2022 AMC 10B Problem 12

On a single roll of two dice there are $36$ outcomes, and six of them ($1{+}6, 2{+}5, 3{+}4, 4{+}3, 5{+}2, 6{+}1$) sum to $7$. So the probability of a $7$ is $\frac16$, and of no $7$ is $\frac56$.

The rolls are independent, so the probability of never seeing a $7$ in $n$ rolls is $\left(\frac56\right)^n$. We need
$$1 - \left(\tfrac56\right)^n > \tfrac12 \quad\Longleftrightarrow\quad \left(\tfrac56\right)^n < \tfrac12.$$

Compute: $\left(\frac56\right)^2 = \frac{25}{36} \approx 0.69$; $\left(\frac56\right)^3 = \frac{125}{216} \approx 0.58$; $\left(\frac56\right)^4 = \frac{625}{1296} \approx 0.48$.

The first power below $\frac12$ is $n = 4$.

The answer is $\boxed{\textbf{(C)}\ 4}$.

"At least once" is almost always handled through its complement "never," which for independent repetitions is a single power. Then the question is just where a geometric sequence crosses a threshold; a couple of multiplications settle it. Compare $125$ with $108$ and $625$ with $648$ to avoid decimal approximations.

• Computing n/6 > 1/2 (adding probabilities) and answering n = 4 for the wrong reason, or stopping at n = 3 because (5/6)^3 looks close to 1/2.
• Using $\frac{1}{6}$ as the probability of not rolling a $7$, which gives a nonsense answer of $n = 1$ or $2$.