2022 AMC 10B Problem 12

To compute this, we can also find the least $n$ such that the probability of not rolling a $7$ is less than $\dfrac 12.$ Each roll has an independent probability of $\dfrac 16$ of getting $7,$ so it has a $\dfrac 56$ probability of not landing on $7.$ Thus, the probability of none of the rolls being $7$ is $\left(\dfrac 56\right)^n.$ We must find the least $n$ such that $\left(\dfrac 56\right)^n < \dfrac 12.$ If $n=3,$ then the probability is $\dfrac{125}{216},$ which is greater than $\dfrac 12.$ If $n=4,$ then the probability is $\dfrac{625}{1296},$ which is less than $\dfrac 12.$ This makes the answer $4.$ Thus, the answer is C .