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This question already has an answer here:

You roll a fair, six-sided die 4 times. what is the probability of getting at least 2 sixes?

How do you answer this question using combination rules?

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marked as duplicate by user147263, Chappers, N. F. Taussig, Lord_Farin, Did Sep 15 '15 at 17:09

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    $\begingroup$ Welcome to math.SE. To be able to better help you, please describe what you have done yourself to solve the problem. $\endgroup$ – DirkGently Sep 15 '15 at 15:35
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We can begin with noting that $(5+1)^4=6^4$, and that $6^4$ is the total number of outcomes. The expansion of $(5+1)^4$ is:

$$\dbinom{4}{0}5^4+\dbinom{4}{1}5^3+\dbinom{4}{2}5^2+\dbinom{4}{3}5+\dbinom{4}{4}$$

The $(k+1)^{th}$ term represents the number of ways we can throw exactly $k$ sixes, so we need to sum $6\cdot25+4\cdot5+1=150+20=1=171$.

The probability is then $\dfrac{171}{1296}\approx0.132$.

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