Discrete Probability DistributionsDiscrete Uniform DistributionSlide 3Slide 4Bernoulli DistributionSlide 6Binomial DistributionSlide 8Slide 9Slide 10Slide 11Slide 12Slide 13Multinomial DistributionSlide 15Slide 16Hypergeometric DistributionSlide 18Slide 19Slide 20Slide 21Slide 22Slide 23Multivariate HypergeometricSlide 25Geometric DistributionSlide 27Slide 28Negative Binomial DistributionSlide 30Slide 31Poisson DistributionSlide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 391Discrete Probability DistributionsDiscrete Uniform DistributionBinomial & Multinomial DistributionsHypergeometric Distribution Geometric DistributionNegative Binomial DistributionPoisson Distribution & Poisson Process2Discrete Uniform DistributionConsider a r.v. X that takes on k possible values: x1, x2, …, xk•The p.m.f. depends on the parameter kExpected ValueVariance kXxxxxkkxf ,,,for 1);(21kiixkXE11][21221])[(1])[(1)( XExkXExkXVkiikii3Discrete Uniform DistributionExample: Roll a six-sided die.6,5,4,3,2,1for 61)(  xxfX27621)654321(61][ XE12351214718244969127)654321(61)(2222222XV4Discrete Uniform Distributionxf(x)0 1 2 3 4 5 6 71/65Bernoulli DistributionConsider a r.v. X with exactly two possible outcomes•0 = “failure” or 1 = “success”Define p = P[success] = P[X = 1]•The p.m.f. depends on the parameter pExpected valueVariance1,0for )1();(1xpppxfxxXpXE ][)1()( ppXV 6Bernoulli DistributionExample: Toss a coin•0 = “tails” or 1 = “heads”•p = 1/21,0for 21212121;1xxfxxX21][ XE412121)( XV7Binomial DistributionOne iteration of a Bernoulli experiment is called a Bernoulli trialConduct n independent Bernoulli trialsLet r.v. X be the # of successes in n trials•The p.m.f. depends on the parameters n and pExpected valueVariancenxppxnpnxbxnx,,2,1,0for )1(),;( npXE ][)1()( pnpXV 8Binomial DistributionCalculating probabilities using Table A.1•The tabulated c.d.f. isCalculating probabilities using Excel function•p.m.f.: BINOMDIST(x, n, p, 0)•c.d.f. : BINOMDIST(x, n, p, 1)xtpntbxXPpnxB0),;(][),;(),;1(),;(]1[][][][][pnaBpnbBaXPbXPaXPbXPbXaP9Binomial DistributionExample: Count # of defectives•0 = “nondefective” or 1 = “defective”•n = 10 manufactured parts, p = P[defective] = 0.1E[X] = (10)(0.1) = 1 defectiveV(X) = (10)(0.1)(0.9) = 0.92 10 210[ 2] (2;10,0.1) (0.1) (0.9)2(2;10,0.1) (1;10,0.1)0.9298 0.7361 0.1937P X bB B-� �= = =� �� �= -= - =10Binomial DistributionExample: Mass production process•Estimated 6% defective rate.•Randomly sample 50 units each week.(a) What is the r.v. and its distribution? X = # defectives in 50 units ~ Bin(n = 50, p = 0.06)(b) What is the probability of 3 defectives?3 50 3[ 3] (3;50,0.06)50(0.06) (1 0.06)3P X b-= =� �= -� �� �11Binomial DistributionExample: Mass production process (cont’d) X = # defectives in 50 units ~ Bin(n = 50, p = 0.06)(c) What is the probability of more than 3?303500[ 3] 1 [ 3] 1 (3;50,0.06)1 ( ;50,0.06)501 (0.06) (1 0.06)tt ttP X P X Bb tt=-=> = - � = -= -� �= - -� �� ���12Binomial DistributionExample: n = 10, p = 0.513Binomial DistributionExample: n = 10, p = 0.314Multinomial DistributionEach independent trial can result in k possible outcomes E1, E2, … , Ek •Outcome Ei has probability pi•Xi = # times outcome Ei occurs•n = total # of trials•The joint p.m.f. iswhere andnxkii1kxkxkkkppxxnppnxxf 11111,,),,,;,,(11kiip15Multinomial DistributionExample: An urn contains 12 colored balls, of which 5 are white, 4 are black, and 3 are red. Sample 7 balls with replacement. •n = 7 trials, k = 3 possible outcomes•E1 = white ball, with probability p1 = 5/12•E2 = black ball, with probability p2 = 4/12•E3 = red ball, with probability p3 = 3/12where x1 + x2 + x3 = 7321123124125,,7123,124,125,7;,,321321xxxxxxxxxf16Multinomial DistributionMachine Breakdown Example•S = {E = electrical, M = mechanical, U = misuse}•P(E) = 0.2, P(M) = 0.5, P(U) = 0.3 X1 = # breakdowns due to electrical failure X2 = # breakdowns due to mechanical failure X3 = # breakdowns due to operator misuseWhat is the probability of 2 electrical failures, 4 mechanical failures, and 3 misuses?( )( ) ( ) ( )2 4 31 2 39, , ;9,0.2,0.5,0.3 0.2 0.5 0.32,4,3f x x x� �=� �� �17Hypergeometric DistributionWe have N items, of which k are “successes”Sample n items without replacementLet r.v. X be the # of successes in the sample•The p.m.f. depends on the parameters N, n, and kExcel function•p.m.f.: HYPGEOMDIST(x, n, k, N)nxnNxnkNxkknNxh ,,2,1,0for ),,;( 18Hypergeometric DistributionExpected valueVarianceBinomial Approximation when n is small compared to N•h(x; N, n, k)  b(x; n, p) with•E[X] = np•V(X) =NnkXE ][NkNknNnNXV 11)(Nkp  )1()1(1pnppnpNnN19Hypergeometric DistributionExample: Batch of 100 parts has 10 defectives. Sample 5 parts without replacement.•N = 100, n = 5, and k = 10Binomial Approximation0702.05100390210)10,5,100;2(]2[  hXP0729.09185.09914.0)1.0,5;1()1.0,5;2()1.0,5;2()10,5,100;2( BBbh20Hypergeometric DistributionExample (cont’d): N = 100, n = 5, and k = 10Binomial Approximationdefectives 5.0100)10)(5(][ NnkXE4318.010010110010)5(11005100)( XV45.0)9.0)(1.0)(5()1()(  pnpXV1.010010pdefectives 5.0)1.0)(5(][ npXE21Hypergeometric DistributionExample: Mass production process•A production batch has 30 units.•Assume there are 5 bad units in the batch.•Randomly sample 4 units.(a) What is