • there is a mathematical notation for this (and your calculator has it)

Factorial Notation • the notation is called factorial • n!= n x (n – 1) x (n – 2) x … x 2 x 1 • for example: • 3! = 3 x 2 x 1 = 6 • 5! = 5 x 4 x 3 x 2 x 1 = 120 • If we have 10 books to place on a shelf, how many possible ways are there to arrange them? • 10! = 3628800 ways

Permutations • Suppose we have a group of 10 people and we want to choose a president, vice-president and treasurer? • in this case we are choosing people for a particular order • however, we are only choosing 3 of the 10 • for the first person, we can choose among 10 • for the second person, we can choose among 9 • for the third person, we can choose among 8 • so there are 10 x 9 x 8 ways

Permutation Notation • a permutation is an ordered arrangement of objects selected from a set • written P(n,r) or sometimes nPr • it is the number of possible permutations of r objects from a set of n objects

Choosing 3 people from 10… • we get 720 possible arrangements

Permutations When Some Objects Are Alike • suppose you are creating arrangements and some objects are alike? • for example, the word ear has 6 or 3! arrangements (aer, are, ear, era, rea, rae) • but the word eel has repeating letters and only 3 arrangements (eel, ele, lee) • how do we calculate arrangements in these cases?

Permutations With Some Objects Alike • to perform this calculation we divide the number of possible combinations by the combinations of objects that are similar • n is the number of objects • a, b, c are objects that occur more than once

So back to our problem • combinations of the letters in the word eel • what would be the possible arrangements of 8 socks if 3 were red, 2 were blue, 1 black, one white and one green?

Arrangements With Replacement • suppose you were looking at arrangements where you replaced the object after you had chosen it • if you draw two cards from the deck, you have 52 x 51 possible arrangements • if you draw a card, replace it and then draw another card, you have 52 x 52 possible arrangements • replacement increases the possible arrangements

Permutations and Probability • if you have 10 different colored socks in a drawer, what is the probability of choosing a red, green and blue sock? • probability is the number of possible outcomes you want divided by the total number of possible outcomes • you need to divide the number of possible arrangements of red, green and blue socks by the total number of ways that 3 socks can be pulled from the drawer

The Answer • so we have 1 chance in 120 or 0.833% probability

Exercises • Page 255 #1, 4, 5, 7, 11, 12

Counting Techniques and Probability Strategies - Combinations Chapter 4.7 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U Author: G. Greer (ideas from K. Myers)

When Order is Not Important • a combination is an unordered selection of elements from a set • there are many times when order is not important • suppose Mr. Greer has 10 players and must choose a starting line of 5 players for his basketball team • order of players is not important • we use the notation C(n,r) where n is the number of elements in the set and r is the number we are choosing

Combinations • a combination of 5 players from 10 is calculated the following way, giving 252 ways for Mr. Greer to choose his starting lineup

An Example of a Restriction on a Combination • suppose that one of Mr. Greer’s players is the superintendent’s son, and so must be one of the 5 starting players • here there are really only 4 choices from 9 players • so the calculation is C(9,4) = 126 • so now there are 126 possible combinations for the starting lineup

Combinations from Complex Sets • if you can choose of 1 of 3 meat dishes, 3 of 6 vegetables and 2 of 4 desserts for a meal, how many possible combinations are there? • combinations of meat dishes = C(3,1) = 3 • combinations of vegetables = C(6,3) = 20 • combinations of desserts = C(4,2) = 6 • possible combinations = • C(3,1) x C(6,3) x C(4,2) = 3 x 20 x 6 = 360 • you have 360 possible dinner combinations, so you had better get eating

Calculating the Number of Combinations • suppose you are playing coed volleyball, with a team of 4 men and 5 women • the rules state that you must have at least 3 women on the floor at all times • how many combinations of team lineups are there? • you need to take into account team combinations with 3, 4, or 5 women

Solution 1: Direct Reasoning • in direct reasoning, you determine the number of possible combinations of suitable outcomes and add them • find the combinations that have 3, 4 and 5 women and add them

Solution 2: Indirect Reasoning • in indirect reasoning, you determine the total possible combinations of outcomes and subtract unsuitable combinations • find the total combinations and subtract those with 0, 1, or 2 women

Finding Probabilities Using Combinations • what is the probability of drawing 10-J-Q-K-A from the same suit in a deck of cards? • there are C(52,5) ways to draw 5 cards • there are 4 ways to draw this type of straight • P = 4 / C(52,5) = 1/649740 • you will likely need to play a lot of poker to get one of these hands

Finding Probability Using Combinations • what is the probability of drawing 4 of a kind? • there are 13 different cards that can be used to make up the 4 of a kind, and the last card can be any other card remaining

Probability and Odds • these two terms have different uses in math • probability involves comparing the number of favorable outcomes with the total number of possible outcomes • if you have 5 green socks and 8 blue socks in a drawer the probability of drawing a green sock is 5/13 • odds compare the number of favorable outcomes with the number of unfavorable • with 5 green and 8 blue socks, the odds of drawing a green sock is 5 to 8 (or 5:8)

Exercises • Page 262 # 3, 5, 7, 9, 13, 14

References • Wikipedia (2004). Online Encyclopedia. Retrieved September 1, 2004 from http://en.wikipedia.org/wiki/Main_Page

Finding Probability Using Tree Diagrams and Outcome Tables