My Math Forum (http://mymathforum.com/math-forums.php)
-   Probability and Statistics (http://mymathforum.com/probability-statistics/)
-   -   Probability problem [help!!!!] (http://mymathforum.com/probability-statistics/41244-probability-problem-help.html)

 p99410 February 4th, 2014 06:11 PM

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... :oops:

 greg1313 February 4th, 2014 06:48 PM

Re: Probability problem [help!!!!]

I get 9/43. Oops! Make that 5/24.

 soroban February 4th, 2014 08:55 PM

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.

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

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

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

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

$\text{But this is }wrong!$

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

$\text{There are only }45\text{ favorable outcomes.}
\text{Greg1313's answer is correct.}$

 p99410 February 5th, 2014 07:17 AM

Re: Probability problem [help!!!!]

Thanks! I got it. :D

 All times are GMT -8. The time now is 01:07 AM.