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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. Thirtyfour 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?
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. Thirtyfour 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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[P(\text{took medicine}\text{no cough})=\frac{P(\text{took medicine AND no cough})}{P(\text{no cough})}\] 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
 one year ago
Best ResponseYou've already chosen the best response.046 reported coughing and 54 did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, 54 people of the total 100 reported no coughing, so \(\dfrac{54}{100}=P(\text{no cough})\). What about the other group?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no cough would be whose left, right? so that's 46

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, you already determined the number of nocoughs from the whole group. Now we're interested in the nocoughs from the group that got the treatment.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Close, 36 is the number people that got treatment and DID report coughing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442799451505:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was a bit confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because the total people who didnt cough and the total who did take the meds would equal 124... right?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.3dw:1442799654988:dw \[\large p(med/no\ cough)=0.7\times\frac{34}{70}=0.34\] \[\large P(no\ med/no\ cough)=0.3\times\frac{20}{30}=0.2\] \[\large P(medno\ cough)=\frac{0.34}{(0.34+0.2)}=0.63\]

kropot72
 one year ago
Best ResponseYou've already chosen the best response.3@tekman1298 An alternative approach, using a probability tree.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woah no way... i guessed right xD i decided to ask here for a double check

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thx, time for another?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.3Cool. I have to log out now, please post as a new question.
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