5.1 Basic Probability Ideas

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Presentation transcript:

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: 0 ≤ p(A) ≤ 1 p(a) = 0  A is impossible, the event will never happen p(a) = 1  A is a certain event, the event must happen 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: P(A) + p(AC) = 1 P(AC) = 1 – p(A) P(A) = 1 – p(AC) Odds in favor of event A – n(A):n(AC) or n(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 B G Simple probability tree: Boy branches root G Girl

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  B Example: (rolling a die – getting an even number or a perfect square) 2 4 6 5 3 1

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  B Example: (rolling a die – getting an even number and a perfect square) 2 4 6 5 3 1

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: P(A  B) = 0 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(AB) When outcomes are equally likely: n(AB) n(B) P(AB) = Conditional Probability Formula (outcomes not necessarily equally likely) P(AB)P(B) P(AB) =

5.4 Conditional Probability and Independence Multiplication Principal: P(A  B) = P(B)  P(AB) 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(AB) = P(A) or P(BA) = P(B) Multiplication Principle for Independent Events: A & B are independent events  P(A  B) = P(A)  P(B)

P(A)  P(BA) + P(AC)  P(BAC) 5.5 Bayes’ Formula Bayes formula for 2 cases: P(A)  P(BA) P(AB) = P(A)  P(BA) + P(AC)  P(BAC)

P(A1)  P(BA1) + P(A2)  P(BA2) + … + P(An)  P(BAn) 5.5 Bayes’ Formula Bayes formula for n disjoint events: P(Ai)  P(BAi) P(AiB) = P(A1)  P(BA1) + P(A2)  P(BA2) + … + P(An)  P(BAn)

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. n! (n – r)! nPr =

5.6 Permutations and Combinations Combinations: The number of subsets of r items from a set of n items. Note: Order does not matter. n! (n – r)!  r! nCr =

5.6 Permutations and Combinations - summary of counting formulas Without replacement With replacement (order matters) Order does not matter (subsets) Order matters (arrangements) Multiplication principal Permutation Combination

5.7 Probability and Counting Formulas Example: A bag contains 4 red marbles and 3 blue marbles. Find the probability of selecting: Two red marbles Two blue marbles A red marble followed by a blue marble # ways to pick 2 red # ways to pick any 2 marbles = 4C2 7C2 P(2 red) = = 6/21 = 2/7

5.7 Probability and Counting Formulas # ways to pick 2 blue # ways to pick any 2 marbles P(2 blue) = = 3C2 7C2 = 3/21 = 1/7 chance of picking red on first  chance of picking blue on second P(red then blue) = = 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 Experiment The same trial is repeated n times. There are only 2 possible outcomes for each trial – success or failure 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  p note: The most likely outcome “tends to be close to” the expected value.