Probability of failure = q = 1 – pTherefore, p = 1/6Pr(X = 3) = 0.1339Binomial Distributionn independent Bernoulli trials—each with only two possible outcomesProbability of success at each trial is some constant, pProbability of failure = q = 1 – p Example: In throwing a fair die, let a “success” be a 6 on the top face.Therefore, p = 1/6q = 1 – p = 1 – 1/6 = 5/6In 5 trials, what is the probability of 5 successes?00012860777616155.p*p*p*p*pp In 5 trials, what is the probability of 5 failures?4018806555.q What is the probability of 2 successes and 3 failures?SSFFF FSSFF FFSFSSFSFF FSFSF FFFSSSFFSF FSFFSSFFFS FFSSFThis is the number of ways of selecting 2 outcomesout of 5 trials—order doesn’t matter101*2*1*2*31*2*3*4*53!2!5!25Therefore, the probability of 2 successes and 3 failures in 5 trials is16075021612536110656110253232.****qpGeneralize: Probability of k successes in n trials (order does not matter) is knkqp!k!knn!k)(XPrwhere k = 0, 1, 2, … , nand q = 1 – pThe distribution is the number of successes in n statistically independent trials, where the probability of success on each trial = p, is called the Binomial Distribution. The probability mass function isknkqpknk)(XPr, k = 0, 1 ,2, …, nFor this examples in 5 trials No. ofsuccesses Probability050qp055065611**0.4019141qp154165*61*50.4019232qp253265*61*100.1608323qp352365*61*100.0322414qp451465*61*50.0032505qp550565*61*10.0001p = 1/6, q = 5/6Note: Total = 1.0000Example: Suppose in a population. 25% of the individuals have type B blood. For a sample of 20 individuals, what is the probability that 3 have type B blood?          13390007516940015625011407525123171718192075253320203173173..*.*..**!!***..!!!)(XPrBinomial Tablespp 811-815n = 2 – 20p = 0.05, 0.10, … 0.50For this examples, n = 20, X = 3, p = 0.25Pr = 0.1339What is the probability that 3- 5 individuals inclusive have type B blood?Pr(X = 3) = 0.1339Pr(X = 4) = 0.1897Pr(X = 5) = 0.2023Pr (3  X  5) = 0.1339 + 0.1897 + 0.2023 = 0.5259What is the probability that 5 or more individuals have Type B blood?Pr(X  5) = 1 – Pr (X < 5) = 1 – Pr (X  4) = 1 – 0.1897 – 0.1339 – 0.0669 – 0.0211 – 0.0032= 0.5852Example: Suppose the probability of hypertension in 50-59 year old women is 0.40. What is the probability in a sample of 14 women of the appropriate age that 7 are hypertensive?n= 14, k = 7, p = 0.40Pr = 0.1574What is the probability that 5- 7 women inclusive are hypertensive?k = 5, Pr = 0.2066k = 6, Pr = 0.2066k = 7, Pr = 0.1574 0.5706What is the probability that 3 or more are hypertensive?Pr (X  3) = Pr (X = 3) + Pr (X = 4) + … + Pr (X = 14)= 1 – Pr (X = 0) – Pr (X = 1) – Pr (X = 2)= 1 – 0.0008 – 0.0073 –0.0317= 1 – 0.0398= 0.9602What happens if p > 0.50?Note:knnknTherefore, to use the tables,Let X be a Binomial random variable with parameters n and pLet Y be a Binomial random variable with parameters n and q = 1 - pThen,k)n(Ypqknnqpknk)(XkknknkPrPrIf p > 0.5, then q < 0.5. We can still use the tables except we look up the column for q and use the row for n - k instead of k.Example: Suppose the probability of getting a passing grade on an exam is 0.90. What is the probability that in a sample of 12 students, 9 will pass?n = 12, k = 9, p = 0.90      08520913121991299339.....)(XPrExample: In a survey of students pursuing a Master’s degree, 70% stated they expect to be promoted within 1 year after receiving their degree. For a sample of 15 students, find the probability that 6 will be promoted.n = 15, X = 6, p = 0.70      011607030915307061566996.....)(XPrSuppose we can’t use the tables.In the 1970’s approximately 26% of the US adults were overweight. Suppose we select a sample of 25 US adults, find the probability that the numberof overweight people in the sample will be:- Exactly 3- Between 4 and 5, inclusive- More than oneExpected Value and VarianceDef: n*pqpknk)x(XxE(X)knknkRiii01PrSimilarly, n*p*qqpknnpkVar(X)knknk02Note: Variance is greatest when p = 0.50. The variance decreases as p moves away from 0.50in either direction.Using EXCEL:In the 1970’s approximately 26% of the US adults were overweight. Suppose we select a sample of 25 US adults, find the probability that the numberof overweight people in the sample will be:- Exactly 3- Between 4 and 5, inclusive- More than