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Probability and combinatorics. Notes from www.math.uah.edu/stat/ Virtual Laboratories in Probability and Statistics. Outcome: . #spots: . Head . 1 . 2 . Tail . 3 . 4 . 5 . 6 . Probability: . Probability: . 1/6 . 1/2 . 1/6 . 1/2 . 1/6 . 1/6 . 1/6 . 1/6 .
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Probability and combinatorics Notes from www.math.uah.edu/stat/ Virtual Laboratories in Probability and Statistics
Outcome: #spots: Head 1 2 Tail 3 4 5 6 Probability: Probability: 1/6 1/2 1/6 1/2 1/6 1/6 1/6 1/6 Discrete probability distributions List of mutually exclusive and exhaustive outcomes of some process and their probabilities Example - 1 coin toss Example - 1 fair die throw This is an example of a discrete uniform distribution
Atoms 1 2 . . . N Probability: 1/N 1/N 1/N 1/N 1/N 1/N Uniform distribution and counting • P(A) = N(A)/N • e.g. A = get less than 3 when you roll a dice • P(A) = P(1) + P(2) = 2/6 • So calculation of probabilities counting • But counting can be complicated. Need rules
Multiplication rule for ordered events • Flying from LA to NY and back • How many airline combinations are possible? • E.g. (American, United), (United, American), (United, United) • These are called ordered pairs • Rule: If the first element of the pair can happen in n1 ways and for each of these ways, the second element can happen n2ways, then the total no. of possibilities is n1 n2
Probabilities using product rule • A license number consists of two letters (uppercase) followed by five digits. • How many different license numbers are there? • If a license number is chosen at random, find the probability that the digits are all less than 5. • Total license nos. = 26*26*105 • Valid license nos.=26*26*45 • P(all digits < 5) = Valid license nos./Total license nos. • = (4/10)5 = 0.01024
Obstetricians and paediatricians • Family needs to choose from 2 clinics: • Clinic 1 has 2 obs and 2 paed • Clinic 1 has 2 obs and 3 paed • How many combos are possible? • Unequal nos. of possibilities
Tree diagrams P1 P2 Total ways = 2+2+3+3 P1 O1 O2 P1 O3 P3 P4 P5 O4 P3 P4 P5
Permutations • Sample space S with n elements • Want to choose an ordered sequence of k • No of ways possible= nPk • nPk= n(n-1)(n-2)…(n-k+1) • In statistics this is called sampling without replacement • A special form of pigeonhole principle
Arrangement problem • Twelve books, consisting of 5 math books, 4 science books • and 3 history books are randomly arranged on a bookshelf. • How many outcomes are there? • Find the probability that the books of each type are together. • Find the probability that the math books are together.
Combinations • Consider a sample space S with n elements. A combination of size k from S is an (unordered) subset{x1, x2,..., xk}of S with k (distinct) elements • C(n,k) = nPk/k! • This is also sampling without replacement, but order is ignored
Games of chance A poker hand consists of 5 cards dealt without replacement and without regard to order from a deck of 52 cards. • Show that the number of poker hands is 2,598,960. • Find the probability that a random poker hand is a full house (3 cards of one kind and 2 of another kind) • Find the probability that a random poker hand has 4 of a kind. 0.001440576 0.000240096
Independence • Events X and Y are said to be independent if the probability that X occurs is not affected by whether or not Y has occurred. • Examples of independent events: • It can be shown that independence implies: • P(X and Y) = P(X)P(Y) • (and not otherwise) • This is called the multiplication rule for independent events.
Example: 2 security guards • Guard 1: answers pager 80% of the time • Guard 2: answers pager 50% of the time • What is the probability • i) they both respond to a call? • ii) at least one responds to a call? • X= event guard 1 answers; Y= guard 2 answers • What can we say about X and Y?
DISCRETE DISTRIBUTIONS Example - Family of 3 children. Let X = number of girls Possible values: X = 3 GGG X = 2 GGB GBG BGG X = 1 BBG BGB GBB X = 0 BBB Assume the 8 outcomes are equally likely so that x 0 1 2 3 P(X = x) 1/8 3/8 3/8 1/8 P(X x)
Probability Distribution of a Discrete r.v. • The probabilities may be written as: • P(Xi=xi) is also referred to as the density function f(x) • The cumulative distribution function (c.d.f.) is defined as
Example - Bernoulli trials Each trial is an 'experiment' with exactly 2 possible outcomes, "success" and "failure" with probabilities p and 1-p. Let X = 1 if success, 0 if failure Probability distribution is x 0 1 P(X = x) p 1-p • Results for Bernoulli trials can be simulated using S-PLUS • e.g. simulate results of a drug trial drug, success (cure) has probability p = 0.3 for each patient, 100 patients in trial. • result _ rbinom(100, size=1, prob=p) • result is a 100 vector that looks like 1,0,0,1,0,1,…...
Example-Binomial Experiment • Generalisation of Bernoulli trials • X ~ Bin(n,p) X = # of successes in n Bernoulli trials • e.g. X = # of heads in 10 tosses of a coin, n = , p = • e.g. X = # of boys in a family of 5 children, n= , p = • e.g. X = # of sixes in 100 rolls of a dice, n= , p= • possible values for X = • probability distribution for X (q = 1-p) • P(X = k) =
Shape of the Binomial Distribution • The shape of the binomial distribution depends on the values of n and p. • probdistr_ dbinom(x=0:n, size=n, prob=p)
Example - 2 dice are thrown Let X denote the sum of the results. Outcomes: 11 21 31 41 51 61 12 22 32 42 52 62 13 23 33 43 53 63 14 24 34 44 54 64 15 25 35 45 55 65 16 26 36 46 56 66 Assume the 36 outcomes are equally likely so each has probability = 1/36 Possible values of X are 2, 3, ... , 12 e.g. P(X = 4) = P(1,3 or 2,2 or 3,1) = 3/36 . The probability distribution is x 2 3 4 . . . 10 11 12 P(X=x) 1/36 2/36 3/36 . . . 3/36 2/36 1/36