The 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 270 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 = 120.1P (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.2P (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.3So 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= .0064.4AssumptionsThis formula is known as the binomial distribution. The essentialconditions for a binomial distribution are:1. Each outcome of the experiment has exactly two possibleoutcomes, which we label “success” and “failure” thoughthey could have many different interpretations (alive or dead,rain or no rain, flight arrives on time or not, etc.). Datahaving this form are called binary data.2. The experiment is repeated a number of times (n) and thedifferent experiments are independent.3. The probability of “success” is some number p (between 0and 1) and is the same for all the experiments.5Here is another example.A tennis player serves her first serve in 70% of the time. Assumeeach serve is independent of all the others. She serves the ballsix times. What is the probability that she gets(a) All 6 serves in?(b) Exactly 4 serves in?(c) At least 4 serves in?(d) No more than 4 serves in?6Solution:(a) (0.7)6= .118.(b)6!4!2!× (0.7)4× (0.3)2=6×5×4×3×2×14×3×2×1×2×1× (0.7)4× (0.3)2= 15 × .2401 × .09 = .324.(c) The probability of exactly 5 is6!5!1!× (0.7)5× 0.3= 6 × 0.16807 × 0.3 = .303. So the probability of at least 4is .324+.303+.118=.745.(d) The probability of at least 5 is .303+.118=.421 so the prob-ability of not more than 4 is 1–.421=.579.7The Normal approximationAlthough the binomial formula is faster to calculate than tryingto count all possibilities, it would still be hard to use for largesamples, say n = 100. In this case, we use an alternative ap-proach based on approximating the binomial distribution by anormal distribution.Suppose we have a binomial distribution with n trials and prob-ability of success p, and n is some large number (say, 100). Inthis situation, we are not usually interested in the exact numberof successes, but in the probability that the number will be moreor less than some given number.8Example 1 (from text): In a certain week in 1997, the police ata certain location in Philadephia made 262 car stops. Of these,207 drivers were African American. Among the whole populationof Philadelphia, 42.2% are African American. Does this prove thepolice were guilty of “racial profiling”, i.e. deliberately stoppingdrivers because they were African Americans?Assuming the traffic stops are independent and the proportionof African Americans driving at this particular location is thesame as the proportion in the whole city, this corresponds to therandom variable X (number of African Americans among thosestopped) having a binomial distribution with n = 262, p = 0.422.The question is, what is the probability that X ≥ 207 if thebinomial distribution is correct?9Note: In this case it wouldn’t make sense to try to calculate theprobability that X is exactly 207. What we’re really concernedabout is that the number is so large, so a natural question is“what is the probability that the number would have been aslarge as this by chance?” That leads us to consider X ≥ 207rather than X = 207.10A Key Formula (page 274)The binomial distribution for n trials with probability p of successon each trial has mean µ and standard deviation σ given byµ = np,σ =qnp(1 − p).11The solution proceeds by several steps:Step 1: Calculate the mean of X. This is given by the formulaµ = np = 262 × 0.422 = 110.6.Step 2: Calculate the standard deviation of X. This is given bythe formulaσ =qnp(1 − p) =√262 × 0.422 × 0.518 = 7.99.Step 3: Convert the given x value (207) to z. Soz =x − µσ=207 − 110.67.99= 12.07.Step 4: Calculate the probability associated with this z value.12The only problem with step 4 is: the number’s off the chart!The regular table only goes up to 3.49. In fact, if you look atthe little table in the bottom corner of page A2, you can seesome further numbers:z Probability3.5 .9997674.0 .99996834.5 .99999665.0 .999999713Even at z = 5, the probability to the left of z (i.e. less than 5)is more than .999999, which means that the probability to theright of z is less than .0000001. Replace z = 5 by z = 12, andthe probability of that is much smaller again.13Conclusion. The probability that we could have got this result(207 African Americans out of 262) by chance is so small that itis effectively 0. This seems to be completely convincing evidencethat the police were engaging in the practice of racial profiling.However, there are other possible explanations — for example,perhaps the proportion of African Americans driving past thisparticular checkpoint was much greater than 42.2%.Further Discussion. It is possible to compute the exact probabil-ity that X ≥ 207} in this example: the answer is 4.9 ×10−34. Togive an idea of how small a probability that is, it is roughly equiv-alent to the probability that your favorite baseball team win theWorld Series 23 times in succession! [On the assumption thatthere are 30 Major League Baseball teams, that any one of themis equally likely to win in a given year, and that results