# Probability and Expected Value problem

• May 29th 2008, 05:09 AM
892king
Probability and Expected Value problem
Here's the question...

The blood samples of k (k > 1) randomly chosen people is tested for
the presence of virus V . Suppose that 5% of the population has this
virus, and that presence of the virus among different people is independent.
separately” method, the following alternative method is proposed:
Mix a little of each of the k blood samples and test for the presence of
virus V in the mixture.
• If virus V is not present - done! (since that means that none of
the k people has the virus)
• If virus V is present, go back and test each of the k blood samples
separately. Let the random variable Y be the number of tests
needed for the alternative methods.
(a) Find PY (y) and E(Y ).
(b) For what values of k is the alternative method more cost effective?
(in the sense that less tests are required, on average)

Thank you
• May 29th 2008, 05:32 AM
CaptainBlack
Quote:

Originally Posted by 892king
Here's the question...

The blood samples of k (k > 1) randomly chosen people is tested for
the presence of virus V . Suppose that 5% of the population has this
virus, and that presence of the virus among different people is independent.
separately” method, the following alternative method is proposed:
Mix a little of each of the k blood samples and test for the presence of
virus V in the mixture.
• If virus V is not present - done! (since that means that none of
the k people has the virus)
• If virus V is present, go back and test each of the k blood samples
separately. Let the random variable Y be the number of tests
needed for the alternative methods.
(a) Find PY (y) and E(Y ).
(b) For what values of k is the alternative method more cost effective?
(in the sense that less tests are required, on average)

Thank you

The probabilities are:

\$\displaystyle P(y=1)=0.95^k\$

\$\displaystyle P(y=(k+1))=1-0.95^k\$

So computing the expectation is now trivial.

RonL