To determine whether or not they have a certain disease, 90 people are to have t

Question

To determine whether or not they have a certain disease, 90 people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to group the people in groups of 10. The blood samples of the 10 people in each group will be pooled and analyzed together. If the test is negative, one test will suffice for the 10 people (we are assuming that the pooled test will be positive if and only if at least one person in the pool has the disease); whereas, if the test is positive each of the 10 people will also be individually tested and, in all, 11 tests will be made on this group. Assume the probability that a person has the disease is 0.04 for all people, independently of each other. Compute the expected number of tests necessary for each group. Expected number for each group: Compute the expected total number of tests necessary for the entire population of 90 people Expected total:

Explanation / Answer

First find the probability that one group out of the ten will test positive.

Let X be the number of people in the group of 10 who have the disease.
X has the binomial distribution with n = 10 trials and success probability p = 0.04

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen
from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n , p )

the mean of the binomial distribution is n * p = 0.4
the variance of the binomial distribution is n * p * (1 - p) =0.384
the standard deviation is the square root of the variance = ( n * p * (1 - p)) =0.61968
The test will be positive if X 1

P( X 1) = 1 - P(X = 0) = 1 -0.664833= 0.33517

== ==
For each group the number of tests = 3.3517+1=4.3517 ..........(1) Ans
Let Y be the number of groups that will test positive. Y is a binomial random variable with n = 10 trials
and successes probability 0.33517.

the expectation of Y is: 3.3517

Let T be the number of tests administered

T = 9 * Y + 9

E(T) = 9* E(Y) + 9
E(T) = 30.1653 + 9
E(T) =39.1653

We expect 39.1653~~40 tests to be given to this group off 90 people................(2) Ans

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