Probability and Statistical Inference. Robert Bartoszynski

      to be read “ factorial.” We have therefore

(3.3)

      For a reason that will become apparent later, we adopt the convention

      (3.4)

      Example 3.4

      The letters I, I, I, I, M, P, P, S, S, S, S are arranged at random. What is the probability that the arrangement will spell MISSISSIPPI?

      Solution

      We can solve this problem treating the choices of consecutive letters as “operations.” The first operation must give the letter M; hence, there is only one way of choosing it. The next letter (out of the remaining 10) must be an I, and it can be selected in four ways. Proceeding in this way, the sequence of consecutive 11 choices leading to the word MISSISSIPPI can be performed in ways, which equals . On the other hand, the total number of ways one can perform the operations of consecutively choosing letters from the set is . Consequently, the required probability equals

(3.5)

      In this solution, the letters are regarded as distinguishable, as if we had four letters , labeled and , and similarly for the other letters. In this case, the numerator and denominator are, respectively, the number of ways one can order the set of distinguishable letters so as to form the word MISSISSIPPI and the total number of orderings. Alternatively, one can regard the identical letters as indistinguishable, and in this case, we have only one way of ordering them so as to spell the required word, and a total of distinguishable ways of ordering these letters. Indeed, the denominator here represents the number of ways of permuting letters so as to leave the arrangement invariant. Now,

which is the same as (3.5).

      Example 3.5 Birthday Problem

      The following problem has a long tradition and appears in many textbooks. If randomly chosen persons attend a party, what is the probability that none of them will have a birthday on the same day?

      Solution

Here we make the following assumption: (1) all years have 365 days (i.e., leap years are disregarded), (2) each day is equally likely to be a birthday of a person (i.e., births occur uniformly throughout the year), and (3) no twins attend the party. To compute , we must find the number of all possible ways in which birthdays can be allocated to people, and the number of such allocations in which birthdays do not repeat. The first number is (by Theorem 3.2.3), while the second number is (assuming ; if , we must have at least one birthday repeating, so ). Thus, for , we have

      As a first approximation, neglecting all products which have denominators of order or higher, we can take

(3.6)

      This approximation works quite well for small . Now since , we have

(3.7)

      It is interesting that for a repeated birthday is about as likely as no repetition. The smallest for which is less than 0.01 is 56.

      Problems

      1 3.2.1 A certain set contains distinct elements. Find if the number of: (i) All possible permutations of length 2 equals 90. (ii) Permutations of length 3 is 10 times larger than the number of permutations of length 2.

      2 3.2.2 A skyscraper is 40 stories tall. Five people enter the elevator on the first floor. Assuming each person is equally likely to get off at any of the 39 floors what is the probability that all people will get off at different floors? Find the exact value, and then derive and compute the approximations analogous to (3.6) and (3.7).

      3 3.2.3 A two letter code is to be formed by selecting (without replacement) the letters from a given word. Find the number of possible codes if the word is: (i) CHART. (ii) ALOHA. (iii) STREET.

      4 3.2.4 Determine the number of 0s at the end of 16! and 27!.

      5 3.2.5 Seated at random in a row of seats are people, among them John and Mary. Find the probability that: (i) John sits next to Mary. (ii) John sits next to Mary on her right. (iii) John sits somewhere to the right of Mary. (iv) John and Mary sit exactly two seats apart.

      6 3.2.6

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