2022 AMC 10B Problems/Problem 12

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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 $\frac{1}{2}$?

$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution 1 (Complement)

Rolling a pair of fair $6$-sided dice, the probability of getting a sum of $7$ is $\frac16:$ 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 $1-\frac16=\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\left(\frac56\right)^n.$

Therefore, we have $1-\left(\frac56\right)^n>\frac12,$ or \[\left(\frac56\right)^n<\frac12.\] Since $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ the least integer $n$ satisfying the inequality is $\boxed{\textbf{(C) } 4}.$


Solution 2 (99% Accurate Guesswork)

Let's try the answer choices. We can quickly find that when we roll $3$ dice, either the first and second sum to $7$, the first and third sum to $7$, or the second and third sum to $7$. There are $6$ ways for the first and second dice to sum to $7$, $6$ ways for the first and third to sum to $7$, and $6$ ways for the second and third dice to sum to $7$. However, we overcounted (but not by much) so we can assume that the answer is $\boxed {\textbf{(C) }4}$


Solution 3

We can start by figuring out what the probability is for each die to add up to $7$ if there is only $1$ roll. We can quickly see that the probability is $\frac16$ , as there are $6$ ways to make $7$ from $2$ numbers on a die, and there are a total of $36$ ways to add $2$ numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for $n$ rolls would be $\frac16$$n$. The smallest number that satisfies this is $\boxed {\textbf{(C) }4}$


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See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions

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