Probability Distribution

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/