# Probability exercise error?

I have a problem with an exercise: Find the probability that when throwing three dice, the six will occur on one of them (no matter which one), if on the facets of the remaining two dice, two different numbers will outcome.

The answer given to this exercise is 1/2. However I think it is $\frac{60}{6^3}=\frac{5}{18}$

Is there an error in an answer?

EDIT: I have edited the exercise to be exactly as written in the book (translated to english).

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what do you mean with no two identical outcomes occuring? all thre dies being different? –  dREaM Feb 24 '13 at 17:11
Did you mean: "what's the probability that when we throw three dice we get a result of 6 in one of them and not two of the dice show the same points"? –  DonAntonio Feb 24 '13 at 17:11
yes. Once there is "6" and two other dice have different numbers of dots. Also different from 6. –  Misery Feb 24 '13 at 17:12
Your wording is odd, @Misery, but I think you're right in your calculation. –  DonAntonio Feb 24 '13 at 17:15
Could you write the question exactly as given in the exercise? Your question is not very clear and with slightly different interpretations you could get different answers. For example, if the question was 'given that no two outcomes are identical, what is the probability that one of them is $6$', then the answer would be $1/2$. –  polkjh Feb 24 '13 at 17:24

number of cases total: $6^3$. number of cases in which there is a 6 and no two equal results: $1*5*4*3$. So the result is $\frac{60}{216}$ I agree with you.

If the 6 is in the first roll then there are 20 other combinations, if the 6 is in the second the same, and if it is in the third the same, you don't repeat any of them so you add them to get 60. I can't see how this is wrong.

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I read the question to ask the chance we get $6ab$, with $a \ne 6, b \ne 6, a \ne b$. It can't be $\frac 12$ as the chance of getting at least one $6$ is $\frac {91}{216} \lt \frac 12$ before we worry about $a$ and $b$

For the specific answer, there are $5$ ways to choose $a$, $4$ ways to choose $b$, and $6$ orders for the numbers, but we have double counted $6ab$ and $6ba$ giving $\frac {6 \cdot 5 \cdot 4}{2 \cdot 216}=\frac {60}{216}=\frac 5{18}$

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With your edit, the question now is - 'given that the outcomes are all distinct, what is the probability that one of them is $6$'. Number of ways in which we can have distinct numbers on the three dice is $6*5*4=120$. And the number of ways in which none of them is $6$ is $5*4*3=60$. So the number of ways in which one of them is $6$ is $120-60=60$ and hence the required probability is $60/120=1/2$

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Thanks, sometimes those exercies in probability books are written in such unclear manner :| –  Misery Feb 25 '13 at 12:38
However in your answer "one of them" became "none of them", but one of them should be equal to 6 –  Misery Feb 25 '13 at 12:46
"None of them is $6$" is the complement of "atleast one of them is $6$". And in this situation it is easier to count the cases in "none of them is $6$". So I counted that and then subtracted from the total to get the number of cases in "atleast one of them is $6$" (they just turn out to be equal in this situation). –  polkjh Feb 25 '13 at 13:51