My Math Forum Probability problem [help!!!!]

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 February 4th, 2014, 06:11 PM #1 Member   Joined: Dec 2013 Posts: 31 Thanks: 0 Probability problem [help!!!!] Three dice are rolled. What is the prob. that the value on two of the dice sum to the value shown on the remaining dice? I got 1/4 over and over but that is not even an option...
 February 4th, 2014, 06:48 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,943 Thanks: 1134 Math Focus: Elementary mathematics and beyond Re: Probability problem [help!!!!] I get 9/43. Oops! Make that 5/24.
February 4th, 2014, 08:55 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Probability problem [help!!!!]

Hello, p99410!

Quote:
 Three dice are rolled. What is the probbability that the value on two of the dice sum to the value shown on the remaining dice? I got 1/4 over and over but that is not even an option.

You probably made the same mistake that I did.

$\text{There are: }\:6^3\,=\,216\text{ possible outcomes.}$

$\text{I found 9 such triples: }\;\begin{Bmatrix}(1,1,2) & (1,4,5) & (2,3,5) \ \\
(1,2,3) & (1,5,6) & 2,4,6) \\ \\
(1,3,4) & (2,2,4) & (3,3,6) \end{Bmatrix}$

$\text{Then I said each triple has }3! = 6\text{ permutations.}
\text{Hence, there are: }\,9\,\cdot\,6 \:=\:54\text{ favorable outcomes.}$

$\text{Therefore, the probability is: }\:\frac{54}{216} \:=\:\frac{1}{4}$

$\text{But this is }wrong!$

$\text{You see, three of the triples: }\:(1,1,2),\:(2,2,4),\:(3,3,6)
\;\;\;\text{ have only }three\text{ permutations.}$

$\text{There are only }45\text{ favorable outcomes.}

 February 5th, 2014, 07:17 AM #4 Member   Joined: Dec 2013 Posts: 31 Thanks: 0 Re: Probability problem [help!!!!] Thanks! I got it.

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