View
216
Download
1
Category
Tags:
Preview:
Citation preview
5.1 Basic Probability Ideas
• Definition: Experiment – obtaining a piece of data
• Definition: Outcome – result of an experiment• Definition: Sample space – list of all possible
outcomes of an experiment• Definition: Event – collection of outcomes from
an experiment (a simple event is a single outcome)• Definition: Equally likely sample space – sample
space in which all outcomes are equally likely
5.1 Basic Probability Ideas
• Definition: Probability – A number between 0 and 1 (inclusive) that indicates how likely an event is to occur
• Definition: n(A) – number of simple events in A• Definition: Theoretical probability of A – p(A)
= n(A)/n(S) = (# outcomes of A) ÷ (# outcomes in the equally likely sample space)
5.1 Basic Probability Ideas
• Law of Large Numbers – as the number of experiments increases without bound, the proportion of a certain event approaches a theoretical probability
• The Law of Large Numbers does not say: If you throw a coin and get heads 10 times, that your probability of getting tails increases. P(heads) stays at 1/2
5.1 Basic Probability Ideas
• Empirical Probability – relative frequency of an event based on past experience
p(A) = (# times A has occurred) ÷ (# observations)
5.1 Basic Probability Ideas
• Properties of Probabilities:1. 0 ≤ p(A) ≤ 1
2. p(a) = 0 A is impossible, the event will never happen
3. p(a) = 1 A is a certain event, the event must happen
4. Let a1, a2, a3,…. an be all events in a sample space, then p(a1) + p(a2) + … + p(an) = 1
5.2 Rules of Probability
• Definition: Complement of an event A – The event that A does not occur denoted AC
• Properties:1. P(A) + p(AC) = 1
2. P(AC) = 1 – p(A)
3. P(A) = 1 – p(AC)
• Odds in favor of event A – n(A):n(AC) orn(A) ÷ n(AC)
5.2 Rules of Probability
• Example: p(A) = 1/3 then p(AC) = 1 – p(A) = 2/3
odds in favor of A = p(A):p(AC) = 1/3:2/3 = 1:2
odds against A = p(AC):p(A) = 2/3:1/3 = 2:1• Probabilities from odds in favor –
odds in favor = s:f (successes to failures)p(A) = s/(s + f)p(AC) = f/(s + f)
5.2 Rules of Probability
• Joint probability tables – displays possible outcomes and their likelyhood of occurrence
• Example: Given the following table of data:
Coke Pepsi
Male 13 31
Female 22 14
5.2 Rules of Probability
• Probability table for example (total of 80 people in the sample):
Coke Pepsi
Male 13/80 = .1625 31/80 = .3875
Female 22/80 = .275 14/80 = .175
5.2 Rules of Probability
• Probability trees are useful when events do not have the same probability (there is no equally likely sample space)
• Problem solutions involving trees can become long if many branches are to be calculated (similar to the brute force method in section 4.5)
5.3 Probabilities of Unions and Intersections
• Definition: The union of two events A and B is the event that occurs if either A or B or both occur in a single experiment. The union of A and B is denoted A BExample: (rolling a die – getting an even number or a perfect square)
1
2 4
3
6
5
5.3 Probabilities of Unions and Intersections
• Definition: The intersection of two events A and B is the event that occurs if both A and B occur in a single experiment. The intersection of A and B is denoted A BExample: (rolling a die – getting an even number and a perfect square)
1
2 4
3
6
5
5.3 Probabilities of Unions and Intersections
• Definition: mutually exclusive or disjoint events – events for which A B = (where represents an event with no elements)
• If A and B are mutually exclusive, then:
1. P(A B) = 0
2. P(A B) = P(A) + P(B)
• Union Principle of Probability:
P(A B) = P(A) + P(B) - P(A B)
5.4 Conditional Probability and Independence
• Definition: The conditional probability of A given B is the probability of A occurring given that B has already occurred – denoted P(AB)When outcomes are equally likely:
P(AB) = n(AB)n(B)
• Conditional Probability Formula (outcomes not necessarily equally likely)
P(AB) =P(AB)
P(B)
5.4 Conditional Probability and Independence
• Multiplication Principal:P(A B) = P(B) P(AB)
• Tree diagrams – useful for conditional probability because each section of a branch is a probability conditional by the previous branches
5.4 Conditional Probability and Independence
• Independence: Two events A and B are said to be independent if the occurrence of A does not affect P(B) and vice versa.
A & B are independent if: P(AB) = P(A) or P(BA) = P(B)
• Multiplication Principle for Independent Events:
A & B are independent events P(A B) = P(A) P(B)
5.5 Bayes’ Formula
• Bayes formula for n disjoint events:
P(AiB) = P(Ai) P(BAi)
P(A1) P(BA1) + P(A2) P(BA2) + … + P(An) P(BAn)
5.6 Permutations and Combinations
• Multiplication Principle – given a tree with the number of choices at each branch being m1, m2, m3, … mn, then the number of possible occurrences is:
m1 m2 m3 … mn
5.6 Permutations and Combinations
• Permutations: The number of arrangements of r items from a set of n items.
Note: Order matters.
nPr =n!
(n – r)!
5.6 Permutations and Combinations
• Combinations: The number of subsets of r items from a set of n items.
Note: Order does not matter.
nCr =n!
(n – r)! r!
5.6 Permutations and Combinations- summary of counting formulas
With replacement(order matters)
Without replacement
Order matters(arrangements)
Order doesnot matter(subsets)
Multiplicationprincipal
Permutation Combination
5.7 Probability and Counting Formulas
• Example: A bag contains 4 red marbles and 3 blue marbles. Find the probability of selecting:
a. Two red marbles
b. Two blue marbles
c. A red marble followed by a blue marble
P(2 red) = # ways to pick 2 red
# ways to pick any 2 marbles= 4C2
7C2
= 6/21 = 2/7
5.7 Probability and Counting Formulas
P(2 blue) = # ways to pick 2 blue
# ways to pick any 2 marbles
= 3C2
7C2
= 3/21 = 1/7
P(red then blue) = chance of picking red on first chance of picking blue on second
= 4/7 3/6 = 2/7
5.7 Probability and Counting Formulas
• Birthday Problem: Suppose there are n people in a room. Find the formula for the probability that at least two people have the same birthday.
Note: P(at least two birthdays the same) = 1 – P(no two birthdays the same)
# ways for n people to have birthdays = 365n
# ways for for n birthdays without repeats = 365Pn
answer = 1 – (365Pn 365n)
5.8 Expected Value
• Expected Value – for a given sample space with disjoint outcomes having probabilities p1, p2, p3, … pn and a value (winnings) of x1, x2, x3, … xn , then the expected value of the sample space is:
x1 p1 + x2 p2 + x3 p3 +…. xn pn
• Definition: A game is said to be fair if the cost of participating equals the expected winnings.– Expected winnings < cost unfair to participant– Expected winnings > cost unfair to organizers
5.9 Binomial Experiments
• Definition: Binomial Experiment1. The same trial is repeated n times.
2. There are only 2 possible outcomes for each trial – success or failure
3. The trials are independent so the probability of success or failure is the same for each trial.
5.9 Binomial Experiments
• Binomial Probabilities:P(x successes) = nCr px (1-p)n-x with x = 0,1,2,…,n
where n is the number of trials, p is the probability of success, and x is the number of successful trials
• Expected value of a binomial experiment = n pnote: The most likely outcome “tends to be close to” the expected value.
Recommended