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If you throw up 3 dice and want to calculate the probability of at least one six, my math book uses the complement rule as follows:

1- (5/6)^3

which is clear for me, but i tried it another way:

at least one six means either one, two, or three times a six, so i got this:

(3C1 * 15C2 + 3C2 * 15C1 + 3C3 )/ 18C3

So I noted there are 6 times 3, thus 18 elements with 3 dice. Picking 3 from 18 are all possible combinations. In the numerator I put the three possibilities, so one, two, or three times a 6.

But i compared the outcome and it wasnt equal to that of the book.

Can anyone explain why this is wrong?

Thanks in advance, also how can i write mathematical expressions on this site?

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    $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Dec 5, 2020 at 18:31
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    $\begingroup$ I don't know where the $18$ comes from. There are $6^3$ possible outcomes when you throw three dice. $\endgroup$
    – lulu
    Dec 5, 2020 at 18:33
  • $\begingroup$ Can you please explain what $15C2$ means here? So is it possible I choose two numbers from the same dice and the last dice shows nothing? $\endgroup$
    – Math Lover
    Dec 5, 2020 at 18:34
  • $\begingroup$ i think this method im using is just wrong, isnt it? $\endgroup$
    – excellence
    Dec 5, 2020 at 18:35
  • $\begingroup$ it means picking 2 elements from 15 $\endgroup$
    – excellence
    Dec 5, 2020 at 18:35

2 Answers 2

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$^{18}C_3$ or ${18 \choose 3}=816$ is not the number of equally likely ways of throwing $3$ dice. Since each die can take six equally likely values, $6^3=216$ is. Similarly with the rest of your calculation.

So the alternative calculation should be $$\frac{^3C_1\,5^2 + \, ^3C_2\,5 + \,^3C_3}{6^3}$$ and this is the same as $1-\left(\frac 56\right)^3=\frac{91}{216}$

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  • $\begingroup$ thank you very much, thats was a perfect explanation $\endgroup$
    – excellence
    Dec 5, 2020 at 19:21
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A correct approach if you want to avoid the complement is $$\frac{\sum_{k=1}^3 \binom{3}{k}5^{3-k}}{6^3} = \frac{\binom{3}{1}5^2+\binom{3}{2}5^1+\binom{3}{3}5^0}{6^3} = \frac{91}{216}$$

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