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May 1st, 2013, 06:22 AM  #1 
Senior Member Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11  hard probability
You have 5 numbers with a random value between 1 and 5. What is the chance that the product of all those 5 numbers is smaller then 100? I have no clue on how to start this /: 
May 1st, 2013, 01:46 PM  #2  
Senior Member Joined: Feb 2010 Posts: 714 Thanks: 151  Re: hard probability Quote:
 
May 1st, 2013, 05:12 PM  #3  
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53  Re: hard probability Quote:
 
May 1st, 2013, 05:28 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond  Re: hard probability
Or "You have ..."

May 1st, 2013, 09:50 PM  #5 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications  Re: hard probability
Hi [color=#0000BF]gelatine1[/color], I get by the following analysis and a 'brute force' ruby script check. The number of permutations of objects with a number of objects indistinguishable from each other is: . If there is also a number of objects indistinguishable from each other the formula is: In the following, the notation , etc., without any parenthesis represents the digits themselves. It does not represent a count. is the number of combinations of choosing out of objects per the formula: Also, a number in parenthesis such as (3) means 3 combinations. This is used when the notation cannot be used because some of the combinations result in a product >= 100. Let's start from the '1' digit(s) and work up: number of combinations for all 1s. for 4 1s and any other digit. for 3 1s and 2 of any other digit. for 3 1s and 2 other single digits. for 2 1s, and 3 of another digit. The other digit can be 2,3, or 4, but not 5, hence the (3). for 2 1s, 2 of another digit, and one of a third digit. The (11) is represented by 2,2,3; 2,2,4; 2,2,5; 3,3,2; 3,3,4; 3,3,5; 4,4,2; 4,4,3; 4,4,5; 5,5,2; and 5,5,3. 5,5,4 gives a product = 100 and is not counted. for 2 1s and any 3 other single digits. for one 1 and 4 2s or 3s (4 4s or 5s not allowed). for one 1 and 2 each of 2,3 or 2,4. Note that 3,2 and 4,2 create duplicate patterns to 2,3 and 2,4 so they are not counted. for one 1, 3 of another digit, and one of a third digit. 2,2,2,3; 2,2,2,4; 2,2,2,5; and 3,3,3,2 are the (4) for one 1, two of another digit, and two other single digits. The (6) are 2,2,3,4; 2,2,3,5; 2,2,4,5; 3,3,2,4; 3,3,2,5; and 4,4,2,3. for all 2s. for 4 2s and any other digit: 3; 4; or 5. Note that four 2s and one 1 have been counted above. for 3 2s and 2 3s. for 3 2s and one 3 and one 4. The sum is 1337. As mentioned above, I have checked the analysis with a script. Hopefully, someone will post an easier way of solving the problem. 
May 3rd, 2013, 07:28 PM  #6  
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324  Quote:
 
May 3rd, 2013, 09:39 PM  #7 
Senior Member Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11  Re: hard probability
yes


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