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February 8th, 2018, 10:32 PM  #1 
Newbie Joined: Feb 2018 From: Canada Posts: 28 Thanks: 2  Probability problem
A doctor is using an adaptive response technique in their clinical trial to decide whether patients get drug A or a placebo. They start with one red ball and one green ball in a bucket. For patient one, they will pull out a ball and look at the colour. If it's a red ball, the patient gets drug A. If it's a green ball, the patient gets the placebo. If the patient receives a placebo, the doctor will do nothing to the bucket before selecting for the next patient. If the patient receives drug A and has a positive outcome, they will add a red ball to the bucket. If the patient receives drug A and have a negative outcome, they will add a green ball to the bucket. The second patient will have their treatment decided based on what is in now in the bucket and we will add balls or not according to the same rule. This continues for patient 3, etc... Suppose there is a constant probability of 0.60 that any patient who receives drug A will have a positive outcome. What is the probability patient 1 gets Drug A with a positive treatment, patient 2 gets a placebo, and patient 3 gets Drug A? I don't understand the question well enough. Can someone help me with a direction? Does the question wants me to find the probability of patient 3 gets drug A and patient 2 gets a placebo and patient 1 has a positive outcome given taking drug A? Thank you. Last edited by Shanonhaliwell; February 8th, 2018 at 10:39 PM. 
February 8th, 2018, 10:51 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,852 Thanks: 959 
this is just an exercise in chasing down a path in a probability tree. $P[p_1\text{ gets drug A with + results.}] = \left(\dfrac 1 2\right)\left(\dfrac 3 5\right) = \dfrac {3}{10}$ Now we add a red ball to the bucket due to the success so that there are now 2 red balls and 1 green ball. $P[p_2 \text{ gets a placebo.}] = \dfrac 1 3$ We do nothing due to $p_2$ receiving a placebo so $P[p_3 \text{ gets drug A.}] = \dfrac 2 3$ We multiply all these probabilities together to get the probability of the chain so $P[p_1 =A^+,~p_2 = P,~p_3=A] = \dfrac{3}{10}\dfrac{1}{3}\dfrac{2}{3} = \dfrac{6}{90} = \dfrac{1}{15}$ 

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