Huge Multi Part Prob + Stats question from past paper

[hb2325]

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](d) 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. I am 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!

 

[mmwave]

a) To find the probability that a randomly sampled page typed by secretary B contains more than 1 misprint, we can use the Poisson distribution formula: P(X>k) = 1 - P(X<=k), where X is the number of misprints on a page and k is the number of misprints we are interested in (in this case, k=1). Therefore, P(X>1) = 1 - P(X<=1) = 1 - (P(X=0) + P(X=1)). We can calculate these probabilities using the Poisson distribution formula with Mean U=1.2. So, P(X>1) = 1 - ((e^-1.2* 1.2^0)/0! + (e^-1.2* 1.2^1)/1!) = 1 - (0.3012 + 0.3614) = 0.3374 or 33.74%.

b) To calculate the overall proportion of pages produced by the department that contain no misprints, we need to find the weighted average of the proportions for secretary A and B. So, the overall proportion will be (0.75*P(X=0) for secretary A) + (0.25*P(X=0) for secretary B). Using the Poisson distribution formula with Mean U=0.3 for secretary A and Mean U=1.2 for secretary B, we get (0.75*(e^-0.3* 0.3^0)/0! + 0.25*(e^-1.2* 1.2^0)/0!) = 0.75*(0.7408) + 0.25*(0.3012) = 0.5556 or 55.56%.

c) To find the probability that a page with 2 misprints was typed by secretary A, we can use Bayes' Theorem: P(A|B) = P(B|A)*P(A)/P(B), where A is the event that the page was typed by secretary A and B is the event that the page has 2 misprints. We know that P(B|A) = P(X=2) = (e^-0.3* 0.3^2)/2! = 0.0443 and P(A) = 0.75. To find P(B