# Roll a fair six-sided die 4 times. What is the probability of getting at least 2 sixes? [duplicate]

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?

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

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 $\dfrac{171}{1296}\approx0.132$.