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October 24th, 2011, 07:26 AM  #1 
Newbie Joined: Oct 2011 Posts: 3 Thanks: 0  Simple probability question, where am I going wrong?
Okay, so I know I posted a question before this however, after trying several permutations of this question I'm just not really sure whats going wrong. Basically a shop has 20 customers, the chance of 1 customer buying a car is 0.3 and they are all independent. What is the least amount of stock needed at the start of the day so that probability shortage will be at most 0.025? Based on what I can see, the average number of cars bought a day is 6. I tried to look at this from the point of a binomial distribution based upon the following info: n=20 x= 0.025 p= 0.3 I think something has gone wrong somewhere because whenever I calculate it based on this assumption I get a value I'm sure is too small to be right. Could anyone explain or hint towards the correct process to work this out? 
October 24th, 2011, 12:56 PM  #2 
Newbie Joined: Oct 2011 Posts: 3 Thanks: 0  Re: Simple probability question, where am I going wrong?
Bump for answering.

October 24th, 2011, 01:11 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,763 Thanks: 697  Re: Simple probability question, where am I going wrong?
Use the binomial as follows: Let Pk = prob. exactly k cars are bought. Then add P20 + P19 + ... until sum > .025. The stop point is the amount of stock needed. 
September 29th, 2013, 08:44 PM  #4 
Newbie Joined: Sep 2013 Posts: 3 Thanks: 0  Re: Simple probability question, where am I going wrong?
Can someone please further explain the solution? Someone suggested letting pk = probability that exactly k computers are bought. Then adding P20 + P19, etc. until the sum > 0.025. Does this mean assuming k, the number of successes, is 20 then 19, etc. when n trial is 20 and p probability is 0.3? Also, can it be explained how to solve this using a binomial distribution table, if possible, and what the formula would be? P(X <\>\=????  n=20, p=0.3)? THANK YOU in advance for help. Have spent quite a bit of time trying to figure this out, but still to no avail. 
September 30th, 2013, 01:31 AM  #5 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Simple probability question, where am I going wrong?
First, as the expected number of cars sold is 6, it will be quicker to calculate the probability of the number of cars sold starting at 0 and working upwards until the total probability exceeds .975. In general the probability P(n) of selling precisely n cars is: This is because (0.3)^n is the probability that n customers buy a car, 0.7^(20n) is the probability the remaining 20n customers do not buy a car and the binomial coefficient is the number of different ways this can happen. For example: Is the probability that the first n customers buy a car and the remaining 20n do not. It is also the probability that the first customer does not buy a car, the next n do, and the remainder don't. And so on for every possibility that results in precisly n cars being sold. The number of such possibilities is the number of ways to choose n things from 20, which is Now, all you have to do is calculate P(0) + P(1) etc, until the total probability exceeds 0.975. 
September 30th, 2013, 12:24 PM  #6 
Newbie Joined: Sep 2013 Posts: 3 Thanks: 0  Re: Simple probability question, where am I going wrong?
Thank you, Pero, for your response. That did clear up part of the confusion; however, I'm still not 100% sure of why I'm looking to add the probabilities. Essentially, if looking at a binomial distribution table or calculating using the formula provided, I find the following: the probability of demand of.... 0 cars is 0.001 1 car is 0.007 2 cars is 0.028 3 cars is 0.072 4 cars is 0.130 5 cars is 0.179 6 cars is 0.192 7 cars is 0.164 8 cars is 0.114 9 cars is 0.065 10 cars is 0.031 When adding up the probablites, I get 0.983 (or what I understand to be the probability for 10 cars). I guess I could have also referred to a cumulative binomical distribution chart and, if i look for n = 20, p = .3, and x = 10(10 cars), I would get the same number, 0.983. So, if the question is asking what is the least amount of stock needed to have on hand so that the probability of shortage (demand exceeding supply) will be at most 0.025, I'm not sure the logic behind it. I now know that the demand for 10 computers is 0.983, which exceeds 0.975 as mentioned in Pero's reply (I'm assuming you just took 1  0.025). again, not sure the reasoning behind the answer, which I suppose is 10 computers. Can someone please explain? Thank you, again! 
September 30th, 2013, 12:44 PM  #7 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Simple probability question, where am I going wrong?
The probability of demand of.... 0 cars is 0.001 1 car is 0.007 (so, up to 1 car = 0.00 2 cars is 0.028 (up to 2 cars = 0.036) 3 cars is 0.072 (up to 3 cars = 0.09 4 cars is 0.130 (up to 4 cars = 0.22 5 cars is 0.179 (up to 5 cars = 0.417) 6 cars is 0.192 (up to 6 cars = 0.609) 7 cars is 0.164 (up to 7 cars = 0.773) 8 cars is 0.114 (up to 8 cars = 0.887) 9 cars is 0.065 (up to 9 cars = 0.952) 10 cars is 0.031 (up to 10 cars = 0.983) So, if you have 10 cars, you will only run out of stock on 0.017 days. The probabilities you calculated on the left are for precisely that number of cars. The probabilities on the right are for up to and including that number of cars. 
September 30th, 2013, 01:52 PM  #8 
Newbie Joined: Sep 2013 Posts: 3 Thanks: 0  Re: Simple probability question, where am I going wrong?
Thank you, Pero! I think I'm starting to see the light now. So, essentially, the demand for cars beyond 11 is relatively low. The likelihood of shortage below 0.025 is only when you exceed 10 cars. This is the 0.017 (1  0.983) with .983 being the cumulative probability of demand of 10 cars. Is that deduction correct? 
September 30th, 2013, 10:37 PM  #9  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Simple probability question, where am I going wrong? Quote:
 

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