A university department has 2 secretaries (labelled A and B) who do all the word processing required by the department. The number of misprints on a randomly sampled page typed by secretary i (i=A,B) has a Poisson distribution with Mean Ui independent of the number of misprints on any other page, where Mean Ua = 0.3 and Mean Ub = 1.2. Assume that each page is typed entirely by a single secretary. Of the typing required by the department, 75% is done by secretary A and 25% by secretary B. (a) Find the probability that a randomly sampled page typed by secretary B contains more than 1 misprint. [2 marks] (b) Calculate the overall proportion of pages produced by the department that contain no misprints. [3 marks] (c) Suppose that a randomly sampled page produced by the department is found to contain 2 misprints. Given this information, calculate the probability that this page was typed by secretary A, Hence which secretary is most likely to have typed the page concerned? [5 marks] A book typed entirely by secretary A consists of 200 pages. (i) Let X be the number of pages in the book that contain no misprints. Name the (exact) distribution of X. Find approximately the probability that at least 150 pages in the book are without misprints. [5 marks] (ii) Find approximately the probability that the book contains at most 50 misprints in total. Attempt At soultions: a) Using P(x=K) = (e^-u* u^k)/k! , I get the probability for X=0, X=1 add them and subtract from 1? b) 0.25* Prob X=0 from part 1 + same thing for Sec A * 0. 75 c) Conditional prob. Im happy with this one. d) i) I think this is a normal distribution. Now if its a normal distribution, mean number of errors per page = 0.3 so number of errors in 200 page will be 60. But how do I find The prob of 150+151...200 :S I am very confused and would love some help as my exam is on wednesday! Thanks!!