课程名称︰机率导论
课程性质︰必修
课程教师︰陈  宏
开课学院:理学院
开课系所︰数学系
考试日期(年月日)︰2013/03/28
考试时限(分钟):2:15-3:10pm
试题 :
                             Introductory Probability
                                      Quiz 1
                        Thursday 2:15-3:10pm, March 28th, 2013
1. (20points) Let X_n have a geometric distribution with p = λ/n in which
   P(X = x) = p(1-p)^x an x = 0,1,…..
   (a) (10 points) Compute P(X_n/n > x).
   (b) (10 points) determine lim  P(X_n/n > x) in terms of λ.
                            n→∞
2. (20 points) The symmetric difference between two events, A and B say, is
                          c           c
   defined to be A△B = (A ∩B)∪(A∩B ). Show that P(A△B)=P(A)+P(B)-2P(A∩B).
3. (25 points) Prove the following version of Stirling's
                          n!
   formula: lim  ────────── exists ane it is a finite number.
           n→∞  n^[(n+1)/2]exp(-n)
                            n!
   Write d_n = log(──────────).
                    n^[(n+1)/2]exp(-n)
   (a) (15 points) Compute d_n - d_n+1 and show that
                                0 <  d_n - d_n+1 < 1/12n - 1/[12(n+1)].
       Hint: The series expansion of the function log[(1+t)/(1-t)] near t=0
             might help.
   (b) (10 points) Show that the lim d_n exists and finite.
                                n→∞
4. (20 points) Let X be a uniformly chosen number from the set {1,2,3} and Y is
   an independent random number uniformly chosen from {1,2}.
   (a) (10 points) Find the distribution of Z = XY. (i.e. find the probability
       mass function.)
   (b) (10 points) Find the distribution of W = cos(2πZ/3).
5. (15 points) An insurance company insures 3000 people, each of whom has a
   1/1000 chance of an accident in one year. Use the Poisson approximation to
   compute the probability there will be at most 2 accidents.
6. (20 points) Let A_1,A_2,…,A_n be events. Show that
    n
   Σ P(A_i)  -   Σ    P(A_i∩A_j)  +    Σ     P(A_i∩A_j∩A_k)
   i=1         1≦i<j≦n              1≦i<j<k≦n
                                     n           n
                               ≧ P(∪  A_i) ≧ Σ A_i -    Σ   P(A_i∩A_j).
                                    i=1         i=1     1≦i<j≦n