A medical researcher must choose one of 14 patients to recei

This is an " or" and an "and" operatiob. In prpbabilty, or is equivalent to adding and "and" to multiplying. So the answer is the probability of getting the first drug (1/14) OR (+) the probability of getting the second drug which is equivalent to dont get the first drug (13/14) AND (*) getting the second drug (1/13). So the operation is: (1/14) + (13/14)*(1/13) = 1/14 + 1/14 = 1/7

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@Capricorn 369,


We only need to find the probability that Donald receives either Progaine or Ropecia. By multiplying by 1/12 you are trying to take into account the placebo round. However, whether Donald is around to be chosen for the placebo does not affect the probability of him getting a dosage of either Progaine or Ropecia.

Had the question been asking for the probability of Donald receiving a placebo the answer would be (13/14)(12/13)(1/12) = 1/14.

Hope that helps!

@ Chris - got it, thanks.
So the operation should be : (1/14) + (13/14)*(1/13) = 1/14 + 1/14 = 1/7

What you say?

There's no need for any multiplications here. The chance anyone gets Progaine is 1/14, so the chance Don gets Progaine is 1/14. The chance anyone gets Ropecia is 1/14, so the chance Don gets Ropecia is 1/14. So the chance he gets one of the two is 1/14 + 1/14 = 1/7.

Or you can just imagine lining the people up at random, and giving the first two people in line Progaine and Ropecia. The chance Don is among the first two people is 2/14 = 1/7.

My approach was same as the popular one here. I got the right answer, but the official explanation was as given by IStewart. I still can't get this explanation of 1/14+1/14. Still, the other way i.e. giving the first two people in line Progaine and Ropecia. The chance Don is among the first two people is 2/14 = 1/7 makes some sense.

1/7 must be right answerr

well here is an alternate solution for those people who still wants to follow the multiplication method.

For the sake of simplicity lets name three medicines as A,B and C.

Now out of 14 people, anyone can be selected for the medicine A in 14C1 ways
out of remaining 13 people, anyone can be selected for medicine B in 13C1 ways
Lastly from remaining 12 people, anyone can be selected for medicine C in 12C1 ways

hence total no. of ways for selecting 3 person for medicine A,B and C is 14C1x13C1x12C1 = 14x13x12

Now donald can get either medicine A or medicine B
Case 1 if he gets medicine A, then person for medicine B can be selected in 13C1 ways and person for medicine C can be selected in 12C1 ways hence total no. of ways = 1x13C1x12C1

Case 2 if he gets medicine B, then person for medicine A can be selected in 13C1 ways and person for medicine C can be selected in 12C1 ways hence total no. of ways= 1x13C1x12C1

hence total no. of favorable ways= case1 +case2 = 2x13x12

hence required probability = (2x13x12)/(14x13x12) =1/7

i hope it helps

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?

Donald to receiver either Prograine or Ropecia must be among first two chosen patients and as there are 14 patients then the probability of this is simply 2/14=1/7.

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