anonymous
  • anonymous
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?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[P(\text{took medicine}|\text{no cough})=\frac{P(\text{took medicine AND no cough})}{P(\text{no cough})}\] Of the people that received the treatment, how many reported coughing afterward? Of the total 100 people in the study, how many reported coughing?
anonymous
  • anonymous
46?
anonymous
  • anonymous
right?

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anonymous
  • anonymous
46 reported coughing and 54 did
anonymous
  • anonymous
didnt*
anonymous
  • anonymous
Right, 54 people of the total 100 reported no coughing, so \(\dfrac{54}{100}=P(\text{no cough})\). What about the other group?
anonymous
  • anonymous
no cough would be whose left, right? so that's 46
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh 36
anonymous
  • anonymous
Close, 36 is the number people that got treatment and DID report coughing.
anonymous
  • anonymous
|dw:1442799451505:dw|
anonymous
  • anonymous
so 46
anonymous
  • anonymous
70+54=124
anonymous
  • anonymous
i was a bit confused
anonymous
  • anonymous
because the total people who didnt cough and the total who did take the meds would equal 124... right?
kropot72
  • kropot72
|dw:1442799654988:dw| \[\large p(med/no\ cough)=0.7\times\frac{34}{70}=0.34\] \[\large P(no\ med/no\ cough)=0.3\times\frac{20}{30}=0.2\] \[\large P(med|no\ cough)=\frac{0.34}{(0.34+0.2)}=0.63\]
kropot72
  • kropot72
@tekman1298 An alternative approach, using a probability tree.
anonymous
  • anonymous
woah no way... i guessed right xD i decided to ask here for a double check
anonymous
  • anonymous
thx, time for another?
kropot72
  • kropot72
Was your guess 0.63?
anonymous
  • anonymous
yes xD
kropot72
  • kropot72
Cool. I have to log out now, please post as a new question.
anonymous
  • anonymous
ok thx man
kropot72
  • kropot72
You're welcome :)

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