
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 4th, 2016, 01:23 AM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  a candy bar company
A candy bar company is having a contest. On the inside of each package, N, U, or T is printed in ratios 3:2:1. How many packages should you buy to spell NUT? My attempt: P(getting an N) = 3/6 P(getting U) = 2/6 P(getting T) = 1/6 That means, on average, if you buy 6 packages you'll get 3 N's, 2 U's, and 1 T. We can spell NUT. So we should buy at least 6 packages in order to spell NUT. Am I correct? 
February 4th, 2016, 02:54 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
You would need "at least" three to have a possibility of spelling "NUT" just because "NUT" has three letters. You might have to buy a large number (n+ m+ 1 where m in the number with "U" and m is the number with "T") to be certain of getting "NUT". Neither of those has anything to do with "on the average".

February 4th, 2016, 06:22 PM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85 
This is hard. It would be easy if the probabilities were equal. In order to have at least a 50 percent (0.5) probability of getting each letter, here's how many trials you need: N: 1 trial U: 2 trials T: 4 trials Although it won't calculate the expected number of boxes, you could use a random number generator at Free Online Random Number Generator and Checker You can tell it to generate 1,000 (fewer if you want) random numbers between 1 and 6 with "Open Sequence" selected. Let N = 1, 2, or 3, U = 4 or 5, and T = 6. Count how many trials it takes in order to get at least one number from 1 to 3, at least one 4 or 5, and at least one 6. Once you have all three of those, you have 1 success, and you can repeat the process for as many random numbers as you want. If you calculate the average number of trials needed for 1 success using 1,000 random numbers, I'd expect it to be close to the actual probability. If you note how many trials it took for each success and type those numbers into a spreadsheet or calculator, you can sort the numbers and then find the median amount of trials. I might do the Random Numbers for you, but not now. 
February 5th, 2016, 02:56 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85 
My 1,000 random numbers found 126 successes from the first 995 numbers, with the last 5 numbers not having a 6 so not being another success. 1,000/126 = about 7.9365 numbers per success. 17 successes (13.5 percent) were from the minimum 3 candy bars. The median amount of candy bars for the successes was 7. The maximum amount of numbers for a success was 27. Here is how many times each letter was the last prize: N: 7 times from 1 + 2 times from 2 + 6 times from 3 = 15 times U: 16 times from 4 + 16 times from 5 = 32 times T: 79 times from 6 = 79 times The mean of the 126 last numbers in each success is 5.13. 
February 9th, 2016, 03:04 PM  #5 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
The table below could be useful under these conditions: a) You buy one bar in store A, one bar in store B and one bar in store C. b) The bars in stores A, B and C were distributed in the manner N:U:T = 3:2:1. So, if you were to purchase one bar off three different stores, there'd be a 36/216 chance of you being able to produce the word NUT. 36/216 = 1/6. Logically speaking, we could say that in order to win (virtually guarantee that win under the conditions which have been set), you should buy a bar off three separate stores 216 times. I'd also advise you to not shop at the same store twice and make sure that noone buys candy bars in any of these stores (as purchases could fudge the probabilities). If these bars cost 0.50 dollars each, you'd spend 0.50 dollars x 3 x 216 = 324.00 dollars plus money on travel expenses to ensure you win... We shouldn't forget to mention your time, as time is money. We'd better hope that there's a heck of a lot of money up for grabs and that the towns you shop in are going to be virtually empty for possibly a few months. You'd want to make a sound profit in return for all the effort. *Please note (disclaimer): This is simply a model, and we know that in the real world  things are far more chaotic. I don't think I'd be able to give you a precise answer as to how many bars you should purchase, where you should purchase them and when you should purchase them. Last edited by perfect_world; February 9th, 2016 at 04:00 PM. 
February 9th, 2016, 04:12 PM  #6 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
That's 648 stores (3 x 216) you should visit in total. Hope I haven't made myself sound like a nut. I wish you the best of luck in trying to win the ultimate prize.
Last edited by perfect_world; February 9th, 2016 at 04:17 PM. 

Tags 
bar, candy, company 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Candy Probability Distribution  MelissaLangley  Advanced Statistics  1  October 29th, 2012 08:30 PM 
candy for the answer  scsi  Algebra  7  September 17th, 2011 04:51 AM 
candy bars  e81  Algebra  7  May 21st, 2011 09:09 AM 
Boxes of Candy  symmetry  Algebra  4  June 8th, 2007 06:30 PM 
Mixing Candy  symmetry  Algebra  1  June 7th, 2007 07:13 AM 