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anonymous

  • one year ago

Bobby is testing the effectiveness of a new cough medication. There are 100 people with a cough in the study. Seventy patients received the cough medication, and 30 other patients did not receive treatment. Thirty-four of the patients who received the medication reported no cough at the end of the study. Twenty of the patients who did not receive medication reported no cough at the end of the study. What is the probability that a patient chosen at random from this study took the medication, given that they reported no cough?

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  1. anonymous
    • one year ago
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    \[P(\text{took medicine}|\text{no cough})=\frac{P(\text{took medicine AND no cough})}{P(\text{no cough})}\] Of the people that received the treatment, how many reported coughing afterward? Of the total 100 people in the study, how many reported coughing?

  2. anonymous
    • one year ago
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    46?

  3. anonymous
    • one year ago
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    right?

  4. anonymous
    • one year ago
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    46 reported coughing and 54 did

  5. anonymous
    • one year ago
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    didnt*

  6. anonymous
    • one year ago
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    Right, 54 people of the total 100 reported no coughing, so \(\dfrac{54}{100}=P(\text{no cough})\). What about the other group?

  7. anonymous
    • one year ago
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    no cough would be whose left, right? so that's 46

  8. anonymous
    • one year ago
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    No, you already determined the number of no-coughs from the whole group. Now we're interested in the no-coughs from the group that got the treatment.

  9. anonymous
    • one year ago
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    oh 36

  10. anonymous
    • one year ago
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    Close, 36 is the number people that got treatment and DID report coughing.

  11. anonymous
    • one year ago
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    |dw:1442799451505:dw|

  12. anonymous
    • one year ago
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    so 46

  13. anonymous
    • one year ago
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    70+54=124

  14. anonymous
    • one year ago
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    i was a bit confused

  15. anonymous
    • one year ago
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    because the total people who didnt cough and the total who did take the meds would equal 124... right?

  16. kropot72
    • one year ago
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    |dw:1442799654988:dw| \[\large p(med/no\ cough)=0.7\times\frac{34}{70}=0.34\] \[\large P(no\ med/no\ cough)=0.3\times\frac{20}{30}=0.2\] \[\large P(med|no\ cough)=\frac{0.34}{(0.34+0.2)}=0.63\]

  17. kropot72
    • one year ago
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    @tekman1298 An alternative approach, using a probability tree.

  18. anonymous
    • one year ago
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    woah no way... i guessed right xD i decided to ask here for a double check

  19. anonymous
    • one year ago
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    thx, time for another?

  20. kropot72
    • one year ago
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    Was your guess 0.63?

  21. anonymous
    • one year ago
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    yes xD

  22. kropot72
    • one year ago
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    Cool. I have to log out now, please post as a new question.

  23. anonymous
    • one year ago
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    ok thx man

  24. kropot72
    • one year ago
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    You're welcome :)

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