Probabilities/Combinations: At a certain clinic, 29% of those who come in for testing

roxanh

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Problem #4:
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, 29% 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 0.29 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.
P(HIV+) = .29 P(HIV-) = .71
a) What are the possible values of X?
S = {0,1,2,3}
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?
P(X=0) = P(NNN) = 1/8
P(X=1) = P(NNP or NPN or PNN) = P(NNP) + P(NPN) + P(PNN) = 3/8
P(X=2) = P(NPP or PNP or PPN) = P(NPP) + P(PNP) + P(PPN) = 3/8
P(X=3) = P(PPP) = 1/8


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

I have an issue understanding how to incorporate the .29 into this problem, is it part C?
 
Yes, the 0.29 does come into play in part C. You've correctly identified that 1 out of the 8 arrangements has all negative patients (i.e. X=0), and that 1 arrangement has three positive patients (X=3) but is the probability of each of those really 1/8? As another way of thinking about it, your answer says that the probably of picking three negative patients is the same as the probability of picking three positive patients. Is that correct? Why or why not? What does that suggest about how you might proceed?
 
Response

Yes, the 0.29 does come into play in part C. You've correctly identified that 1 out of the 8 arrangements has all negative patients (i.e. X=0), and that 1 arrangement has three positive patients (X=3) but is the probability of each of those really 1/8? As another way of thinking about it, your answer says that the probably of picking three negative patients is the same as the probability of picking three positive patients. Is that correct? Why or why not? What does that suggest about how you might proceed?


I see what you mean there, X= 3 should not have the same probability as X=0, would the probability of those actually be the .29?
 
Yes, the 0.29 does come into play in part C. You've correctly identified that 1 out of the 8 arrangements has all negative patients (i.e. X=0), and that 1 arrangement has three positive patients (X=3) but is the probability of each of those really 1/8? As another way of thinking about it, your answer says that the probably of picking three negative patients is the same as the probability of picking three positive patients. Is that correct? Why or why not? What does that suggest about how you might proceed?

should the probability of those be .29? then? or the subtraction of 1-.29? I'm a bit confused here sorry.
 
should the probability of those be .29? then? or the subtraction of 1-.29? I'm a bit confused here sorry.

The problem tells you that 0.29 is the probability of any one patient being HIV positive, and from that you can deduce that 0.71 (1 - 0.29) is the probability of a patient being negative. But you need to know the probability of three patients being negative. If you're having trouble figuring it out, you might try considering a different, simpler, exercise. For instance, suppose you have a coin and a six-sided dice. You know the probability of the coin coming up heads (1/2) and you know the probability of the dice rolling a 3 (1/6), but what's the odds of both happening? Now, let's suppose you also have a 4-sided dice. The probability of it rolling a 4 is 1/4. So, what's the odds of getting a Heads, a 3 on the six-sided dice, and a 4 on the four-sided dice? How did you calculate these probabilities? Can you see how to apply this to your actual problem?
 
The problem tells you that 0.29 is the probability of any one patient being HIV positive, and from that you can deduce that 0.71 (1 - 0.29) is the probability of a patient being negative. But you need to know the probability of three patients being negative. If you're having trouble figuring it out, you might try considering a different, simpler, exercise. For instance, suppose you have a coin and a six-sided dice. You know the probability of the coin coming up heads (1/2) and you know the probability of the dice rolling a 3 (1/6), but what's the odds of both happening? Now, let's suppose you also have a 4-sided dice. The probability of it rolling a 4 is 1/4. So, what's the odds of getting a Heads, a 3 on the six-sided dice, and a 4 on the four-sided dice? How did you calculate these probabilities? Can you see how to apply this to your actual problem?


I see what you did there......I believe i figured it out ....
I put the probabilites of N's and P's multiplied the probability and made sure all the probilities added 1. or 100%

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Problem #4:
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.are the 8 possible arrangements? What is the probability of each of these 8 arrangements?
P(X=0) = P(NNN) = 1/8
P(X=1) = P(NNP or NPN or PNN) = P(NNP) + P(NPN) + P(PNN) = 3/8
P(X=2) = P(NPP or PNP or PPN) = P(NPP) + P(PNP) + P(PPN) = 3/8
P(X=3) = P(PPP) = 1/8
I find this entire thread totally confusing, even the replies. All of that is wrong.
This a a very simple binomial probability. If \(\displaystyle p\) is the probability of success then the probability of \(\displaystyle K\) successes in \(\displaystyle N\) independent trials is \(\displaystyle \mathcal{P}(X=K)=\binom{N}{K}(p)^K(1-p)^{N-K}\)

So the above #3 should be: \(\displaystyle \mathcal{P}(X=2)=\binom{3}{2}(0.29)^2(0.71)^{1}\)
 
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