Introduction to Probability and Statistics for Engineers and Scientists

164 Chapter 5: Special Random Variables

For an illustration of the use of random numbers, suppose that a medical center is
planning to test a new drug designed to reduce its users’ blood cholesterol levels. To test
its effectiveness, the medical center has recruited 1,000 volunteers to be subjects in the
test. To take into account the possibility that the subjects’ blood cholesterol levels may be
affected by factors external to the test (such as changing weather conditions), it has been
decided to split the volunteers into 2 groups of size 500 — atreatmentgroup that will be
given the drug and acontrolgroup that will be given a placebo. Both the volunteers and
the administrators of the drug will not be told who is in each group (such a test is called
adouble-blind test). It remains to determine which of the volunteers should be chosen
to constitute the treatment group. Clearly, one would want the treatment group and the
control group to be as similar as possible in all respects with the exception that members
in the first group are to receive the drug while those in the other group receive a placebo;
then it will be possible to conclude that any difference in response between the groups is
indeed due to the drug. There is general agreement that the best way to accomplish this is
to choose the 500 volunteers to be in the treatment group in a completely random fashion.
That is, the choice should be made so that each of the

( 1000

500

)
subsets of 500 volunteers is
equally likely to constitute the control group. How can this be accomplished?

*EXAMPLE 5.4d Choosing a Random SubsetFrom a set of n elements — numbered
1, 2,...,n— suppose we want to generate a random subset of sizekthat is to be chosen
in such a manner that each of the

(n

k

)
subsets is equally likely to be the subset chosen.
How can we do this?
To answer this question, let us work backwards and suppose that we have indeed
randomly generated such a subset of sizek. Now for eachj=1,...,n,weset

Ij=

{

1 if elementjis in the subset

0 otherwise

and compute the conditional distribution ofIjgivenI 1 ,...,Ij− 1. To start, note that the
probability that element 1 is in the subset of sizekis clearlyk/n(which can be seen either by
noting that there is probability 1/nthat element 1 would have been thejth element chosen,
j=1,...,k; or by noting that the proportion of outcomes of the random selection that

results in element 1 being chosen is

(

1

1

)(

n− 1

k− 1

)

/

(n

k

)

=k/n). Therefore, we have that

P{I 1 = 1 }=k/n (5.4.1)

To compute the conditional probability that element 2 is in the subset givenI 1 , note
that ifI 1 =1, then aside from element 1 the remainingk−1 members of the subset
would have been chosen “at random” from the remainingn−1 elements (in the sense that

* Optional.