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 May 1st, 2013, 05:22 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, 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.

May 1st, 2013, 04:12 PM   #3
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Re: hard probability

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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...

 May 1st, 2013, 04:28 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond Re: hard probability Or "You have ..."
 May 1st, 2013, 08:50 PM #5 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: hard probability Hi [color=#0000BF]gelatine1[/color], I get $\ \frac{1337}{5^5}=\frac{1337}{3125} \$ by the following analysis and a 'brute force' ruby script check. The number of permutations of $\ n \$ objects with a number $\ n_1 \$ of objects indistinguishable from each other is: $P=\frac{n!}{n_1!} \$. If there is also a number $\ n_2 \$ of objects indistinguishable from each other the formula is: $P=\frac{n!}{n_1! \cdot n_2!}$ In the following, the notation $\ 111 \$, etc., without any parenthesis represents the digits themselves. It does not represent a count. ${n \choose k} \$ is the number of combinations of choosing $\ k \$ out of $\ n \$ objects per the formula: ${n \choose k}=\frac{n!}{(n-k)!k!}$ Also, a number in parenthesis such as (3) means 3 combinations. This is used when the $\ {n \choose k} \$ 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: $11111= 1 \$ number of combinations for all 1s. $1111{4 \choose 1} \cdot \frac{5!}{4!}=20 \$ for 4 1s and any other digit. $111{4 \choose 1} \cdot \frac{5!}{3! \cdot 2!}=40 \$ for 3 1s and 2 of any other digit. $111{4 \choose 2} \cdot \frac{5!}{3!}=120 \$ for 3 1s and 2 other single digits. $11(3) \cdot \frac{5!}{2! \cdot 3!}=30 \$ for 2 1s, and 3 of another digit. The other digit can be 2,3, or 4, but not 5, hence the (3). $11(11) \cdot \frac{5!}{2! \cdot 2!}=330 \$ 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. $11{4 \choose 3} \cdot \frac{5!}{2!}=240 \$ for 2 1s and any 3 other single digits. $1(2) \cdot \frac{5!}{4!}=10 \$ for one 1 and 4 2s or 3s (4 4s or 5s not allowed). $1(2) \cdot \frac{5!}{2! \cdot 2!}=60 \$ 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. $1(4) \cdot \frac{5!}{3!}=80 \$ 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) $1(6) \cdot \frac{5!}{2!}=360 \$ 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. $22222=1 \$ for all 2s. $2222(3) \cdot \frac{5!}{4!}=15 \$ for 4 2s and any other digit: 3; 4; or 5. Note that four 2s and one 1 have been counted above. $222(1) \cdot \frac{5!}{3! \cdot 2!}=10 \$ for 3 2s and 2 3s. $222(1) \cdot \frac{5!}{3!}=20 \$ 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, 06:28 PM   #6
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Quote:
 Originally Posted by gelatine1 . . . 5 numbers . . .
Do you mean whole numbers (integers)?

 May 3rd, 2013, 08:39 PM #7 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: hard probability yes

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