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 and put the ball back in the bucket. 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 3 gets a placebo? This is my answer. I would appreciate if someone can take a look at it for me. Thank you. I did draw out all the cases for 3 patients but I don't know how to show here. My answer has 9 cases. P(p3=placebo) = P(p3=placebo and (p2=A and +) and (p1=A and +)) + P(p3=placebo and (p2=A and ) and (p1=A and +)) + P(p3=placebo and (p2=green) and (p1=A and +)) + P(p3=placebo and (p2=A and +) and (p1=A and )) + P(p3=placebo and (p2=A and ) and (p1=A and )) + P(p3=placebo and (p2=green) and (p1=A and )) + P(p3=placebo and (p2=A and +) and p1=green) + P(p3=placebo and (p2=A and ) and p1=green) + P(p3=placebo and (p2=green) and p1=green) + = (3/10 * (2/3 * 0.6) * 1/4) + (3/10 * (2/3 * 0.4) * 2/4) + (3/10 * 1/3 * 1/3) + ((0.5 * 0.4) * (1/3 * 0.6) * 2/4) + ((0.5 * 0.4) * (1/3 * 0.4) * 3/4) + ((0.5 * 0.4) * 2/3 * 1/3) + (0.5 * (0.5 * 0.6) * 2/3) + (0.5 * (0.5 * 0.4) * 2/3) + (0.5 * 0.5 * 0.5) = 0.47389 
the expression seems correct looking into the way the results are drawn but am trying something different here for now 
0.47389 is the value I come up with as well. 
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