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anonymous

  • one year ago

I need someone to explain this to me, I really don't understand... Using Baye's Theorem: It is known that 2% of the population has a certain allergy. A test correctly identifies people who have the allergy 98% of the time. The test correctly identifies people who do not have the allergy 94% of the time. A doctor decides that anyone who tests positive for the allergy should begin taking anti-allergy medication. Do you think this is a good decision? Why or why not? I know it's pretty long and a little bit of reading but it's on paper so I had to copy it, I wouldn't have if I didn't need the

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  1. anonymous
    • one year ago
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    this is often very confusing one good way to approach this is to use some actual numbers and see what happens

  2. anonymous
    • one year ago
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    it makes "baye's theorem" far more undersantable

  3. anonymous
    • one year ago
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    Okay but how do I choose my numbers?

  4. anonymous
    • one year ago
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    i would pick a large number for the population, so that we get all whole number answers (although once we get the answer, we can redo it without using that crutch) lets say the population is 1000 people ok?

  5. anonymous
    • one year ago
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    okay

  6. anonymous
    • one year ago
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    and if 2% of the population suffers from the allergies, how many people exactly will have it? i.e what is 2% of 1000 ?

  7. anonymous
    • one year ago
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    Umm...40? I'm not sure I did this right

  8. anonymous
    • one year ago
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    yes that is right

  9. anonymous
    • one year ago
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    oops sorry that is wrong

  10. anonymous
    • one year ago
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    dang it

  11. anonymous
    • one year ago
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    2% of 1,000 is \(.02\times 1000=20\)

  12. anonymous
    • one year ago
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    that is why i picked a nice round number like 1000 so the it would be easy to compute the percents

  13. anonymous
    • one year ago
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    That what I did but my calculator said i was 40...oh well what do I have to do next?

  14. anonymous
    • one year ago
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    ok so 20 people have the allergy, how many do not?

  15. anonymous
    • one year ago
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    By the way that already helps a lot

  16. anonymous
    • one year ago
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    20 out of 1000 which is... 20/10000=0.2%?

  17. anonymous
    • one year ago
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    hold on i think i have confused you

  18. anonymous
    • one year ago
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    we have a population of 1,000 2% have the allergy, so 20 have the allergy if out of 1,000 people, 20 have the allergy, how many (not what percent) do not have the allergy

  19. anonymous
    • one year ago
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    ooooh, sorry with the 10000 I got confused with another problem on my paper sorry. Umm so 980 don't have the allergy?

  20. anonymous
    • one year ago
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    right

  21. anonymous
    • one year ago
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    so we know how many have it, and how many do not now lets see how many people will test positive for it

  22. anonymous
    • one year ago
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    of the 20 people that have it, the test is 98% accurate what is 98% of 20?

  23. anonymous
    • one year ago
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    btw i hope it is clear that you do \(.98\times 20\)

  24. anonymous
    • one year ago
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    19.6?

  25. anonymous
    • one year ago
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    right

  26. anonymous
    • one year ago
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    yeah I'm good with the percentage....most of the time

  27. anonymous
    • one year ago
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    i guess we have to work with the decimal, should have chosen a population of 10,000 then it would be whole numbers but too late now

  28. anonymous
    • one year ago
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    now we know \(980\) so NOT have the allergy, lets see how many of those will also test positive

  29. anonymous
    • one year ago
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    for them the test is 96% accurate so 96% of them will test negative, but that means 6% of them will test positive what is 6% of 980?

  30. anonymous
    • one year ago
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    oops bad math

  31. anonymous
    • one year ago
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    4% will test positive what is 4% of 980

  32. anonymous
    • one year ago
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    39.2?

  33. anonymous
    • one year ago
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    yeah i get that too

  34. anonymous
    • one year ago
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    ok so now how many people total (it is a decimal) will test positive?

  35. anonymous
    • one year ago
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    0.04 people? I'm confused

  36. anonymous
    • one year ago
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    we computed two numbers of people that test positive right? the ones with the allergy, 19.6 and the ones without the allergy, 39.2 what is the total?

  37. anonymous
    • one year ago
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    oh, 58.8?

  38. anonymous
    • one year ago
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    right, that is the total number that test positive

  39. anonymous
    • one year ago
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    out of those, we know that 19.6 actually have the allergy

  40. anonymous
    • one year ago
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    okay

  41. anonymous
    • one year ago
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    so the question is, IF you test positive, what is the probabilty you actually have the allergy that is the number of people who test positive and have the allergy, divided by the total number of people who test postive ie\[\frac{ 19.6}{58.8}\]

  42. anonymous
    • one year ago
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    or whatever you get when you write that as a decimal, i get \(0.337\) rounded, a little more than a third

  43. anonymous
    • one year ago
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    okay I get 0.33333etc.

  44. anonymous
    • one year ago
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    oh maybe i put the wrong numbers in yeah you are right

  45. anonymous
    • one year ago
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    one third in other words

  46. anonymous
    • one year ago
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    would you like to redo this using "baye's formula" ? will will do pretty much the same arithmetic, and get the same answer

  47. anonymous
    • one year ago
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    yeah, it would help a lot, my teacher just throws a formula and exercises at us

  48. anonymous
    • one year ago
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    ok do you have the formula you are supposed to use? i can make a guess if you like

  49. anonymous
    • one year ago
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    yeah I have it

  50. anonymous
    • one year ago
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    go ahead and write it, we can walk through it slowly

  51. anonymous
    • one year ago
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    It's \[P(A|B)=\frac{P(A|B)*P(A)}{ P(B)}\]

  52. anonymous
    • one year ago
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    ok good

  53. anonymous
    • one year ago
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    I have it right in front of me

  54. anonymous
    • one year ago
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    first off, it this is going to make sense, you need to know what \(P(A|B)\) means do you know what it means?

  55. anonymous
    • one year ago
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    oh crap your definition is wrong, but we will get to that in a second

  56. anonymous
    • one year ago
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    It means probability of A given B

  57. anonymous
    • one year ago
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    ok good

  58. anonymous
    • one year ago
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    and the definition (not baye's theorem) is \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

  59. anonymous
    • one year ago
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    My teacher made us study about a formula like that, but I really didn't get a thing

  60. anonymous
    • one year ago
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    baye's theorem, at least one form, is not quite what you wrote, the condition is switched on the right it is \[P(A|B)=\frac{P(B|A)*P(A)}{ P(B)}\]

  61. anonymous
    • one year ago
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    ok lets see if we can get it

  62. anonymous
    • one year ago
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    oh yeah my mistake

  63. anonymous
    • one year ago
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    ok now the numerator in your fraction \[P(B|A)P(A)\] is just another way to find \(P(A\cap B)\) the probability that both A and B occur

  64. anonymous
    • one year ago
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    That's where I get extremely confused

  65. anonymous
    • one year ago
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    lets use our example you want the probability you have the allergy GIVEN you test positive

  66. anonymous
    • one year ago
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    okay

  67. anonymous
    • one year ago
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    we can put A = you have the allergy, B = you test positive then we are trying to find \(P(A|B)\) the probability you have the allergy given that you test positive

  68. anonymous
    • one year ago
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    let me know if i lost you, we need to go slow i am sure

  69. anonymous
    • one year ago
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    okay so far I think I'm okay...I think so for now

  70. anonymous
    • one year ago
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    ok to compute that we need the numerator \(P(A\cap B) \) i.e. the probability you have it AND test positive we were not told that number

  71. anonymous
    • one year ago
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    so far so good

  72. anonymous
    • one year ago
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    but we were told \(P(B)\) the probabily you have the disease, that was \(2\%=0.02\)

  73. anonymous
    • one year ago
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    okay

  74. anonymous
    • one year ago
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    and we were also told \(P(B|A)\) the probability you have the disease given that you test positive that was the \(98\%=0.98\)

  75. anonymous
    • one year ago
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    to find the numerator, multiply those together, i.e. the probability you have the disease AND test positive is the probability you have the disease times the probability you test positive GIVEN that you have the disease, i.e. \[.02\times.98=.0196\]

  76. anonymous
    • one year ago
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    you notice that is the same numerator we had before, just with the decimal moved over three places, because we did not start out with a population of 1,000 we just went straight for the number

  77. anonymous
    • one year ago
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    so far so good?

  78. anonymous
    • one year ago
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    Wait why did you change the 98% to a decimal. to make it easier to work with?

  79. anonymous
    • one year ago
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    when doing any work in math, you use numbers, not percents \(98\%=0.98\) as a number percents are nice to look at, but think about what you did when you wanted to find 98% of 20 you multiplied 20 by .98 right, not by 98

  80. anonymous
    • one year ago
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    ready to find the denominator?

  81. anonymous
    • one year ago
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    okay

  82. anonymous
    • one year ago
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    that is \(P(B)\) the probability you test positive remember how we found that before? we had to add two numbers, they were \(19.6\) and \(39.2\) they will be the same this time, but with the decimal in a different place

  83. anonymous
    • one year ago
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    the fist one is the \(.00196\) we already found, that is the number of people who test positive given that they have the disease

  84. anonymous
    • one year ago
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    so you divided it by 1000 right?

  85. anonymous
    • one year ago
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    we also need the number of people who test positive who do NOT have the disease

  86. anonymous
    • one year ago
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    well, yes, but we just found again computing \(.02\times .98=.00196\)

  87. anonymous
    • one year ago
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    to compute the probability that you test positive if you do not have the disease we used \(.04\times .980=.0392\)

  88. anonymous
    • one year ago
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    okay

  89. anonymous
    • one year ago
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    I'm sorry I really want to learn but I have to go to bed

  90. anonymous
    • one year ago
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    now add together to get the probability you test positive, that is \(.00196+.00392=.00588\)

  91. anonymous
    • one year ago
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    you did help a lot already though I'll make sure I have a lot of time next time I ask a question though

  92. anonymous
    • one year ago
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    yeah it is late you can look at this later and see that we are really using the same numbers as before

  93. anonymous
    • one year ago
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    okay

  94. anonymous
    • one year ago
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    later

  95. anonymous
    • one year ago
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    later!

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