A medical researcher must choose one of 14 patients to receive an experimental
medicine called Progaine. The researcher must then choose one of
the remaining 13 patients to receive another medicine, called Ropecia.
Finally, the researcher administers a placebo to one of the remaining 12
patients. All choices are equally random. If Donald is one of the 14 patients,
what is the probability that Donald receives either Progaine or Ropecia?
The answer given is 1/14+1/14 = 1/7, which i feel as incorrect, since the chances of getting ropecia is only 1/13. Could anyone help understanding this. Thanks!
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yuvarajait
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well my advice is don't get confused with the wording,each patient has an equal chance of being selected for any of the three experimental medicine..!!!yuvarajait wrote:A medical researcher must choose one of 14 patients to receive an experimental
medicine called Progaine. The researcher must then choose one of
the remaining 13 patients to receive another medicine, called Ropecia.
Finally, the researcher administers a placebo to one of the remaining 12
patients. All choices are equally random. If Donald is one of the 14 patients,
what is the probability that Donald receives either Progaine or Ropecia?
The answer given is 1/14+1/14 = 1/7, which i feel as incorrect, since the chances of getting ropecia is only 1/13. Could anyone help understanding this. Thanks!
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HSPA
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A) selected first = 1/14
B) getting selected for the second = 13/14* 1/13 = 1/14
Here 13/14 stands for not getting selected for first medicine Progaine and the next 1/13 is getting selected for Ropecia.
P(AUB) = P(A) + P(B) = 1/14+ 1/14 = 1/7
B) getting selected for the second = 13/14* 1/13 = 1/14
Here 13/14 stands for not getting selected for first medicine Progaine and the next 1/13 is getting selected for Ropecia.
P(AUB) = P(A) + P(B) = 1/14+ 1/14 = 1/7
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You can also solve the question using the complement.yuvarajait wrote:A medical researcher must choose one of 14 patients to receive an experimental
medicine called Progaine. The researcher must then choose one of
the remaining 13 patients to receive another medicine, called Ropecia.
Finally, the researcher administers a placebo to one of the remaining 12
patients. All choices are equally random. If Donald is one of the 14 patients,
what is the probability that Donald receives either Progaine or Ropecia?
The answer given is 1/14+1/14 = 1/7, which i feel as incorrect, since the chances of getting ropecia is only 1/13. Could anyone help understanding this. Thanks!
P(Progaine or Ropecia) = 1 - P(neither Progaine nor Ropecia)
P(neither Progaine nor Ropecia) = P(not getting selected for Progaine AND not getting selected for Ropecia)
= P(not getting selected for Progaine) x P(not getting selected for Ropecia)
= (13/14) x (12/13)
= 12/14
= 6/7
So, P(Progaine or Ropecia) = 1 - P(neither Progaine nor Ropecia)
= 1 - 6/7
= 1/7
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Ian Stewart
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I posted about a similar problem a moment ago, but you can also just imagine lining up your 14 people, giving the 1st person in line Progaine, the 2nd person Ropecia, and the 3rd person the placebo. The question then becomes 'if you line up 14 people, including Don, what is the probability Don is one of the first 2 people in line?' There are 14 places to put Don, 2 of which are in the first two positions in line, so the answer is 2/14 = 1/7.yuvarajait wrote:A medical researcher must choose one of 14 patients to receive an experimental
medicine called Progaine. The researcher must then choose one of
the remaining 13 patients to receive another medicine, called Ropecia.
Finally, the researcher administers a placebo to one of the remaining 12
patients. All choices are equally random. If Donald is one of the 14 patients,
what is the probability that Donald receives either Progaine or Ropecia?
The answer given is 1/14+1/14 = 1/7, which i feel as incorrect, since the chances of getting ropecia is only 1/13. Could anyone help understanding this. Thanks!
Note that the chance of getting Ropecia is *not* 1/13; it is still 1/14. It is only 1/13 if you *know* you were not selected to receive Progaine.
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