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What is the product of these two probabilities

  1. Apr 8, 2012 #1
    Took Drug Didn't Total
    Cholesterol ≤ 200 60 50 110
    Cholesterol > 200 40 50 90
    100 100 200


    We’ll call event A “cholesterol ≤ 200” and event B “took the drug”…..

    What is the probability a person’s cholesterol ≤ 200? P(A)
    What is the probability a person took the drug? P(B)
    What is the product of these two probabilities (multiply the above together)?
    P(A) * P(B)
    Given that a person took the cholesterol drug, what is the probability that their cholesterol ≤ 200? P(A|B)
    Does P(A) * P(B) = P(A|B)?
    Are these events independent?
    What does this tell us about the effectiveness of the drug?

    * * Help me please!
     
    Last edited: Apr 8, 2012
  2. jcsd
  3. Apr 8, 2012 #2
    Re: Independent??

    Let's do the first one. What's the probability that somebody's cholesterol is ≤200

    For this: what is the total population?? Which ones are ≤200 ???
     
  4. Apr 8, 2012 #3
    Re: Independent??

    Total of everyone is 200. 110 are equal to or greater than 200
     
  5. Apr 8, 2012 #4
    Re: Independent??

    You probably meant to say smaller or equal than 200 :tongue2:

    Anyway. The P(A) is now easy: it's the percentage of people with cholesterol ≤200. So it's just 110/200.

    Can you find P(B) is the same fashion?? That is:
    - What's the number of people that took the drug
    - What's the total population
    - What's the percentage?
     
  6. Apr 8, 2012 #5
    Re: Independent??

    Yeah yeah, lol. You know what I meant :)
    40/90
    200
    40/200

    right??
     
  7. Apr 8, 2012 #6
    Re: Independent??

    Hmmm, not sure what you mean with those numbers...

    How many people took the drug?? 40/90 doesn't make much sense, you will want an integer as an answer.

    The bottom row gives us the information here. It tells us that 100 people took the drug, 100 people didn't and 200 people is the total population.

    So what's the percentage of people that took the drug??
     
  8. Apr 8, 2012 #7
    Re: Independent??

    50% took it
     
  9. Apr 8, 2012 #8
    Re: Independent??

    OK, that's good!!

    So the answer to the first question is 110/200 and the second is 1/2.

    So you can multiply these together and that gives the third question.

    For the fourth, you're interested in all the people that actually took the drug. So the first column interests us here. So of all the people who took the drug, we have that

    60 people have cholesterol ≤200
    40 people have cholesterol >200
    and 100 people in total took the drug.

    Sooo, what's the percentage of people that took the drug with cholesterol ≤200??
     
  10. Apr 8, 2012 #9
    Re: Independent??

    3/10, I think, right?
     
  11. Apr 8, 2012 #10
    Re: Independent??

    Did you do 60/200??

    In this case, you calculated the people who tool the drug AND have cholesterol ≤200 and you divided it by the total population.

    However, you are now interested ONLY in the people who took the drug. So the people who did not took the drug don't interest us and thus shouldn't be counted in the population.

    So our population now is not 200 but 100.
     
  12. Apr 8, 2012 #11
    Re: Independent??

    Ok, so it's 60/100??
     
  13. Apr 8, 2012 #12
    Re: Independent??

    Good!!

    Now, we will want to know if the event are independent or not. Can you give me the definition of independence??
     
  14. Apr 8, 2012 #13
    Re: Independent??

    The probability has to add up to 1. Yes, I did look it up :) So.... I'm gonna say it's not independent....
     
  15. Apr 8, 2012 #14
    Re: Independent??

    Huh?? You're right that they are not independent. But your reason is not correct. That the probabilities add up to 1 has nothing to do with it here :frown:

    Usually, we say that A is independent from B if P(A|B)=P(A). Translated in words, we have that the chance that A happens is the same as that chance that A happens given B. So it is irrelevant that B is true, because the chance that A happens remains the same anyway.

    Now, is P(A|B)=P(A) here??
     
  16. Apr 8, 2012 #15
    Re: Independent??

    I read it in my school book....
     
  17. Apr 8, 2012 #16
    Re: Independent??

    I don't doubt it, but it's strange. Can you quote what's in your book??
     
  18. Apr 8, 2012 #17
    Re: Independent??

    No, its not
     
  19. Apr 8, 2012 #18
    Re: Independent??

    i can try to send a link... may just be directly linked to my account
     
  20. Apr 8, 2012 #19
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