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May 1st, 2013, 05:22 AM   #1
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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 /:
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May 1st, 2013, 12:46 PM   #2
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Re: hard probability

Quote:
Originally Posted by gelatine1
You 5 numbers with a random value between 1 and 5.
What is the chance that the product of these numbers is smaller then 100?
I have no clue on how to start this /:
Any random number between 1 and 5 multiplied by any random number between 1 and 5 will give a product that is between 1 and 25. The probability that the product is smaller than 100 is one.
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May 1st, 2013, 04:12 PM   #3
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Re: hard probability

Quote:
Originally Posted by mrtwhs
Quote:
Originally Posted by gelatine1
You 5 numbers with a random value between 1 and 5.
What is the chance that the product of these numbers is smaller then 100?
I have no clue on how to start this /:
Any random number between 1 and 5 multiplied by any random number between 1 and 5 will give a product that is between 1 and 25. The probability that the product is smaller than 100 is one.
I think he's saying "you multiply 5 numbers with a random value...
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May 1st, 2013, 04:28 PM   #4
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Re: hard probability

Or "You have ..."
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May 1st, 2013, 08:50 PM   #5
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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.
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May 3rd, 2013, 06:28 PM   #6
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Quote:
Originally Posted by gelatine1
. . . 5 numbers . . .
Do you mean whole numbers (integers)?
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May 3rd, 2013, 08:39 PM   #7
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Re: hard probability

yes
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