FrequencyExample: For drug studyWhere X = random variableDISCRETE PROBABILITYDISTRIBUTIONSDef: Random Variable—a numeric function that assigns probabilities to different events in a sample space.1. Discrete—specific values with specified probabilitiesExample: Prevalence of Prescription andNonprescription Drug Use during Pregnancy amongWomen Who Delivered Infants at a Large EasternHospitalNumber ofDrugs Used(X)Frequency ProbabilityDistributionPr(X = r)0 1425 0.34051 1351 0.32282 793 0.18953 348 0.08324 156 0.03735 58 0.01396 54 0.0128Total 4185 1.00002. Continuous—possible values cannot be enumeratedDef: Probability Mass Function or Probability Distribution—This is a rule to assign possible values for a discrete Random Variable, X. Pr (X = x) or Pr(X = r). All values of X that have positive probability. 0  Pr(X = r)  1Probability Distribution for X, the Number ofSpots on the Top Face of a Fair DieNumber of SpotsrPr(X = r)1 1/62 1/63 1/64 1/65 1/66 1/6 1Pr r)(XExpected Value of a Discrete RandomVariableE(X) or --obtained by multiplying each possible value by the probability of occurrence and then summing. Riii)x(XxμE(X)1PrWhere the xi’s are the possible values of the random variableExample: What is the expected value for thenumber of spots facing up when throwing a fairsix-sided die?Fair DiexiPr(X = xi) xi * Pr(X = xi)1 1/6 1/62 1/6 2/63 1/6 3/64 1/6 4/65 1/6 5/66 1/6 6/6E(X).)x(Xxiii53621Pr61Example: Drug Use StudyxiPr(X = xi) xi * Pr(X = xi)0 0.3405 0.00001 0.3228 0.32282 0.1895 0.37903 0.0832 0.24964 0.0373 0.14925 0.0139 0.06956 0.0128 0.0768E(X).)x(Xxiii2469160PrWe expect patients to take about 1.25 drugs on theaverage.Variance of a Discrete Random Variable—Population Variance )x(XPrμ)(xσVar(X)ii22or 22μ)x(XxVar(X)iiPrVar(X)σsd(X) Example: Variance for Single Diexi(xi - )2 * Pr(X = xi)1 (1 – 3.5)2 * 1/6 = 1.04172 (2 – 3.5)2 * 1/6 = 0.37503 (3 – 3.5)2 * 1/6 = 0.04174 (4 – 3.5)2 * 1/6 = 0.04175 (5 – 3.5)2 * 1/6 = 0.37506 (6 – 3.5)2 * 1/6 = 1.0417Var(X) = 2.91687079191682 ..sd(X) Using alternative formula91672251269125126366256166964615361661561461361261153Pr22222222612....)()()()()()().()x(XxVar(X)iiiExample: For drug study296116799167991246910128601395037340832318952322813405024691222222226022..sd(X)Therefore,..)(.)(.)(.)(.)(.)(.)(..)x(XxVar(X)iiiPrNote: About 95% of the probability mass is within 2 standard deviations of the mean of a random variable.Cumulative Distribution FunctionF(X) = Pr(X  x)Where X = random variable x = specific valueExample: Fair die—X = number of spots on topfaceX f(X) = Pr(X = x)F(X) = Pr(X  x)1 1/6 1/62 1/6 2/6 = 1/33 1/6 3/6 = 1/24 1/6 4/6 = 2/35 1/6 5/66 1/6 1Example: Drug Study—X = Number of drugstakenX f(X) =Pr(X = x)F(X) =Pr(X  x)0 0.3405 0.34051 0.3228 0.66332 0.1895 0.85283 0.0832 0.93604 0.0373 0.97335 0.0139 0.98726 0.0128 1.0000Questions from Drug Study1. What is the probability that a randomly selected woman used 2 or fewer drugs?2. What is the probability that a randomly selected woman used more than 2 drugs?3. What is the probability that a randomly selected woman used 5 or more drugs?4. What is the probability that a randomly selected woman used between 3 and 5 drugs, inclusive?PERMUTATIONSA permutation is the number of ways of selecting k items out of n, when the order is important.Example: Number of ways of arranging 2 peoplewhen 5 are available. Let’s call the 5 people A B C D EA B A C A D A EB A B C B D B EC A C B C D C ED A D B D C D EE A E B E C E DFor n = 5 people, k = 2 are chosen and order is importantThere are n people available for the first position.There are n-1 people available for the second position.k)!(nn!1)k(n*...*2)(n*1)(n*nPknwhere ! = factorialn! = n * (n – 1) * (n – 2) * … * 2 * 1For this problem,021*2*31*2*3*4*53!5!2)!(55!P25Example: Suppose we have 6 people and want toselect all 6 and order matters.7201*2*3*4*5*60!6!6)!(66!P66Def: 0! = 1 1! = 1Find:5P38P220P12COMBINATIONSNumber of ways of selecting k objects out of n when order does not matter. i.e., A B C or B A C is OK. Same three objectsare chosen.(k)!k)!(nn!knCknwhere k = 0, 1, 2, …, nknnknorCCTherefore,151*2*1*2*3*41*2*3*4*5*62)!2!(66!C151*2*3*4*1*21*2*3*4*5*64)!4!(66!C201*2*3*1*2*31*2*3*4*5*63)!3!(66!Ck-nnkn264636if n  k, and n, k are non-negative