The Binomial DistributionExample. Suppose we have a large number of cards each markedwith one of the letters A,B,C,D,E. I draw five cards (indepen-dently) from this deck and ask someone to guess which letter ison each of them. What is the probability that they get exactlytwo of them correct?Context: This is a model for an ESP experiment (p. 292 oftext). If someone guesses the right answer more often than theyshould do by chance, this is sometimes taken as evidence of ESP.However, maybe they just got lucky — we can assess this betterif we know the probabilities of different outcomes that mightoccur by chance.1In this case let’s label the possible outcomes of the individualexperiments either S (success) or F (failure).How many ways can we do this five times and get exactly twoS’s?SSFFF SFSFFSFFSF SFFFSFSSFF FSFSFFSFFS FFSSFFFSFS FFFSSThere are 10 ways to do it. Each of these has the same proba-bility, which is 0.2 × 0.2 × 0.8 × 0.8 × 0.8 = .02048.Therefore the answer is 10 × .02048 = .2048.2This is not an especially small probability. But suppose the per-son got it right 4 times out of 5. In this case the possible ordersareSSSSFSSSFSSSFSSSFSSSFSSSSFive possible outcomes, each with a probability 0.2 × 0.2 × 0.2 ×0.2 × 0.8 = .00128.The overall probability is 5 × .00128 = .0064.This seems small enough to be suspicious — maybe he or shereally does have ESP!3The MathSuppose there are n experiments, and the probability that some-one gets the right answer on any given experiment is p. So in thefirst example above, n = 5 and p = 0.2. Let X be the numberof correct results — this is a random variable (discrete). Theformula (page 293 of the text) saysP (x) =n!x!(n − x)!px(1 − p)n−x.Here n! is a special symbol known as factorial n. Factorial meansyou multiply together all the numbers from 1 to n. So5! = 1 × 2 × 3 × 4 × 5 = 120.4P (x) =n!x!(n − x)!px(1 − p)n−x.The second part of the formula gives the probability of a specificsequence of S’s and F’s — so if there are exactly x S’s andn − x F’s, we calculate the probability as p × p × p × ... (x times)multiplied by (1 − p) × (1 − p) × (1 − p) × ... (n − x times) for ananswer of px(1 − p)n−x.5P (x) =n!x!(n − x)!px(1 − p)n−x.The first part of the formula is the number of possible ways ofarranging the S’s and F’s. For example, with n = 5 and x = 2we findn!x!(n − x)!=5!2!3!=5 × 4 × 3 × 2 × 12 × 1 × 3 × 2 × 1=12012= 10exactly as we got before by counting. However if n and x werelarge it would be very tedious to write out all the possible com-binations of S’s and F’s, whereas this formula is relatively easyto apply.6So in the case n = 5, x = 2, p = 0.2 we haveP (x) =5!2!3!× 0.22× 0.83= 10 × 0.04 × 0.512= 0.2048.Or with n = 5, x = 4, p = 0.2 we haveP (x) =5!4!1!× 0.24× 0.81=5 × 4 × 3 × 2 × 14 × 3 × 2 × 1 × 1× 0.0016 × 0.8= 5 × 0.0016 × 0.8=
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