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Exam P Practice Question 238
Question: In a large population of patients, 20% have early stage cancer, 10% have advanced stage cancer, and the other 70% do not have cancer. Six patients from this population are randomly selected. Calculate the expected number of selected patients with advanced stage cancer, given that at least one of the selected patients has early stage cancer.
(A) 0.403 (B) 0.500 (C) 0.547 (D) 0.600 (E) 0.625 Solution: https://imgur.com/aiDBihS Here's my confusion. I understand the solution mathematically. I understand how they calculate it and the formulas they're using. However, my intuitive understanding of the situation would lead me to answer B. The way I see it, if we pick 6 patients, and at least one has early stage cancer, then since this is a large population, we can consider this the same as randomly picking 5 patients, since the 6th patient is set. Then of course we would expect 5*0.1 = 0.5 patients with advanced stage cancer. This is obviously wrong. So what is it that shifts the expected value? Why isn't it effectively the same as picking 5 random patients? Thank you. 
#2




That is a more complicated version of a fairly common puzzler.
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#3




Imagine the probability for early stage cancer was 0.9 so that there are only two cases: early with 0.9 and advanced with 0.1. Then the probability of having at least 1 early stage of the 6 selected is nearly 1. This means the expected value of # of advanced stage given at least 1 early has to be nearly equal to the unconditional expected value of the # of advanced stage people and we know that’s 6*0.1=0.6. It really depends on how likely early stage is. If early stage was nearly impossible, the answer would be close to 0.5.
You could answer this question without doing any work just by observing that the answer must be between 0.5 and 0.6. 
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