Date: 12/22/97 at 19:15:33 From: Smith, Cassandra Subject: Probability distribution I have written to you before with much success. I am once again stumped. I have tried several times to answer this question and even though the answers are available to me, I am unable to solve it. A county containing a large number of rural homes is thought to have 60% of those homes insured against fire. Four homeowners are chosen at random and x are found to be insured against fire. Find the probability distribution for x. What is the probability that at least 3 of the 4 are insured? I am able to find p(0) = 0.0256 and the p(4) = 0.1296 The answers to p(1) = 0.1536 p(2) = 0.3456 p(3) = 0.3456 Probability of 3 out of 4 = 0.4572 I have tried to work towards these answers without any success. Please show me how to solve this problem. Thank you.
Date: 01/28/98 at 13:54:14 From: Doctor Sonya Subject: Re: Probability distribution What you have described is known as a Bernoulli trial. You use a Bernoulli trial when you want to know the number of successes in n trials. For example, if I have a probablity of 0.34 of winning a game, I could use Bernoulli trials to tell me my probability of winning three out of five. "What does this have to do with insurance?" you may ask. Well, if my game is picking a house, and I win that game if the house is insured, we can use Bernoulli trials to find the probability of "winning" 0, 1, 2, 3, or 4 times. This is also the probability that 0, 1, 2, 3, or 4 of the houses are insured. One thing that has to be true about our game before we can use Bernoulli trials is that each play must be independent. This means that one house having insurance has nothing to do with its neighbor also having insurance. I'm sure there's a chapter about Bernoulli trials in your textbook if you want more information. Let's say I have n trials (or n plays of a game), with a probability p for success. Then the probability that I will win EXACTLY k of these n trials is given by P(X = k) = (n choose k) * (p^k) * (1 - p)^(n-k) So for our problem, if n = 4, and k = 3, and p = .6 we have P(X = 3) = (4 choose 3) * (.6^3) * (.4)^1 = (4) * (.6^3) * .4 = (4) * (.216) * .4 = .3456 However, this isn't all. Your problem asked for the probability of "at least 3" being insured. So if 4 of the 4 get insurance, then the problem is also solved. I'll let you use the Bernoulli trials for P(X=4). (Remember that k=4.) Thus the probability that at least 3 of the four houses are insured is: P(X=4) + P(X=3) and you'll see that these are the answers you're looking for. Good luck, and don't hesitate to write back with more questions. -Doctor Sonya, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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