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October 16th, 2016, 11:55 PM   #1
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Suppose we observe 84 alcoholics with cirrhosis of the liver, of whom 29 have hepatomas - that is liver-cell carcinoma. Suppose we know, based on large sample, that the risk of hepatoma among alcoholics without cirrhosis of the liver is 24%.

5.50 what is the probability that we observe exactly 29 alcoholics with cirrhosis of the liver who have hepatomas if the true rate of hepatoma among alcoholics (with or without cirrhosis of the liver) is .24?

5.51 What is the probability of observing at least 29 hepatomas among the 84 alcoholics with cirrhosis of the liver under the assumptions in problem 5.50?

Last edited by skipjack; October 17th, 2016 at 05:34 AM.
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October 17th, 2016, 11:10 AM   #2
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Both questions can be answered exactly using the binomial distribution.
The probability of exactly k concurrences of an event in a sample of size n is given by $\displaystyle \binom {n}{k}p^k(1-p)^{n-k}$ where p is the probability of one occurrence.
For your problems, n=84 and p=0.24. For 5.50 the probability is given for k=29. For 5.51 the probability is $\displaystyle \sum_{k=29}^{84}\binom {n}{k}p^k(1-p)^{n-k}$.
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