f*********n
发帖数: 4 | 1 《Statistical Inference》(Casella & Berger) 第一章的一道习题(1.20)
My telephone rings 12 times each week, the calls being randomly distributed
among the 7 days. What is the probability that I get at leaset one call each
day?
Tha answer provided in the book is 0.2258. I couldn't find a way to get
that number. The analysis I did seems to be over-counting. However, the
monte carlo simulation I did gave the number 0.1630 (I attached the code here
, in c++), which is really strange.
Any help would be greatl | l******n
发帖数: 9344 | 2 Let P_i denote the probability that I have no call on the ith day
Then
P(UP_i)=C(7,1)(6/7)^12-C(7,2)(5/7)^12+C(7,3)(4/7)^12-C(7,4)(3/7)^12+C(7,5)(2
/7)^12-C(7,6)(1/7)^12
=.7715
So 1-P(UP_i)=.2285
distributed
each
here
【在 f*********n 的大作中提到】
: 《Statistical Inference》(Casella & Berger) 第一章的一道习题(1.20)
: My telephone rings 12 times each week, the calls being randomly distributed
: among the 7 days. What is the probability that I get at leaset one call each
: day?
: Tha answer provided in the book is 0.2258. I couldn't find a way to get
: that number. The analysis I did seems to be over-counting. However, the
: monte carlo simulation I did gave the number 0.1630 (I attached the code here
: , in c++), which is really strange.
: Any help would be greatl
| f*********n
发帖数: 4 | 3 Thanks, but I don't quite understand this. Maybe that's where I
misunderstood, but I think
P(n) := C(7,n)(1-n/7)^12
is the probability of "having no calls for at least n days". Then the
probability of "having at least one call each day" would be 1-P(1), which is
clearly not the answer.
Could you explain where I thought wrong? And why add and substract
alternatively in your solution? Many thanks. | f*********n
发帖数: 4 | 4 I think I understand now, P(n) is not the probability of "having no calls for at least n days", it indeed overcounted.
I think the alternating add and substract comes from the formular
P( U A_i ) = \sum_i{P(A_i)} - \sum_{i,j} P( A_i intersects A_j) + .......
I didn't find this formular in the first chapter of Casella&Beger's book, I guess the authors must have intended to
let their readers discover it by themselves.
is
【在 f*********n 的大作中提到】
: Thanks, but I don't quite understand this. Maybe that's where I
: misunderstood, but I think
: P(n) := C(7,n)(1-n/7)^12
: is the probability of "having no calls for at least n days". Then the
: probability of "having at least one call each day" would be 1-P(1), which is
: clearly not the answer.
: Could you explain where I thought wrong? And why add and substract
: alternatively in your solution? Many thanks.
|
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