Course Notes - Department of Mathematics and Statistics

Pr(D ∩ T)Pr(D| T) =Pr(T)P r(D| T) = 0.0000950.0599P r(D | T) = 0.00158So from this result Fred should not be very worried at all, as thereis only a 0.2% chance of him actually having the disease given apositive test result. Note this very different to the 95% chance ofreturning a positive test given you have the disease. Having a betterunderstanding of conditional probability can help Fred sleep a lot easier.Now to the second problem of the day, at first Fred thought it wasamazing two people out of twenty people shared a birthday, but afterremembering his Statistics class he decided to work out the probabilityof this occuring. If he could work out the probability that no one ofthe 23 shared a birthday he could work out, the probability 2 or morepeople did, by subtracting that answer from 1. (For simplicity sake wewill ignore leap years).To calculate the probability of people not sharing the same birthday,we assume each of the 23 people are an independant event so we cancombine them using the simplified multiplication rule. The probabilitythat person 1 does not share a birthday is equal to 365365since there isno one for him to clash with, the probability that person 2 does notshare a birthday is equal to 364365, since the birthday of the first personis no excluded from the possible birthdays, the probability that person3 does not share birthday becomes 363365and so on and so forth. Sothe probability that no one shares a birthday in the 23 people equals365365 × 365 364 × 363365 . . . × 343365 = 0.4927.58