Chapter 3.1 Probability Students will learn several ways to model situations involving probability, such as tree diagrams and area models. They will formalize methods for computing probabilities of unions, intersections and complements of events. Then they will calculate expected value in games of chance.

3.1.3 Probability Models

Which one to Use Area model or Tree Diagram could work for any probability situation with 2 different situations When you start looking at 3 or more a tree diagram is helpful When you get too many it is helpful to just list everything and write probabilities next to each situation

Tree diagram 3-23 Rock Paper Scissors 3 People hard to make an area model Each branch of the tree is one situation

Area Model 3-26 Shooting one and one free throws Easier to represent each part in an area model Could be done as a tree diagram

List outcomes and write probability Deck of cards Hard to have 52 different branches or 52 different boxes 4 of every kind of card Ace through King 13 of each suit 26 of each color Now you can look at combinations of suit and kind

Homework Pg 158 3-29, 3-30 and 3-32

Extra Combinations and Permutations

How many 7 digit phone numbers are there How would you figure this out? How many numbers could the first number be? How about second? Thirds? What would you then do? Multiply all the numbers across This would give you the total number our outcomes to help do a probability

Combination A grouping of outcomes in which order does not matter ! Means factorial Means repeated multiplication from number down to 1 5! = 5 * 4 * 3 * 2 *1 = 120 Combination represented as C and n is the number of things you are choosing from r is how many you want out of those The division helps to eliminate all the other choices

Permutation Grouping of outcomes, or arrangement, in which order does matter Permutation represented as P Variables mean the same thing You should get a higher number for permutations that combinations because AB is different than BA in a permutation but the same in a combination

Examples How many ways can we group the students in class? Is this a combination or permutation? If we run a race how many different outcomes of 1st, 2nd and 3rd can we get? Is this a combination or permtutaion

Homework Worksheet

3.1.4 Unions, Intersections and Complements

Intersection Is the set in which both the first event and the second event occur Use numbers {1,2,3,4,5,6,7,8,9,10} P(prime) P(even) Intersection of 2 events Event A = {2,3,5,7} Event B = {2,4,6,8,10} Intersection of A and B is {2} 2 is in both events it is both prime and even Dependent events the intersection is P(A)*P(B)=P(A and B) or the intersection

Intersection Example 2 cards are drawn without replacement from a standard deck of cards. What is the probability that both with be a 5. P(A) first card is a 5, 4 fives in a standard deck P(B) second card is a 5, since one 5 is gone no on 3 fives in a deck, minus 1 card P(A)=4/52 P(B)=3/52 P(A and B)=4/52 * 3/52 = 1/121

Intersection Example 2 Rolling a fair die and you get an even number and it is divisible by 4 P(A) it is even {2,4,6) P(B) it is divisible by 4 {4} The intersection of the two events, or what is common is 4 P(A and B) = 1/6 Only 1 number is common in the two events out of 6 possible outcomes

Union Probability of A or B happening but not both P(A)+P(B)-P(intersection of A and B) If the two situations have nothing to do with each other then there is no union Rolling a die and flipping a coin If the two situations do have something in common then we need to subtract the intersection or what is common

Union Example Rolling a die and finding the probability of getting an even number and that is a multiple of 3. A is getting an even number = {2,4,6} B is getting a multiple of 3 = {3,6} P(A)= 3/6 P(B)= 2/6 Intersection of A and B, what can happen in both Both have a 6 so P(A and B) = 1/6 Therefore P(A or B) = P(A)+P(B) – P(A and B)=3/6 + 2/6 -1/6=4/6=2/3

Union 2 Example Rolling an odd number or a multiple of 5 P(A) is odd number = {1,3,5} = 3/6 P(B) is a multiple of 5 = {5} = 1/6 P(A and B) common in the 2 = {5} = 1/6 P(A or B)=P(A)+P(B)-P(A and B) = 3/6+1/6-1/6=3/6=1/2

Complements Is the other part of the probability that would help it add to 1 Probability of rolling a 3 on a standard die is 1/6 The compliment would be 1-1/6 = 5/6 Probability of all other things happening

Homework Worksheet on Unions Otherwise called compound events

Average amount paid out or made each turn during a game 3.1.5 Expected Value Average amount paid out or made each turn during a game

3-47 Take a spin What were your answers Does it make sense Expected Value for single player Prob of each thing happening Multiply prob and number of times played Multiply by amount win or los Add all together Fair game would equal 0

3-48 Different Spinner P(4)=1/4 P(100)=1/4 P(0)=1/2 4*1/4=1 4*1/2=2 1*4+1*100+2*0=104 or 26 per turn

Try 3-50 The double spinner Making a table helps Use 64 games – make number of times played a multiple of the denominator, makes math easier Table completed Expected value – don’t need other sections because the do not result in anything If you need to pay, take average amount won and subtract the amount you pay to see if you come out in the positive

Homework Pg 172 3-55 and 3-61

Similar idea for expected value if you play against someone else

Find each persons probability Multiply each persons probability to the number of times they play Multiply answer from 2 to how much they make Subtract what the other person makes (this is what they pay out) Divide answer by number of times played to get the average If only 2 people play one answer should be opposite each other If multiple people play the sum of the expected values should be 0 You can’t make more than what is payed in

Example 1 Work Through this based on the steps discuss to calculate the expected value P(AL winning) = ¼ P(Betty Winning) = ¾ Play 100 times Al: ¼*100=25 Betty: ¾*100=75 Al: 25*4=100 Betty: 75*1=75 AL: 100-75=25 Betty: 75-100=-25 Al: 25/100= ¼ Betty: -25/100=-1/4

I pay you $6 when you hit gray and you pay me $8 when I hit white. Do you think the game is fair? Why? Find the expected value for each of us Go Through Steps to answer questions P(Me)= 7/15 P(You)=8/15 Play 30 times, make it a multiple of the denominator to make math easier Me: 7/15 * 30 = 14 You: 8/15* 30 =16 Me: 14 *8= 112 You: 16 * 6 = 96 Me: 112-96 = 16 You: 96-112=-16 Me: 16/30 = .533 You: -16/30 = -.533 I would get about 50 cents every time we played and you would lose 50 cents. If it was closer to zero we would neither win nor lose