1. Welcome to MM150 Unit 7 Seminar

2. 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.

3. 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.

4. 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

5. 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

6. 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.

7. 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)

8. 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

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

10. 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)

11. 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.

12. 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

13. 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 ...

14. 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/ *1/4 =
= This is the expected growth per day.
There are 31 days in July
(0.4375)(31) = inches is the expected growth in July in Seattle

15. 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.

16. 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

17. 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

18. 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)

19. 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)