Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITYIn calculating the number of permutations of the various objects we have sofar assumed that the objects are sampled without replacement – i.e. once an objecthas been drawn from the set it is put aside. As mentioned previously, however,we may instead replace each object before the next is chosen. The number ofpermutations of k objects from n with replacement may be calculated very easilysince the first object can be chosen in n different ways, as can the second, thethird, etc. Therefore the number of permutations is simply n k . This may also beviewed as the number of permutations of k objects from n where repetitions areallowed, i.e. each object may be used as often as one likes.◮Find the probability that in a group of k people at least two have the same birthday(ignoring 29 February).It is simplest to begin by calculating the probability that no two people share a birthday,as follows. Firstly, we imagine each of the k people in turn pointing to their birthday ona year planner. Thus, we are sampling the 365 days of the year ‘with replacement’ and sothe total number of possible outcomes is (365) k . Now (for the moment) we assume that notwo people share a birthday and imagine the process being repeated, except that as eachperson points out their birthday it is crossed off the planner. In this case, we are samplingthe days of the year ‘without replacement’, and so the possible number of outcomes forwhich all the birthdays are different is365 365!P k =(365 − k)! .Hence the probability that all the birthdays are different is365!p =(365 − k)! 365 . kNow using the complement rule (30.11), the probability q that two or more people havethe same birthday is simply365!q =1− p =1−(365 − k)! 365 . kThis expression may be conveniently evaluated using Stirling’s approximation for n! whenn is large, namelyn! ∼ √ ( n) n2πn ,eto give( ) 365−k+0.5 365q ≈ 1 − e −k .365 − kIt is interesting to note that if k = 23 the probability is a little greater than a half thatat least two people have the same birthday, and if k = 50 the probability rises to 0.970.This can prove a good bet at a party of non-mathematicians! ◭So far we have assumed that all n objects are different (or distinguishable). Letus now consider n objects of which n 1 are identical and of type 1, n 2 are identicaland of type 2, ...,n m are identical and of type m (clearly n = n 1 + n 2 + ···+ n m ).From (30.30) the number of permutations of these n objects is again n!. However,1134