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October 16th, 2016, 10:55 PM  #1 
Newbie Joined: Oct 2016 From: United States Posts: 1 Thanks: 0 
Suppose we observe 84 alcoholics with cirrhosis of the liver, of whom 29 have hepatomas  that is livercell 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 04:34 AM. 
October 17th, 2016, 10:10 AM  #2 
Global Moderator Joined: May 2007 Posts: 6,581 Thanks: 610 
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 $\displaystyle \binom {n}{k}p^k(1p)^{nk}$ 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 $\displaystyle \sum_{k=29}^{84}\binom {n}{k}p^k(1p)^{nk}$. 

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