Welcome to MM150 Unit 7 Seminar

Definitions Experiment - A controlled operation that yields a set of results. Outcomes - The possible results from an experiment. Event - A subcollection of the outcomes of an experiment.

Definitions Continued Empirical Probability - The relative frequency of occurrence of an event and is calculated by ACTUAL OBSERVANCE of an experiment. Theoretical Probability - The determination of probability through a STUDY OF POSSIBLE OUTCOMES that can occur for the experiment.

Definitions Continued Empirical Probability P(E) = number of times event E has occurred total number of time exp. is performed Theoretical Probability (Outcome equally likely) P(E) = number of outcomes favorable to E total number of possible outcomes

Example Let’s use 10 flips of a fair coin, which has 2 outcomes: heads or tails. Empirical: I toss the coin 10 times and flip heads 7 times. P(Head) = 7/10 = 0.7 Theoretical: There are 2 outcomes, 1 of which is favorable to being a head. P(Head) = 1/2 = 0.5

Law of Large Numbers Probability statements apply in practice to a large number of trials, not to a single trial. It is the relative frequency over the long run that is accurately predictable, not individual events or precise totals.

4 Important Facts 1. The probability of an event that cannot occur is 0. It is said to be impossible. 2. The probability of an event that must occur is 1. It is said to be certain. 3. Every probability is a number between 0 and 1 inclusive; that is, 0 ≤ P(E) ≤ 1. 4. The sum of the probabilities of all possible outcomes of an experiment is 1. That is, P(A) + P(not A) = 1 P(not A) = 1 - P(A)

Single Toss of Fair Coin Outcomes - There are 2, heads or tails. P(Head) = 1/2 = 0.5 P(Tail) = 1/2 = 0.5 P(Head) + P(Tail) = 1/2 + 1/2 = 1 P(Tail) = 1 - P(Head) P(Tail) = 1 - 1/2 P(Tail) = 1/2

Single Toss of Fair Coin Outcomes - There are 2, heads or tails. P(Head or Tail) = 2/2 = 1 P(Yellow) = 0/2 = 0

Odds Odds against event P(event fails to occur) = P(failure) P(event occurs) P(success) Odds in favor of event P(event occurs) = P(success) P(event fails to occur) P(failure)

Odds Example Look at page 293 of your text, #49. The odds against the person having A+ blood. Odds against A+ = P(person not having A+) P(person having A+) = 66/34 or 66:34 Now, what if the question was what are the odds for a person having A+ blood. It would be 34/66 or 34:66.

Probability from Odds Again, turn to page 293 of your text, #47. The probability that the person has A+ blood. P(A+ Blood) = # of outcomes favorable to A+ total # of outcomes = 34/100 = 0.34 NOTE Odds for was 34:66; therefore probability of A+ is 34/(34+66) = 34/100

Expected Value E = P1*A1 + P2A2 + ... + PnAn where P1 is the probability the first event will occur and A1 is the net amount won or lost if the first event occurs P2 is the probability the second event will occur and A2 is the net amount won or lost if the second event occurs ...

Expected Value Example Page 301 #16 In July in Seattle, the grass grows 1/2 in. a day on a sunny day and 1/4 in. a day on a cloudy day. In Seattle in July, 75% of the days are sunny and 25% are cloudy. B) Determine the expected total grass growth in the month of July in Seattle. E = 0.75*1/2 + 0.25*1/4 = 0.375 + 0.0625 = 0.4375 This is the expected growth per day. There are 31 days in July (0.4375)(31) = 13.5625 inches is the expected growth in July in Seattle

Definitions Sample Space - All possible outcomes of an experiment. Sample Point - Each individual outcome in the sample space. Counting Principle - If a first experiment can be performed in M distinct ways and a second experiment can be performed in N distinct ways, then the two experiments in that specific order can be performed in M*N distinct ways.

Topping Sauce Sample Space Crust Tree Diagram Example Mathy’s Pizza has a choice of: 3 crusts: hand-tossed, pan, thin 2 sauces: traditional, meat 3 toppings: pepperoni, sausage, green peppers Construct a tree diagram and list the sample space. What is the probability that a student selects meat sauce? Topping Sauce Sample Space Crust e hre hrs hrg hme hms hmg pre prs prg pme pms pmg tre trs trg tme tms tmg s r g h e m s g e r s p g m e s g e r s t g e m s g

With/Without Replacement A bag with 5 red marbles, 2 blue marbles and 1 white marble. With Replacement You pick one marble out of the bag, record it and put it back in the bag. It is available to be picked again. Counting Principle: 8 * 8 = 64 There are 64 outcomes. For example: R1R1 R1R2 R1R3 R1R4 R1R5 R1B1 R1B2 R1W R2R1 R2R2 R2R3 R2R4 R2R5 R2B1 R2B2 R2W ... Without Replacment You pick one marble out of the bag, record it and set it aside (not back in the bag). It is not available to be picked again. Counting Principle: 8 * 7 = 56 There are 56 outcomes. For example: R1R2 R1R3 R1R4 R1R5 R1B1 R1B2 R1W R2R1 R2R3 R2R4 R2R5 R2B1 R2B2 R2W ... Note: doubles missing

Or Problems P(Heart or 6) = P(Heart) + P(6) --- Now...think about this...There is a 6 of hearts that will be counted in the hearts and with the 6s, so it is counted twice. We only want it to be counted once. Therefore we have to subtract 1/52 from the equation above. P(Heart or 6) = P(Heart) + P(6) - 1/52 = 13/52 + 4/52 - 1/52 = 16/52 NOTE: P(Heart ∩ 6) = 1/52, so P(Heart or 6) = P(Heart) + P(6) - P(Heart ∩ 6)

Or Problems P(A or B) = P(A) + P(B) - P(A ∩ B) When A and B are mutually exclusive P(A or B) = P(A) + P(B)