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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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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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