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Probability

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Probability

- 0 ≤ p(A) ≤ 1
- 0 = never happens
- 1 = always happens
- A priori definition

- p(A) = number of events classifiable as A
total number of classifiable events

- A posteriori definition

- p(A) = number of times A occurred
total number of occurrences

So:

p(A)= nA/N = number of events belonging to subset A vs. the total possible (which includes A).

If 6 movies are playing at the theater and 5 are crappy but 1 is not so crappy what is the probability that I will be disappointed?

5/6 or p = .8333

- Analytic (classical) view
- The common approach: if there are 5 bad movies and one good one I have an 83% chance in selecting a bad one.
- Fisher

- Relative Frequency view
- Refers to the long run of events: the probability is the limit of chance i.e. in a hypothetical infinite number of movie weekends I will select a bad movie about 83% of the time
- Neyman-Pearson

- Subjective view
- Probability is akin to a statement of belief and subjective e.g. I always seem to pick a good one.
- Bayesian

Experiment

- Experiment -- a Process that produces outcomes
- More than one possible outcome
- Only one outcome per trial

- Trial -- one repetition of the process
- Event -- an outcome of an experiment
- may be an elementary event, usually represented by an uppercase letter, e.g., A, E1

- Generally we can calculate the probability of one of a set of equally likely events by counting the sample space
- Many problems in probability can be solved in this way
- probability very often makes use of combinatorics (permutation and combination – we’ll talk about this later)

Sample Space -- Set Notation for Random Sample of Two Families

- S = {(x,y) | x }
- x is the family selected on the first draw
- y is the family selected on the second draw

- Concise description of large sample spaces

Sample Space

- The set of all elementary events for an experiment
- Methods for describing a sample space
- Listing
- Venn diagram

Family

Children in Household

Number of Automobiles

Listing of Sample Space

(A,B), (A,C), (A,D),

(B,A), (B,C), (B,D),

(C,A), (C,B), (C,D),

(D,A), (D,B), (D,C)

A

B

C

D

Yes

Yes

No

Yes

3

2

1

2

Sample Space -- Listing Example

- Experiment: randomly select, without replacement, two families from the residents of Denton
- Each ordered pair in the sample space is an elementary event, for example -- (D,C)

Y

X

Venn

Diagram

Venn Diagrams and the union of sets

- Venn diagrams helps us pictorially represent many of the algebraic rules of probability
- The union of two sets contains an instance of each element of the two sets.

Y

X

Venn

Diagram

Intersection of Sets

- The intersection of two sets contains only those elements common to the two sets.

- Mutually exclusive events
- both events cannot occur simultaneously.
- Can’t be a junior and senior

- Complementary events
- Two mutually exclusive events that are all inclusive

- both events cannot occur simultaneously.
- Independent events:
- occurrence of one event has no effect on the probability of occurrence of the other

Y

X

Mutually Exclusive Events

- Events with no common outcomes
- Occurrence of one event precludes the occurrence of the other event

Venn

Diagram

- All elementary events not in the event ‘A’ are in its complementary event.

Sample

Space

A

- Exhaustive sets
- set includes all possible events
- the sum of probabilities of all the events in the set = 1

- Equal likelihood
- roll a fair die each time the likelihood of 1-6 is the same whichever one we get, we could have just as easily have gotten another
- Counter example- put the numbers 1-7 in a hat. What’s the probability of even vs. odd?

- How do we find the probability of one event or another occurring?
- How do we find the probability of one event and another occurring?

- p(A or B) = p(A) + p(B)
- Probability of getting a grape orlemon skittle in a bag of 60 pieces where there are 15 strawberry, 13 grape, 12 orange, 8 lemon, 12 lime?
- p(G) = 13/60p(L) = 8/60
- 13/60 + 8/60 = 21/60 = .35 or a 35% chance we’ll get one of those two flavors when we open the bag and pick one out

Y

X

General Law of Addition

(not necessarily mutually exclusive)

Married (Yes/No)

Yes

No

Total

Children (Yes/No)

.70

.14

.56

Yes

.19

.11

No

.30

.33

1.00

.67

Total

Example: Marriage and Children

S

N

.56

.70

.67

General Law of Addition

Married (Yes/No)

Yes

No

Total

Children (Yes/No)

.70

.14

.56

Yes

.19

.11

No

.30

.33

1.00

.67

Total

If A & B are independent

- p(A and B) = p(A)p(B)
- p(A and B and C) = p(A)p(B)p(C)
- Probability of getting a grape and a lemon after two draws (with replacement) from the bag
- p(Grape)*p(Lemon) = 13/60*8/60 = ~.0288

If events X and Y are not independent then:

- p(X|Y) = probability that X happens given that Y happens
- The probability of X “conditional on” or “given” Y occurs
- It’s our ‘and’ type of question from before so we are going to use multiplication, however we don’t have independent events so it will be a little different

- p(A and B) = p(A)*p(B|A)

- Example: once we grab one skittle we aren’t going to put it back (sampling without replacement) so:
- p(A and B and C) = p(A)*p(B|A)*p(C|A,B)
- Probability of getting grape and lemon on successive turns = p(G)*p(L|G)
- (13/60)(8/59) = .0293

- A conditional probability is one where you are looking for the probability of some event with some sort of information in hand, e.g. the odds of having a boy given that you had a girl already.
- A joint event or probability would be the probability of a combination of events e.g. that you have a boy and a girl for children
- Aside: In this case the conditional would be higher b/c if we knew there was already a girl that means they’re of child-rearing age, probably interested in having kids etc. We have some additional info that would help us if we were just drawing out people randomly from some population.

Conditional Probability

Venn Diagram

N

S

.56

.70

*note before with our previous conditional probability we were dealing with mutually exclusive events i.e. can’t be grape and lemon at same time

- Let’s do a conditional probability: If I have a male, what is the probability of him being in the ‘Other’ category? Formally:
- p(A|B) = p(A and B)/p(B) =
- p(O|M) = p(O and M)/p(M) =
= (.412*.714)/.588= .5

- Easier way by looking at table- there are 10 males and of those 10 (i.e. given that we are dealing with males) how many are “Other”?
- p(O|M) = 5/10 or 50%.

- What is the probability of obtaining a Female Independent from this sample?
- In this case we’re looking for the joint probability of someone who is Female and Independent out of all possible outcomes:
2/17 = 11.8

- a. What is the probability a person is over 25?
- b. What is the probability that people under 25 spend at least 10 hours on the internet?
- c. What is the probability that someone who does not spend 10 hours on the internet each week is over 25?
- d. What is the probability of picking someone who spends less than 10 hours/wk on the internet and is under 25?

- a. 45/110 = .41
- b. 50/65 = .77
- c. 20/35 = .57
- d. 15/110 = .14