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Statistics Review for CM 160/260. Topics. Counting Permutations Combinations Set Theory Probability & Conditional Probability Independence Bayes Theorem Random Variables Discrete vs. Continuous Probability Density Function ( pdf ) Simple discrete, Uniform, Binomial,…
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Topics • Counting • Permutations • Combinations • Set Theory • Probability & Conditional Probability • Independence • Bayes Theorem • Random Variables • Discrete vs. Continuous • Probability Density Function (pdf) • Simple discrete, Uniform, Binomial,… • Expectation & Variance
Reader pg 142–52, 437 Counting • Experiments? • Coin toss, die roll, DNA sequencing • Basic Principle of Counting: • For r number of experiments • First Experiment can have n1 possible outcomes, second experiment n2, third n3,… • Total number of possible outcomes is n1* n2* n3*…* nr
Factorials m! = m(m-1)(m-2)(m-3)…*3*2*1 5! = 5*4*3*2*1 = 120 0! = 1 1! = 1 100! = 100*99*98*…*3*2*1 = ~9 x 10157
Permutations • Groups of Ordered arrangements of things • How many 3 letter permutations of the letters a, b, & c are there? • abc, acb, bac, bca, cba, cab 6 total • Can use Basic Principle of Counting: • 3*2*1 = 6 • General Formula: • n = total number of things • k = size of the groups your taking • k<n • 3!/(3-3)! = 6
Permutations • What if some of the things are identical? • How many permutations of the letters a, a, b, c, c & c are there? Where n1, n2, … nr are the number of objects that are alike 6! / (3!2!) = 60
Combinations • Groups of things (Order doesn’t matter) • How many 3 letter combinations of the letters a, b, & c are there? 1: abc • How many 2 letter combinations of the letters a, b, & c are there? 3: ab, ac, bc • ab = ba; ac = ca; bc = cb *Order doesn’t matter • General Formula: • n = total number of things • k = size of the groups your taking • k<n “n choose k”
Reader pg 129-34 Set Theory • Sample Space of an experiment is the set of all possible values/outcomes of the experiment S = {a, b, c, d, e, f, g, h, i, j} S = {Heads, Tails} S = {1, 2, 3, 4, 5, 6} E = {a, b, c, d} F = {b, c, d, e, f, g} E S F S Ec = {e, f, g, h, i, j} Fc = {a, h, i, j} E F = {a, b, c, d, e, f, g} E F = EF = {b, c, d}
S E F G Venn Diagrams
Reader pg 135 Simple Probability • Frequent assumption: All Outcomes Equally likely to occur • The probability of an event E, is simply: Number of possible outcomes in E Number of Total possible outcomes S = {a, b, c, d, e, f, g, h, i, j} E = {a, b, c, d} F = {b, c, d, e, f, g} P(E) = 4/10 P(F) = 6/10 P(S) = 1 0 < P(E) < 1 P(Ec) = 1 – P(E) P(E F) = P(E) + P(F) – P(EF)
Reader pg 165 Independence • Two events, E & F are independent if neither of their outcomes depends on the outcomes of others. • So if E & F are independent, then: P(EF) = P(E)*P(F) • If E, F & G are independent, then: P(EFG) = P(E)*P(F)*P(G)
EF S E F Conditional Probability • Given E, the probability of F is: • Similarly:
EF S E F Bayes Theorem • Solve for P(EF) to get: • Combine them to get:
S Observed Hidden Bayes Theorem • In terms we use:
S Observed Hidden Bayes Theorem • Also: • In our terms:
Reader pg 9 Bayes Theorem
Random Variables • Definition: A variable that can have different values • Each value has its own probability • X = Result of coin toss • Heads 50%, Tails 50% • Y = Result of die roll • 1, 2, 3, 4, 5, 6 each 1/6
Discrete vs. Continuous • Discrete random variables can only take a finite set of different values. • Die roll, coin flip • Continuous random variables can take on an infinite number (real) of values • Time of day of event, height of a person
Reader pg 438 Probability Density Function • Many problems don’t have simple probabilities. For those the probabilities are expressed as a function… • aka “pdf” Plug a into some function i.e. 2a2 – a3
Some Useful pdf’s • Simple cases (like fair/loaded coin/dice, etc…) • Uniform random variable (“equally likely”) For a = Heads For a = Tails
pdf of a Binomial • You will need this • Where p = P(success) & q = P(failure) • P(success) + P(failure) = 1 • n choose k is the total number of possible ways to get k successes in n attempts
Using the p.d.f. • What is the Probability of getting 3 Heads in 5 coin tosses? (Same as 2T in 5 tosses) n = 5 tosses k = 3 Heads p = P(H) = .5 q = P(T) = .5 • P(3H in 5 tosses) = p3q2 = 10p3q2 = 10*P(H)3*P(T)2 = 10(.5)3(.5)2 = 0.3125
Notice how these are Binomials… • What is the probability of winning the lottery in 2 of your next 3 tries? n = 3 tries k = 2 wins Assume P(win) = 10-7 P(lose) = 1-10-7 • P(win 2 of 3 lotto) = P(win)2P(lose) = 3(10-7)2(1-10-7) = ~ 3*10-14 • That’s about a 3 in 100 trillion chance. Good Luck!
It may be important… • Assume that Kobe stays on a hot streak, shooting a constant 66% (~2/3). What is the probability that he will make 25 of 30 field goals? n = 30 k = 25 P(score) = 2/3 P(miss) = 1/3 • P(25 for 30) = P(score)25P(miss)5 = 142506(2/3)25(1/3)5 = 0.0232
Reader pg 439 Expectation of a Discrete Random Variable • Weighted average of a random variable • …Of a function
Reader pg 439 Variance • Variation, or spread of the values of a random variable • Where μ = E[X]