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From: Natalie Wong on 5 Oct 2007 07:54
I have difficulty in doing my exercise, I don't have any ideas to find out the ans of 10(b)(ii), could you help me?

10. (a) Suppose that each of 1000 people in a company independently has a certain disease
with probability 0.005.
i. If x is the number people having the disease, argue that x is a binomial random
variable.
ii. What is the probability that at least one person has the disease?
iii. What is the probability that more than one person has the disease, if we know that
there is at least one person having the disease?
iv. One of the 1000 people is John, who knows that he has the disease. What does
John think is the probability that more than one person (including himself) has
the disease?
(b) Suppose that the disease will show up in a blood test and there are two methods to
perform the test:
[Method 1] Each person is given the test separately.
[Method 2] Ten groups of 100 people are formed. The blood samples of 100 people in a
group are pooled, mixed, and then tested. If the test is positive, implying that at least
one person in the group has the disease, all 100 people must be tested separately. We
do this for all the ten groups.
i. What is the expected number of blood tests performed for the 1000 people in
method 1?
ii. In method 2, if y is the number of blood tests performed in a single group of 100
people, what is the expected value of y?
iii. Which method gives us the smaller expected number of tests performed?
( Hint: If y1, · · ·, y10 are 10 random variables, then E[y1 + · · · + y10] = E[y1] +
· · · + E[y10]. )

my ans to the parts above 10b)ii):
a) ii) 0.993
iii) 0,967
iv) 0.993
b) i) 1000
From: Ray Vickson on 5 Oct 2007 15:28
On Oct 5, 8:54 am, Natalie Wong <cottonsw...(a)hotmai.com> wrote:
> I have difficulty in doing my exercise, I don't have any ideas to find out the ans of 10(b)(ii), could you help me?
>
> 10. (a) Suppose that each of 1000 people in a company independently has a certain disease
> with probability 0.005.
> i. If x is the number people having the disease, argue that x is a binomial random
> variable.
> ii. What is the probability that at least one person has the disease?
> iii. What is the probability that more than one person has the disease, if we know that
> there is at least one person having the disease?
> iv. One of the 1000 people is John, who knows that he has the disease. What does
> John think is the probability that more than one person (including himself) has
> the disease?
> (b) Suppose that the disease will show up in a blood test and there are two methods to
> perform the test:
> [Method 1] Each person is given the test separately.
> [Method 2] Ten groups of 100 people are formed. The blood samples of 100 people in a
> group are pooled, mixed, and then tested. If the test is positive, implying that at least
> one person in the group has the disease, all 100 people must be tested separately. We
> do this for all the ten groups.
> i. What is the expected number of blood tests performed for the 1000 people in
> method 1?
> ii. In method 2, if y is the number of blood tests performed in a single group of 100
> people, what is the expected value of y?

In a group of 100, 0 additional tests are performed if there are 0
diseases in the group; otherwise, 100 additional tests are performed.
(Note: if the "group test" is of a different sort from the individual
tests, it is necessary to look at the "additional" tests, as I have
done. Otherwise---if they are the same---the numbers of tests are 1
and 101.) This holds for each group.

R.G. Vickson

> iii. Which method gives us the smaller expected number of tests performed?
> ( Hint: If y1, · · ·, y10 are 10 random variables, then E[y1 + · · · + y10] = E[y1] +
> · · · + E[y10]. )
>
> my ans to the parts above 10b)ii):
> a) ii) 0.993
> iii) 0,967
> iv) 0.993
> b) i) 1000


From: matt271829-news on 5 Oct 2007 18:20
On Oct 5, 4:54 pm, Natalie Wong <cottonsw...(a)hotmai.com> wrote:
> I have difficulty in doing my exercise, I don't have any ideas to find out the ans of 10(b)(ii), could you help me?
>
> 10. (a) Suppose that each of 1000 people in a company independently has a certain disease
> with probability 0.005.
> i. If x is the number people having the disease, argue that x is a binomial random
> variable.
> ii. What is the probability that at least one person has the disease?
> iii. What is the probability that more than one person has the disease, if we know that
> there is at least one person having the disease?
> iv. One of the 1000 people is John, who knows that he has the disease. What does
> John think is the probability that more than one person (including himself) has
> the disease?
> (b) Suppose that the disease will show up in a blood test and there are two methods to
> perform the test:
> [Method 1] Each person is given the test separately.
> [Method 2] Ten groups of 100 people are formed. The blood samples of 100 people in a
> group are pooled, mixed, and then tested. If the test is positive, implying that at least
> one person in the group has the disease, all 100 people must be tested separately. We
> do this for all the ten groups.
> i. What is the expected number of blood tests performed for the 1000 people in
> method 1?
> ii. In method 2, if y is the number of blood tests performed in a single group of 100
> people, what is the expected value of y?
> iii. Which method gives us the smaller expected number of tests performed?
> ( Hint: If y1, · · ·, y10 are 10 random variables, then E[y1 + · · · + y10] = E[y1] +
> · · · + E[y10]. )
>
> my ans to the parts above 10b)ii):
> a) ii) 0.993
> iii) 0,967
> iv) 0.993
> b) i) 1000

Bonus question: given n people, and an independent probability, p,
that each has the disease, which testing procedure minimises the
expected number of tests needed to determine which individuals have
the disease? Blood samples can be mixed and tested together in any
combination at any stage.

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