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In addition to a small amount of number crunching, this problem also asks you to

ID: 3128505 • Letter: I

Question

In addition to a small amount of number crunching, this problem also asks you to make sure you can keep track of identifiers such as ‘X’, ‘N’, or ‘P’ as specified in the problem. In other words, you must make sure that you are clear about what each of these variables represents.

At a certain HIV clinic, 32% of the people who come in for testing are found to be HIV positive. (As with problem #1, this is a made-up number – real life findings are, I’m happy to report, not this high!)

Suppose that you choose three patients at random so that each has probability 0.32 of being HIV positive, and that they are independent of each other. Let the ‘X’ represent the number of HIV-positive individuals in your sample of three people.

a)       What are the possible values of X?

b)      Consider these three individuals. There are 8 possible arrangements of HIV positive and HIV negative. For example, we might use the symbol NNP to indicate that first two are HIV negative, and the third is HIV positive. What are the 8 possible arrangements? What is the probability of each of these 8 arrangements?

c)       Think about what we mean when we say ‘X’. Then provide the value of X for each arrangement in question ‘b’. Then calculate the probability of each possible value of X.

d)      What was it important that I specify in the question that “they are independent of each other”?

Explanation / Answer

a) The possible values of X are 0,1,2,3.

b) NNN p=0.314 , NNP p= 0.148, NPN p= 0.148, PNN p=0.148, NPP p= 0.0696, PNP p=0.0696, PPN p=0.0696, PPP p= 0.0327

c)‘X’ represents the number of HIV-positive individuals in your sample of three people.

NNN : X = 0

NNP, NPN, PNN: X=1

NPP, PNP, PPN: X=2

PPP:X=3

P(X=0) = 0.314 , P(X=1) = 0.444 , P(X=2) = 0.2088, P(X=3) = 0.0327

d) When the result of the HIV test of the first person os known the probabilities of the HIV result of the second and third person can change if the events are not independent. Hence, this was important. Without independence, additional information would be needed to calculate the probability for the remaining two persons.

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