Rolling a pair of fair 6 -sided dice, the probability of getting a sum of 7 is $\frac{1}{6}$ : Regardless what the first die shows, the second die has exactly one outcome to make the sum 7 . We consider the complement: The probability of not getting a sum of 7 is $1-\frac{1}{6}=\frac{5}{6}$. Rolling the pair of dice $n$ times, the probability of getting a sum of 7 at least once is $1-\left(\frac{5}{6}\right)^{}n$. Therefore, we have $1-\left(\frac{5}{6}\right)^{}n\textgreater\frac{1}{2}$, or $$ \left(\frac{5}{6}\right)^{}n\textless\frac{1}{2} $$ Since $\left(\frac{5}{6}\right)^{4}\textless\frac{1}{2}\textless\left(\frac{5}{6}\right)^{3}$, the least integer $n$ satisfying the inequality is (C) 4 . 
A pair of fair 6 -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals 7 at least once is greater than $\frac{1}{2}$ ? 
Options:
* $$2$$ 
* $$3$$ 
* $$4$$ 
* $$5$$ 
* $$6$$ 