Suppose we observe 84 alchoholics with cirrhosis of the liver, of whom 29 have hepatomas, that is, a liver-cell carcinoma. Suppose we know, based on a large sample, that the risk of hepatoma among alcoholics without cirrhosis of the liver is 24%.

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?

  • 2
    $\begingroup$ When asking homework questions, you may want to detail what you've thought and indicate precisely where you got stuck in the problem. Doing this will help to get you a better answer and help answerers focus in on the points you are most confused about. $\endgroup$ – WWright Oct 25 '10 at 21:08

When you have $n$ people, with probability $p$ for each of them to have hepatoma, and you want to know the probability that exactly $k$ of them have hepatoma, then the formula is given by:

$${n \choose k}p^k(1-p)^{n-k}$$

In your case, $n$ = 84, $k$ = 29, $p$ = 0.24. This is called the Binomial Distribution. You can learn more about it here: http://en.wikipedia.org/wiki/Binomial_distribution


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy